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# Lecture 29 - Chapter 2: Enriched Categories

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51.
edited May 2018

Try using set union ($$\cup )$$ as your monoidal product, and then apply the definition John gave in the first post. The enriched structure that comes out should be an old friend.

But if we're enriching over the unit monoidal preorder, $$\{\bullet\}$$, there is no "union" -- there's only a single, trivial monoidal preorder structure on $$\textbf{Unit}$$. Any set enriched over $$\textbf{Unit}$$ must necessarily have $$\mathcal{X}(x, y) = \bullet$$, the constant (and only) function into $$\mathbf{Unit}$$.

EDIT: It isn't clear whether you want $$\mathbf{Unit}$$ enriched over some monoidal poset, or whether you want some set enriched over $$\mathbf{Unit}$$. The terminology "$$\bullet$$-enriched category" would seem to refer to the latter.

Comment Source:> Try using set union (\$$\cup )\$$ as your monoidal product, and then apply the definition John gave in the first post. The enriched structure that comes out should be an old friend. But if we're enriching over the unit monoidal preorder, \$$\\{\bullet\\}\$$, there is no "union" -- there's only a single, trivial monoidal preorder structure on \$$\textbf{Unit}\$$. Any set enriched over \$$\\textbf{Unit}\$$ must necessarily have \$$\mathcal{X}(x, y) = \bullet\$$, the constant (and only) function into \$$\mathbf{Unit}\$$. **EDIT:** It isn't clear whether you want \$$\mathbf{Unit}\$$ enriched over some monoidal poset, or whether you want some set enriched over \$$\mathbf{Unit}\$$. The terminology "\$$\bullet\$$-enriched category" would seem to refer to the latter.
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52.
edited May 2018

Any set enriched over $$\textbf{Unit}$$ must necessarily have $$\mathcal{X}(x, y) = \bullet$$, the constant (and only) function into $$\mathbf{Unit}$$.

I agree. Basically such a condition amounts to saying there is exactly 1 morphism from this object $$x$$ to this other object $$y$$ in $$\mathcal{X}$$.

Comment Source:>Any set enriched over \$$\\textbf{Unit}\$$ must necessarily have \$$\mathcal{X}(x, y) = \bullet\$$, the constant (and only) function into \$$\mathbf{Unit}\$$. I agree. Basically such a condition amounts to saying there is exactly **1** morphism from this object \$$x \$$ to this other object \$$y\$$ in \$$\mathcal{X}\$$.
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53.
edited May 2018

Is that not the codiscrete preorder, a.k.a. the complete directed graph, on a set $$\mathrm{Ob}(\mathcal{X})$$?

Comment Source:Is that not the codiscrete preorder, a.k.a. the complete directed graph, on a set \$$\\mathrm{Ob}(\mathcal{X})\$$?
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54.
edited May 2018

Jonathan wrote:

Any set enriched over $$\textbf{Unit}$$ must necessarily have $$\mathcal{X}(x, y) = \bullet$$, the constant (and only) function into $$\mathbf{Unit}$$.

You meant to say any category enriched over $$\textbf{Unit}$$ must have this property. There's no such thing as a set enriched over something. But modulo that slip of the tongue, you're right.

Is that not the codiscrete preorder, a.k.a. the complete directed graph, on a set $$\mathrm{Ob}(\mathcal{X})$$?

Basically yes.

Speaking pedantically, it's not quite right to say a $$\textbf{Unit}$$-category is a preorder, since a preorder is a $$\textbf{Bool}$$-category. But there's a way to turn any $$\textbf{Unit}$$-enriched category into a $$\textbf{Bool}$$-category, and this indeed always gives a codiscrete preorder... just as you suggest.

Here's a really nice way to think about it. Whenever you have a monoidal monotone between monoidal preorders

$$f: \mathcal{V} \to \mathcal{W}$$ you can use it to turn $$\mathcal{V}$$-categories into $$\mathcal{W}$$-categories. If you start with a $$\mathcal{V}$$-category, say $$\mathcal{X}$$, you can turn it into a $$\mathcal{W}$$-category $$\mathcal{X}'$$ by keeping the same set of objects and defining

$$\mathcal{X}'(x,y) = f(\mathcal{X}(x,y)) .$$ You can easily check that this works. This trick is called base change.

Now, there's a unique monoidal monotone $$f: \textbf{Unit} \to \textbf{Bool}$$. Using this, you can use base change to turn any $$\textbf{Unit}$$-category into a $$\textbf{Bool}$$-category, also known as a preorder. You always get a codiscrete preorder this way, and you can get any codiscrete preorder this way.

Having worked through all this, we can relax and sloppily say that $$\textbf{Unit}$$-categories "are" codiscrete preorders, and any category theorist will know what we mean.

I happen to be doing a lot of base change for enriched categories in my work for Pyrofex these days, so this stuff is on my mind.

Base change becomes a lot more fun when we look at some "bases" that are more interesting than $$\textbf{Unit}$$ and $$\textbf{Bool}$$.

Comment Source:Jonathan wrote: > Any set enriched over \$$\\textbf{Unit}\$$ must necessarily have \$$\mathcal{X}(x, y) = \bullet\$$, the constant (and only) function into \$$\mathbf{Unit}\$$. You meant to say any _category_ enriched over \$$\\textbf{Unit}\$$ must have this property. There's no such thing as a set enriched over something. But modulo that slip of the tongue, you're right. >Is that not the codiscrete preorder, a.k.a. the complete directed graph, on a set \$$\\mathrm{Ob}(\mathcal{X})\$$? Basically yes. Speaking pedantically, it's not quite right to say a \$$\\textbf{Unit}\$$-category _is_ a preorder, since a preorder is a \$$\\textbf{Bool}\$$-category. But there's a way to turn any \$$\\textbf{Unit}\$$-enriched category into a \$$\\textbf{Bool}\$$-category, and this indeed always gives a codiscrete preorder... just as you suggest. Here's a really nice way to think about it. Whenever you have a monoidal monotone between monoidal preorders $f: \mathcal{V} \to \mathcal{W}$ you can use it to turn \$$\mathcal{V}\$$-categories into \$$\mathcal{W}\$$-categories. If you start with a \$$\mathcal{V}\$$-category, say \$$\mathcal{X}\$$, you can turn it into a \$$\mathcal{W}\$$-category \$$\mathcal{X}'\$$ by keeping the same set of objects and defining $\mathcal{X}'(x,y) = f(\mathcal{X}(x,y)) .$ You can easily check that this works. This trick is called **base change**. Now, there's a unique monoidal monotone \$$f: \\textbf{Unit} \to \textbf{Bool}\$$. Using this, you can use base change to turn any \$$\\textbf{Unit}\$$-category into a \$$\\textbf{Bool}\$$-category, also known as a preorder. You always get a codiscrete preorder this way, and you can get _any_ codiscrete preorder this way. Having worked through all this, we can relax and sloppily say that \$$\\textbf{Unit}\$$-categories "are" codiscrete preorders, and any category theorist will know what we mean. I happen to be doing a lot of base change for enriched categories in my work for Pyrofex these days, so this stuff is on my mind. Base change becomes a lot more fun when we look at some "bases" that are more interesting than \$$\textbf{Unit}\$$ and \$$\\textbf{Bool}\$$.
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55.
edited May 2018

Speaking pedantically, it's not quite right to say a $$\textbf{Unit}$$-category is a preorder, since a preorder is a $$\textbf{Bool}$$-category.

Right. I think Anindya's answer -- that a $$\textbf{Unit}$$-category is just a set -- is more precise.

You meant to say any category enriched over $$\textbf{Unit}$$ must have this property. There's no such thing as a set enriched over something. But modulo that slip of the tongue, you're right.

Yes, I think the grammatical tense got confused a little bit in my post. From my perspective, if $$\mathcal{V}$$ is a symmetric monoidal poset, a $$\mathcal{V}$$-enriched category is just a set $$S$$ endowed with a map $$\mathcal{X} : S \times S \to \mathcal{V}$$ obeying certain properties. So I'd say we can enrich a set to produce a $$\mathcal{V}$$-category, and what was enriched was the set, but it's the category that is enriched. (So enrichment is a temporal activity carrying sets to $$\mathcal{V}$$-categories.)

Comment Source:[John wrote](https://forum.azimuthproject.org/discussion/comment/18470/#Comment_18470): > Speaking pedantically, it's not quite right to say a \$$\\textbf{Unit}\$$-category _is_ a preorder, since a preorder is a \$$\\textbf{Bool}\$$-category. Right. I think Anindya's answer -- that a \$$\textbf{Unit}\$$-category is just a set -- is more precise. > You meant to say any _category_ enriched over \$$\\textbf{Unit}\$$ must have this property. There's no such thing as a set enriched over something. But modulo that slip of the tongue, you're right. Yes, I think the grammatical tense got confused a little bit in my post. From my perspective, if \$$\mathcal{V}\$$ is a symmetric monoidal poset, a \$$\mathcal{V}\$$-enriched category is just a set \$$S\$$ endowed with a map \$$\mathcal{X} : S \times S \to \mathcal{V}\$$ obeying certain properties. So I'd say we can _enrich_ a set to produce a \$$\mathcal{V}\$$-category, and what _was enriched_ was the set, but it's the category that _is enriched_. (So enrichment is a temporal activity carrying sets to \$$\mathcal{V}\$$-categories.)
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56.
edited May 2018

Ultimately we'll see the historical origin of the term "enriched category": the original examples all arose by equipping categories with extra structure. For example, while a category has a set of morphisms between any pair of objects, people noticed that in many examples these sets are vector spaces, or abelian groups, or topological spaces, etc. So they - meaning especially Max Kelly, the student of Eilenberg who brought category theory to Australia, who sadly is no longer with us - invented "enriched categories" to handle all these examples.

By the time the theory was done, the "thing of morphisms" between a pair of objects could be an object in any monoidal category. But the only monoidal categories we know about so far are monoidal preorders. So for us, each "thing of morphisms" $$\mathcal{X}(x,y)$$ is an element of some monoidal preorder $$\mathcal{V}$$.

This isn't a set at all! So the particular class of enriched categories we're studying now have strayed far from their origins.

For the real deal, see this free book:

(If you think my course is abstract, just wait until you see this!)

Comment Source:Ultimately we'll see the historical origin of the term "enriched category": the original examples all arose by equipping categories with extra structure. For example, while a category has a _set_ of morphisms between any pair of objects, people noticed that in many examples these sets are _vector spaces_, or _abelian groups_, or _topological spaces_, etc. So they - meaning especially [Max Kelly](https://en.wikipedia.org/wiki/Max_Kelly), the student of Eilenberg who brought category theory to Australia, who sadly is no longer with us - invented "enriched categories" to handle all these examples. By the time the theory was done, the "thing of morphisms" between a pair of objects could be _an object in any monoidal category_. But the only monoidal categories we know about so far are monoidal preorders. So for us, each "thing of morphisms" \$$\mathcal{X}(x,y)\$$ is an element of some monoidal preorder \$$\mathcal{V} \$$. This isn't a set at all! So the particular class of enriched categories we're studying now have strayed far from their origins. For the real deal, see this free book: * Max Kelly, _[Basic Concepts of Enriched Category Theory](http://www.tac.mta.ca/tac/reprints/articles/10/tr10.pdf)._ (If you think my course is abstract, just wait until you see this!)
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57.

Jonathan Castello wrote in #47

Puzzle JMC8 (unsolved): What do we get if we enrich over the lattice given by $$\{\bot \le \mathrm{false} \le \top, \bot \le \mathrm{true} \le \top\}$$, with identity $$\mathrm{\bot}$$ and monoidal operator given by join ($$\vee$$)?

The logic you are mentioning appears to be the Dunn/Belnap 4 valued logic. I'm going to call it Belnap4.

Theorem. In a Belnap4-enriched category $$\mathcal{X}$$, then for all $$a,b,c$$ and $$d$$ we have $$\mathcal{X}(a,b) = \mathcal{X}(c,d)$$.

Proof.

The proof runs top down in a way.

First, assume there is some pair $$a$$ and $$b$$ such that $$\mathcal{X}(a,b) = \top$$. We have from the (b) condition of enriched categories that $$\mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \leq \mathcal{X}(c,b)$$. Since $$\otimes = \vee$$ it must be $$\mathcal{X}(c,b) = \top$$. However we have that $$\mathcal{X}(c,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d)$$, hence $$\mathcal{X}(c,d) = \top$$.

So assume for all pairs $$\mathcal{X}(x,y) \neq \top$$. Assume without loss of generality that there is some pair where $$\mathcal{X}(a,b) = \mathtt{true}$$ (the case for $$\mathtt{false}$$ is similar). We have $$\mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \leq \mathcal{X}(c,b)$$, so it must be that $$\mathtt{true} \leq \mathcal{X}(c,b)$$. But by assumption we have that $$\mathcal{X}(c,b) \neq \top$$, to it must be that $$\mathcal{X}(c,b) = \mathtt{true}$$. Using $$\mathcal{X}(c,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d)$$ we have $$\mathcal{X}(c,d) = \mathtt{true}$$.

Since we reasoned without loss of generality, we have for all pairs $$\mathcal{X}(x,y)$$ must not be in $$\{ \top, \mathtt{true}, \mathtt{false} \}$$. So it must be that $$\mathcal{X}(x,y) = \bot$$ for all pairs since its the only value possible.     $$\Box$$

We can recover a preorder. This follows the theorem I attempted to show in #37. That theorem attempts to generalize Theorem 2.30 in Spivak and Fong.

Lemma. In a Belnap4-enriched category, define $$x \preceq_\mathcal{X} y \iff \mathcal{X}(x,y) = \mathcal{X}(x,x)$$ ...then $$\preceq_\mathcal{X}$$ is the trivial or codiscrete preorder.

Proof. Follows immediately from the theorem above     $$\Box$$

Comment Source:[Jonathan Castello wrote in #47](https://forum.azimuthproject.org/discussion/comment/18463/#Comment_18463) > **Puzzle JMC8** (unsolved): What do we get if we enrich over the lattice given by \$$\\{\bot \le \mathrm{false} \le \top, \bot \le \mathrm{true} \le \top\\}\$$, with identity \$$\mathrm{\bot}\$$ and monoidal operator given by join (\$$\vee\$$)? The logic you are mentioning appears to be the [Dunn/Belnap 4 valued logic](https://plato.stanford.edu/entries/logic-manyvalued/#Dun4ValSys). I'm going to call it **Belnap4**. I believe I have an answer to your puzzle. **Theorem**. In a **Belnap4**-enriched category \$$\mathcal{X}\$$, then for all \$$a,b,c\$$ and \$$d\$$ we have \$$\mathcal{X}(a,b) = \mathcal{X}(c,d) \$$. **Proof**. The proof runs *top down* in a way. First, assume there is some pair \$$a\$$ and \$$b\$$ such that \$$\mathcal{X}(a,b) = \top\$$. We have from the (b) condition of enriched categories that \$$\mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \leq \mathcal{X}(c,b)\$$. Since \$$\otimes = \vee\$$ it must be \$$\mathcal{X}(c,b) = \top\$$. However we have that \$$\mathcal{X}(c,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d)\$$, hence \$$\mathcal{X}(c,d) = \top\$$. So assume for all pairs \$$\mathcal{X}(x,y) \neq \top\$$. Assume without loss of generality that there is some pair where \$$\mathcal{X}(a,b) = \mathtt{true}\$$ (the case for \$$\mathtt{false}\$$ is similar). We have \$$\mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \leq \mathcal{X}(c,b)\$$, so it must be that \$$\mathtt{true} \leq \mathcal{X}(c,b)\$$. But by assumption we have that \$$\mathcal{X}(c,b) \neq \top\$$, to it must be that \$$\mathcal{X}(c,b) = \mathtt{true} \$$. Using \$$\mathcal{X}(c,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d)\$$ we have \$$\mathcal{X}(c,d) = \mathtt{true}\$$. Since we reasoned without loss of generality, we have for all pairs \$$\mathcal{X}(x,y)\$$ must *not* be in \$$\\{ \top, \mathtt{true}, \mathtt{false} \\}\$$. So it must be that \$$\mathcal{X}(x,y) = \bot \$$ for all pairs since its the only value possible. &nbsp;&nbsp;&nbsp;&nbsp;\$$\Box\$$ We can recover a preorder. This follows the theorem I attempted to show in [#37](https://forum.azimuthproject.org/discussion/comment/18446/#Comment_18446). That theorem attempts to generalize Theorem 2.30 in Spivak and Fong. **Lemma**. In a **Belnap4**-enriched category, define $$x \preceq_\mathcal{X} y \iff \mathcal{X}(x,y) = \mathcal{X}(x,x)$$ ...then \$$\preceq_\mathcal{X}\$$ is the trivial or *codiscrete* preorder. **Proof**. Follows immediately from the theorem above &nbsp;&nbsp;&nbsp;&nbsp;\$$\Box\$$
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58.
edited May 2018

I'd never heard of $$\textbf{Belnap4}$$, but my knowledge of multi-valued logics is poor. Here's what this little poset looks like:

In this picture, taken from the Stanford Encylopedia of Philosophy, the two intermediate truth values are called $$\varnothing$$, where we are ignorant of whether something is true or false, and $$\{\bot,\top\}$$, where we have contradictory information saying that something is both true and false! $$\top$$ is true and $$\bot$$ is false.

So, in a $$\textbf{Belnap4}$$-category, we can say

• yes, $$x \leq y$$
• no, $$x \nleq y$$
• I don't know whether $$x \leq y$$ or $$x \nleq y$$
• I've got contradictory information suggesting both $$x \leq y$$ and $$x \nleq y$$.

Cool!

There's a nice monoidal monotone $$f: \textbf{Bool} \to \textbf{Belnap4}$$ embedding ordinary Boolean logic in this 4-valued logic, so using "base change" (as explained in comment #54) we can turn any preorder into a $$\textbf{Belnap4}$$-category.

But this is more interesting: are there any monoidal monotones $$g: \textbf{Bool} \to \textbf{Belnap4}$$? If so, we can use them to "crush down" $$\textbf{Belnap4}$$-categories into preorders.

In general it should be lots of fun to combine multi-valued logic with enriched categories as we are doing here.

Comment Source:I'd never heard of \$$\textbf{Belnap4}\$$, but my knowledge of multi-valued logics is poor. Here's what this little poset looks like: <center><a href = "https://plato.stanford.edu/entries/logic-manyvalued/#Dun4ValSys"><img src = "https://plato.stanford.edu/entries/logic-manyvalued/4V-truth.gif"></a> </center> In this picture, taken from the _Stanford Encylopedia of Philosophy_, the two intermediate truth values are called \$$\varnothing\$$, where we are _ignorant_ of whether something is true or false, and \$$\\{\bot,\top\\} \$$, where we have _contradictory information_ saying that something is both true and false! \$$\top\$$ is true and \$$\bot\$$ is false. So, in a \$$\textbf{Belnap4}\$$-category, we can say * yes, \$$x \leq y\$$ * no, \$$x \nleq y \$$ * I don't know whether \$$x \leq y\$$ or \$$x \nleq y \$$ * I've got contradictory information suggesting both \$$x \leq y\$$ and \$$x \nleq y \$$. Cool! There's a nice monoidal monotone \$$f: \textbf{Bool} \to \textbf{Belnap4} \$$ embedding ordinary Boolean logic in this 4-valued logic, so using "base change" (as explained in [comment #54](https://forum.azimuthproject.org/discussion/comment/18470/#Comment_18470)) we can turn any preorder into a \$$\textbf{Belnap4}\$$-category. But this is more interesting: are there any monoidal monotones \$$g: \textbf{Bool} \to \textbf{Belnap4} \$$? If so, we can use them to "crush down" \$$\textbf{Belnap4}\$$-categories into preorders. In general it should be lots of fun to combine multi-valued logic with enriched categories as we are doing here.
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59.
edited May 2018

The logic you are mentioning appears to be the Dunn/Belnap 4 valued logic. I'm going to call it Belnap4.

Yes, exactly! (I didn't know it was called this, so this is an awesome reference for me to explore a bit later. plato.stanford.edu is a wonderful resource.) I like to work with the so-called "information ordering" described on that page. John posted the truth ordering image; here's the information ordering (notice the different convention of symbols from my original problem statement):

This is essentially the most elementary example of the posets I'm interested in. More examples can be found in Lindsey Kuper's & Ryan Newton's paper on "lattice variables" (which is actually the paper that motivated me to become a Ph.D. student!), as well as Vijay Saraswat's concurrent constraint logic programming.

Perhaps I'll cross-post this to the hackers' discussion thread!

Since we reasoned without loss of generality, we have for all pairs $$\mathcal{X}(x,y)$$ must not be in $$\{ \top, \mathtt{true}, \mathtt{false} \}$$. So it must be that $$\mathcal{X}(x,y) = \bot$$ for all pairs since its the only value possible.

Is that the case? It seems that you've shown that $$\mathcal{X}(\cdot, \cdot)$$ must be a(ny) constant function; I didn't see any explicit contradictions arise in your argument to knock out any of the other cases.

At any rate, this is very interesting! It seems my choice of monoidal structure is not clever enough to give any particularly noteworthy enrichments.

Comment Source:[Matthew wrote](https://forum.azimuthproject.org/discussion/comment/18463/#Comment_18463): > The logic you are mentioning appears to be the [Dunn/Belnap 4 valued logic](https://plato.stanford.edu/entries/logic-manyvalued/#Dun4ValSys). I'm going to call it **Belnap4**. Yes, exactly! (I didn't know it was called this, so this is an awesome reference for me to explore a bit later. plato.stanford.edu is a wonderful resource.) I like to work with the so-called "information ordering" described on that page. John posted the truth ordering image; here's the information ordering (notice the different convention of symbols from my original problem statement): <center>![](https://plato.stanford.edu/entries/logic-manyvalued/4V-info.gif)</center> This is essentially the most elementary example of the posets I'm interested in. More examples can be found in Lindsey Kuper's & Ryan Newton's [paper on "lattice variables"](https://dl.acm.org/citation.cfm?id=2502326) (which is actually the paper that motivated me to become a Ph.D. student!), as well as Vijay Saraswat's [concurrent constraint logic programming](https://dl.acm.org/citation.cfm?id=96733). Perhaps I'll cross-post this to the hackers' discussion thread! > Since we reasoned without loss of generality, we have for all pairs \$$\mathcal{X}(x,y)\$$ must *not* be in \$$\\{ \top, \mathtt{true}, \mathtt{false} \\}\$$. So it must be that \$$\mathcal{X}(x,y) = \bot \$$ for all pairs since its the only value possible. Is that the case? It seems that you've shown that \$$\mathcal{X}(\cdot, \cdot)\$$ must be a(ny) constant function; I didn't see any explicit contradictions arise in your argument to knock out any of the other cases. At any rate, this is very interesting! It seems my choice of monoidal structure is not clever enough to give any particularly noteworthy enrichments.
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60.
edited May 2018

I think this ordering

might give a more reasonable way to think of$$\vee$$ as "and" For example, if you think something is "true" ($$\top$$\) and then someone tells you it's "false" ($$\bot$$) you wind up in the contradictory state $$\{ \top, \bot \}$$.

Of course, rotating a diamond 90 degrees gives you another diamond, so abstractly the above poset is isomorphic to this other one:

Also, from a poset-theoretic point of view, it's silly to use $$\top$$ to mean anything other than the top element of a poset, or $$\bot$$ for anything other than the bottom!

So, it's all just about how we interpret the four elements of this poset.

I would usually think of this monoidal poset as the power set of the 2-element set, or equivalently $$\mathbf{Bool} \times \mathbf{Bool}$$. These suggest the interpretations of the 4 elements as:

• true today and true tomorrow
• true today but not true tomorrow
• not true today but true tomorrow
• not true today and not true tomorrow.

So, we could use a category enriched over this poset, made monoidal using $$\wedge$$, to talk about whether various people are more wealthy than others today and/or tomorrow.

Comment Source:I think this ordering <center> <a href = "https://plato.stanford.edu/entries/logic-manyvalued/#Dun4ValSys"> <img src = "https://plato.stanford.edu/entries/logic-manyvalued/4V-info.gif"></a></center> might give a more reasonable way to think of\$$\vee\$$ as "and" For example, if you think something is "true" (\$$\top\$$\\) and then someone tells you it's "false" (\$$\bot\$$) you wind up in the contradictory state \$$\\{ \top, \bot \\} \$$. Of course, rotating a diamond 90 degrees gives you another diamond, so abstractly the above poset is isomorphic to this other one: <center><a href = "https://plato.stanford.edu/entries/logic-manyvalued/#Dun4ValSys"><img src = "https://plato.stanford.edu/entries/logic-manyvalued/4V-truth.gif"></a> </center> Also, from a poset-theoretic point of view, it's silly to use \$$\top\$$ to mean anything other than the top element of a poset, or \$$\bot\$$ for anything other than the bottom! So, it's all just about how we interpret the four elements of this poset. I would _usually_ think of this monoidal poset as the power set of the 2-element set, or equivalently \$$\mathbf{Bool} \times \mathbf{Bool} \$$. These suggest the interpretations of the 4 elements as: * true today and true tomorrow * true today but not true tomorrow * not true today but true tomorrow * not true today and not true tomorrow. So, we could use a category enriched over this poset, made monoidal using \$$\wedge\$$, to talk about whether various people are more wealthy than others today and/or tomorrow.
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61.
edited May 2018

Matthew wrote:

Theorem. In a Belnap4-enriched category $$\mathcal{X}$$, then for all $$a,b,c$$ and $$d$$ we have $$\mathcal{X}(a,b) = \mathcal{X}(c,d)$$.

It's good to emphasize here that you are making $$\mathbf{Belnap4}$$ into a monoidal preorder using the join operation $$\vee$$. Here's another proof of this result. I'll just sketch it:

If we make $$\mathbf{Belnap4}$$ into a monoidal preorder using $$\vee$$, it's isomorphic to $$(\mathbf{Bool},\vee) \times (\mathbf{Bool},\vee)$$ - the product of two copies of the Booleans with their join operation as the monoid structure. Thus, a $$(\mathbf{Belnap4},\vee)$$ -category with some set $$S$$ of objects is the same as two separate $$(\mathbf{Bool},\vee)$$-categories with the same set $$S$$ of objects!

Thus, it suffices to prove that for any $$(\mathbf{Bool},\vee)$$-category $$\mathcal{X}$$, we have $$\mathcal{X}(a,b) = \mathcal{X}(c,d)$$ for all objects $$a,b,c,d$$.

Daniel Wang noted that this is true back in comment #10. It's the last case in his table:

$$\begin{array}{c|c} \text{monoid} & \text{effect of b) on \mathcal{X}(x,y)} \\ \hline (\oplus, \tt{false}) & \text{If \mathcal{X}(x,y), then for all objects z, either \mathcal{X}(x,z) or \mathcal{X}(z,x) or both.} \\ (\wedge, \tt{true}) & \text{If \mathcal{X}(x,y) and \mathcal{X}(y,z), then \mathcal{X}(x,z).} \\ (\leftrightarrow, \tt{true}) & \text{If \mathcal{X}(x,y) = \mathcal{X}(y,z), then \mathcal{X}(x,z).} \\ (\vee, \tt{false}) & \text{Either \mathcal{X}(x,y) = \tt{false} for all objects x,y, or \mathcal{X}(x,y) = \tt{true} for all objects x,y.} \\ \end{array}$$ So far, so good. But I'd prefer to make $$\textbf{Belnap4}$$ into a monoidal preorder using the meet operation $$\wedge$$. If we do this, it's isomorphic to $$(\mathbf{Bool},\wedge) \times \mathbf{Bool},\wedge)$$, so a $$\mathbf{Belnap4}$$ -category is just the same as a set equipped with two independent preorders.

Comment Source:Matthew wrote: > **Theorem**. In a **Belnap4**-enriched category \$$\mathcal{X}\$$, then for all \$$a,b,c\$$ and \$$d\$$ we have \$$\mathcal{X}(a,b) = \mathcal{X}(c,d) \$$. It's good to emphasize here that you are making \$$\mathbf{Belnap4}\$$ into a monoidal preorder using the join operation \$$\vee\$$. Here's another proof of this result. I'll just sketch it: If we make \$$\mathbf{Belnap4}\$$ into a monoidal preorder using \$$\vee\$$, it's isomorphic to \$$(\mathbf{Bool},\vee) \times (\mathbf{Bool},\vee)\$$ - the product of two copies of the Booleans with their join operation as the monoid structure. Thus, a \$$(\mathbf{Belnap4},\vee)\$$ -category with some set \$$S\$$ of objects is the same as two separate \$$(\mathbf{Bool},\vee) \$$-categories with the same set \$$S\$$ of objects! Thus, it suffices to prove that for any \$$(\mathbf{Bool},\vee) \$$-category \$$\mathcal{X}\$$, we have \$$\mathcal{X}(a,b) = \mathcal{X}(c,d) \$$ for all objects \$$a,b,c,d\$$. Daniel Wang noted that this is true back in [comment #10](https://forum.azimuthproject.org/discussion/comment/18390/#Comment_18390). It's the last case in his table: $$\begin{array}{c|c} \text{monoid} & \text{effect of b) on \mathcal{X}(x,y)} \\\\ \hline (\oplus, \tt{false}) & \text{If \mathcal{X}(x,y), then for all objects z, either \mathcal{X}(x,z) or \mathcal{X}(z,x) or both.} \\\\ (\wedge, \tt{true}) & \text{If \mathcal{X}(x,y) and \mathcal{X}(y,z), then \mathcal{X}(x,z).} \\\\ (\leftrightarrow, \tt{true}) & \text{If \mathcal{X}(x,y) = \mathcal{X}(y,z), then \mathcal{X}(x,z).} \\\\ (\vee, \tt{false}) & \text{Either \mathcal{X}(x,y) = \tt{false} for all objects x,y, or \mathcal{X}(x,y) = \tt{true} for all objects x,y.} \\\\ \end{array}$$ So far, so good. But I'd prefer to make \$$\textbf{Belnap4}\$$ into a monoidal preorder using the meet operation \$$\wedge\$$. If we do this, it's isomorphic to \$$(\mathbf{Bool},\wedge) \times \mathbf{Bool},\wedge)\$$, so a \$$\mathbf{Belnap4}\$$ -category is just the same as a set equipped with two independent preorders.
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62.
edited May 2018

With regard to the confusion over terminology, the term "enriched category" is an example of what some people know as the Mathematical Red Herring Principle:

Mathematical Red Herring Principle A “red herring” need not, in general, be either red or a herring.

Just as John said these things are not generally categories, so too there is not necessarily anything you might recognise as enriching going on. Well, maybe it would be fair to think of these as enriched sets :-)

The terminology came about when the early examples of enriched categories that people considered were actually categories which had been 'enriched', but the terminology never changed when people realised that something much more general was going on.

The confusion exists amongst high-level mathematicians as well: I have been to a couple of talks recently in which a geometer or topologist has said "A $$\mathcal{V}$$-category is a category such that..." So don't feel bad if you think that the terminology is daft!

Comment Source:With regard to the confusion over terminology, the term "enriched category" is an example of what some people know as the [Mathematical Red Herring Principle](https://ncatlab.org/nlab/show/red+herring+principle): > **Mathematical Red Herring Principle** A “red herring” need not, in general, be either red or a herring. Just as John said these things are not generally categories, so too there is not necessarily anything you might recognise as enriching going on. Well, maybe it would be fair to think of these as enriched sets :-) The terminology came about when the early examples of enriched categories that people considered were actually categories which had been 'enriched', but the terminology never changed when people realised that something much more general was going on. The confusion exists amongst high-level mathematicians as well: I have been to a couple of talks recently in which a geometer or topologist has said "A \$$\mathcal{V}\$$-category is a category such that..." So don't feel bad if you think that the terminology is daft!
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63.
edited May 2018

I have a puzzle.

Define the category DiscretePosEnrich to be the category of $$\mathcal{V}$$-enriched categories where $$\mathcal{V}$$ is any monoidal preorder over a discrete poset. Morphisms over this category are maps $$\phi : \mathcal{X} \to \mathcal{Y}$$ obeying:

$$\mathcal{X}(a,b) \leq_{\mathcal{X}} \mathcal{X}(c,d) \implies \mathcal{Y}(\phi(a),\phi(b)) \leq_{\mathcal{Y}} \mathcal{Y}(\phi(c),\phi(d))$$

$$\mathcal{X}(a,b) \otimes_{\mathcal{X}} \mathcal{X}(c,d) \leq_{\mathcal{X}} \mathcal{X}(e,f) \implies \mathcal{Y}(\phi(a),\phi(b)) \otimes_{\mathcal{Y}} \mathcal{Y}(\phi(c),\phi(d)) \leq_{\mathcal{Y}} \mathcal{Y}(\phi(e),\phi(f))$$

$$\mathcal{X}(a,b) = I \implies \mathcal{Y}(\phi(a),\phi(b)) = I$$

Let Grp be the category of Groups with group homomorphisms as its morphisms.

Puzzle MD1. Show that there is mapping $$T : \mathbf{DiscretePosEnrich} \hookrightarrow \mathbf{Grp}$$ which is surjective up to isomorphism. Lift this map into a functor.

Comment Source:I have a puzzle. Define the category **DiscretePosEnrich** to be the category of \$$\mathcal{V}\$$-enriched categories where \$$\mathcal{V}\$$ is any monoidal preorder over a discrete poset. Morphisms over this category are maps \$$\phi : \mathcal{X} \to \mathcal{Y} \$$ obeying: \$$\mathcal{X}(a,b) \leq_{\mathcal{X}} \mathcal{X}(c,d) \implies \mathcal{Y}(\phi(a),\phi(b)) \leq_{\mathcal{Y}} \mathcal{Y}(\phi(c),\phi(d)) \$$ \$$\mathcal{X}(a,b) \otimes_{\mathcal{X}} \mathcal{X}(c,d) \leq_{\mathcal{X}} \mathcal{X}(e,f) \implies \mathcal{Y}(\phi(a),\phi(b)) \otimes_{\mathcal{Y}} \mathcal{Y}(\phi(c),\phi(d)) \leq_{\mathcal{Y}} \mathcal{Y}(\phi(e),\phi(f)) \$$ \$$\mathcal{X}(a,b) = I \implies \mathcal{Y}(\phi(a),\phi(b)) = I \$$ Let **Grp** be the category of [Groups](https://en.wikipedia.org/wiki/Abelian_group) with group homomorphisms as its morphisms. **Puzzle MD1**. Show that there is mapping \$$T : \mathbf{DiscretePosEnrich} \hookrightarrow \mathbf{Grp}\$$ which is *surjective* up to isomorphism. Lift this map into a functor.
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64.
edited May 2018

The $$\mathbf{Bool} \times \mathbf{Bool}$$ interpretation seems to extend nicely to let one talk about S indexed preorders as categories enriched in the space of (monotonic?) functions from S to $$\mathbf{Bool}$$.

Comment Source:The \$$\mathbf{Bool} \times \mathbf{Bool} \$$ interpretation seems to extend nicely to let one talk about S indexed preorders as categories enriched in the space of (monotonic?) functions from S to \$$\mathbf{Bool}\$$.
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65.
edited May 2018

red-herring comment Funny. The poset vs. preordered set vs. partially-order set, discussion that happened earlier seems to be a case of the mathematical red herring principle almost happening.

When we start with a set, let's call it $$\mathbb{X}$$ and some relation $$\oplus$$ we do this... $$\mathcal{X} := ( \mathbb{X}, \oplus )$$ ...and then apply some rules governing the behavior of $$\oplus$$ over $$\mathbb{X}$$. What do we call what was just done and what was formed namely $$\mathcal{X}$$?

One more question. Is this a case where all $$\mathcal{V}\text{-categories}$$ are $$\text{categories}$$ but not all $$\text{categories}$$ are $$\mathcal{V}\text{-categories}$$? Surely there will be some relationship between them [maybe I should just be patient and all will be revealed].

Comment Source:[red-herring comment](https://forum.azimuthproject.org/discussion/comment/18489/#Comment_18489) Funny. The poset vs. preordered set vs. partially-order set, discussion that happened earlier seems to be a case of the mathematical red herring principle almost happening. When we start with a set, let's call it \$$\mathbb{X} \$$ and some relation \$$\oplus \$$ we do this... $$\mathcal{X} := ( \mathbb{X}, \oplus )$$ ...and then apply some rules governing the behavior of \$$\oplus \$$ over \$$\mathbb{X} \$$. What do we call what was just done and what was formed namely \$$\mathcal{X} \$$? One more question. Is this a case where all \$$\mathcal{V}\text{-categories} \$$ are \$$\text{categories} \$$ but not all \$$\text{categories} \$$ are \$$\mathcal{V}\text{-categories} \$$? Surely there will be some relationship between them [maybe I should just be patient and all will be revealed].
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66.
edited May 2018

Fredrick: a category is a special case of an enriched category. A category is a $$\mathbf{Set}$$-category, since it has for any pair of objects a set of morphisms.

However, $$\mathbf{Set}$$ is not a monoidal preorder: it's something more general, called a monoidal category!

So, you can only understand enriched categories at a sufficient level of generality to understand what I'm talking about after you understand "categories enriched in monoidal categories". Doing this will require understanding categories first.

So, as a road to first understanding categories, saying "they're $$\mathbf{Set}$$-categories" is useless: you need to understand categories first!

However, it may be reassuring to know that in the end, once we define them in a fully general way, enriched categories are more general than categories.

maybe I should just be patient and all will be revealed.

All will be revealed. Well, a lot will be revealed. Section 4.4.4 of the book will explain "categories enriched in monoidal categories". After you understand that, you'll see that categories are $$\mathbf{Set}$$-categories, and thus a special case of enriched categories.

We're in the foothills of a huge mountain range of abstract ideas. This course will never make anywhere near the top. But until September, we will keep climbing. We'll keep coming back to the ideas we've seen so far, and keep generalizing them, and keep seeing them in new ways.

Comment Source:Fredrick: a category is a special case of an enriched category. A category is a \$$\mathbf{Set}\$$-category, since it has for any pair of objects a _set_ of morphisms. However, \$$\mathbf{Set}\$$ is not a monoidal preorder: it's something more general, called a monoidal category! So, you can only understand enriched categories at a sufficient level of generality to understand what I'm talking about after you understand "categories enriched in monoidal categories". Doing this will require understanding categories first. So, as a road to first understanding categories, saying "they're \$$\mathbf{Set}\$$-categories" is useless: you need to understand categories first! However, it may be reassuring to know that _in the end_, once we define them in a fully general way, enriched categories are more general than categories. > maybe I should just be patient and all will be revealed. All will be revealed. Well, _a lot_ will be revealed. Section 4.4.4 of the book will explain "categories enriched in monoidal categories". After you understand that, you'll see that categories are \$$\mathbf{Set}\$$-categories, and thus a special case of enriched categories. We're in the foothills of a huge mountain range of abstract ideas. This course will never make anywhere near the top. But until September, we will keep climbing. We'll keep coming back to the ideas we've seen so far, and keep generalizing them, and keep seeing them in new ways.
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67.

Fredrick: also read Remark 3.18 in the book!

Comment Source:Fredrick: also read Remark 3.18 in the book!
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68.
edited May 2018

Matthew wrote:

Define a DiscretePos-enriched category to be an enriched category with a discrete poset and an arbitrary symmetric monoid, obeying the rules John wrote in the lecture.

I'm confused by this. So far we've been talking about $$\mathcal{V}$$-enriched categories when $$\mathcal{V}$$ is a monoidal preorder. So, for example, there's a monoidal preorder $$\mathbf{Bool}$$, and that's what lets us talk about $$\mathbf{Bool}$$-enriched categories.

Your phrase "$$\mathbf{DiscretePos}$$-enriched category" would thus suggest that you have in mind some specific monoidal preorder $$\mathbf{DiscretePos}$$. If you did, and I knew what it was, I would instantly know what a $$\mathbf{DiscretePos}$$-enriched category was.

But you seem to be using this phrase in a different way.

And when you say "... to be an enriched category", you're not saying what it's enriched in, so we don't know enough to do anything. An enriched category that's not enriched in something is not a real thing. It's like a son that's not anyone's son.

Comment Source:Matthew wrote: > Define a **DiscretePos**-enriched category to be an enriched category with a discrete poset and an arbitrary symmetric monoid, obeying the rules John wrote in the lecture. I'm confused by this. So far we've been talking about \$$\mathcal{V}\$$-enriched categories when \$$\mathcal{V}\$$ is a monoidal preorder. So, for example, there's a monoidal preorder \$$\mathbf{Bool}\$$, and that's what lets us talk about \$$\mathbf{Bool}\$$-enriched categories. Your phrase "\$$\mathbf{DiscretePos}\$$-enriched category" would thus suggest that you have in mind some specific monoidal preorder \$$\mathbf{DiscretePos}\$$. If you did, and I knew what it was, I would instantly know what a \$$\mathbf{DiscretePos}\$$-enriched category was. But you seem to be using this phrase in a different way. And when you say "... to be an enriched category", you're not saying what it's enriched _in_, so we don't know enough to do anything. An enriched category that's not enriched _in_ something is not a real thing. It's like a son that's not anyone's son.
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69.

I'm confused by this. So far we've been talking about $$\mathcal{V}$$-enriched categories when $$\mathcal{V}$$ is a monoidal preorder. So, for example, there's a monoidal preorder $$\mathbf{Bool}$$, and that's what lets us talk about $$\mathbf{Bool}$$-enriched categories.

I am so sorry for the confusion, I have tried to clean up the question.

I edited the question to define a category DiscretePosEnrich. The idea behind this new category is that consists of $$\mathcal{V}$$-enriched categories for arbitrary $$\mathcal{V}$$. The only constraint is that $$\mathcal{V}$$ must have a discrete poset underlying its monoidal preorder.

I hope that clears this up. I can prove this tomorrow if nobody finds it interesting.

Comment Source:> I'm confused by this. So far we've been talking about \$$\mathcal{V}\$$-enriched categories when \$$\mathcal{V}\$$ is a monoidal preorder. So, for example, there's a monoidal preorder \$$\mathbf{Bool}\$$, and that's what lets us talk about \$$\mathbf{Bool}\$$-enriched categories. I am so sorry for the confusion, I have tried to clean up the question. I edited the question to define a category **DiscretePosEnrich**. The idea behind this new category is that consists of \$$\mathcal{V}\$$-enriched categories for arbitrary \$$\mathcal{V}\$$. The only constraint is that \$$\mathcal{V}\$$ must have a discrete poset underlying its monoidal preorder. I hope that clears this up. I can prove this tomorrow if nobody finds it interesting.
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70.

Define the category DiscretePosEnrich to be the category of $$\mathcal{V}$$-enriched categories where $$\mathcal{V}$$ is any monoidal preorder over a discrete poset. Morphisms over this category are maps $$\phi : \mathcal{X} \to \mathcal{Y}$$ obeying:

$$\mathcal{X}(a,b) \leq_{\mathcal{X}} \mathcal{X}(c,d) \implies \mathcal{Y}(\phi(a),\phi(b)) \leq_{\mathcal{Y}} \mathcal{Y}(\phi(c),\phi(d))$$

$$\mathcal{X}(a,b) \otimes_{\mathcal{X}} \mathcal{X}(c,d) \leq_{\mathcal{X}} \mathcal{X}(e,f) \implies \mathcal{Y}(\phi(a),\phi(b)) \otimes_{\mathcal{Y}} \mathcal{Y}(\phi(c),\phi(d)) \leq_{\mathcal{Y}} \mathcal{Y}(\phi(e),\phi(f))$$

$$\mathcal{X}(a,b) = I \implies \mathcal{Y}(\phi(a),\phi(b)) = I$$

Let Grp be the category of Groups with group homomorphisms as its morphisms.

Puzzle MD1. Show that there is mapping $$T : \mathbf{DiscretePosEnrich} \hookrightarrow \mathbf{Grp}$$ which is surjective up to isomorphism. Lift this map into a functor.

I am afraid this puzzle is a little lame. I'll try to cook up something more fun next time.

However, I did think this was cute when I thought it up so I'll just go over the answer.

First, consider any $$\mathcal{V}$$-enriched category $$\mathcal{X}$$ where $$\mathcal{V}$$ is a monoidal preorder over the discrete poset. This is an object in $$\mathbf{DiscretePosEnrich}$$.

There is a key observation: $$\mathcal{X}$$ has a natural group associated with it

To see this, first see that for any $$x$$ from rule (a) of enriched categories

$$I \leq \mathcal{X}(x,x)$$ But since $$\leq$$ is the discrete poset, it must be that

$$I = \mathcal{X}(x,x)$$ Next, observe that from rule (b) of enriched categories that

$$\mathcal{X}(x,y) \otimes \mathcal{X}(y,x) \leq \mathcal{X}(x,x)$$ But since $$I = \mathcal{X}(x,x)$$ and $$(\leq)$$ is the same as $$(=)$$, we have:

$$\mathcal{X}(x,y) \otimes \mathcal{X}(y,x) = I$$ We can also see that

$$\mathcal{X}(y,x) \otimes \mathcal{X}(x,y) = I$$ In other words, $$\mathcal{X}(y,x)$$ is the inverse of $$\mathcal{X}(x,y)$$.

This relation lifts further - consider any finite sequence $$\langle (a,b), (c,d), \ldots, (y,z)\rangle$$ of pairs of elements of $$\mathcal{X}$$. We can define $$\bigotimes_{\mathcal{X}}$$ to act on sequences as follows:

$$\bigotimes_{\mathcal{X}} \langle (a,b), (c,d), \ldots, (y,z)\rangle = I \otimes \mathcal{X}(a,b) \otimes \mathcal{X}(c,d) \otimes \ldots \otimes \mathcal{X}(y,z)$$ Next, define $$\mathrm{rev}$$ such that:

$$\mathrm{rev}(\langle (a,b), (c,d), \ldots, (y,z)\rangle ) = \langle (z,y), \ldots, (d,c), (b,a)\rangle$$ Then we have:

\begin{align} \bigotimes_{\mathcal{X}} \langle(a,b), (c,d), \ldots, (y,z)\rangle \otimes \bigotimes_{\mathcal{X}} \mathrm{rev}(\langle (a,b), (c,d), \ldots, (y,z)\rangle) = I \end{align} We also have a congruence relation $$\cong_{\mathcal{X}}$$ on finite sequences, defined as:

$$S_1 \cong_{\mathcal{X}} S_2 \iff \bigotimes_{\mathcal{X}} S_1 = \bigotimes_{\mathcal{X}} S_2$$ Let $$[S]$$ the equivalence class for $$S$$ under this congruence.

We now have enough to define the mapping $$T: \mathbf{DiscretePosEnrich} \to \mathbf{Grp}$$. In particular, it maps $$\mathcal{X} \mapsto G_\mathcal{X}$$, where $$G_\mathcal{X}$$ is defined:

• The domain of $$G_\mathcal{X}$$ are just the equivalence classes of finite sequences of pairs of elements of $$\mathcal{X}$$ modulo $$\cong_{\mathcal{X}}$$
• The group operation $$\odot$$ is defined by:

$$[X] \odot [Y] = \left[\bigotimes_{\mathcal{X}} X\otimes \bigotimes_{\mathcal{X}} Y\right]$$ (this can be defined other ways that make it easier to see its well defined, but they may not be so clear)
• Identity is $$[\langle\rangle]$$, the equivalence class generated by the empty sequence
• Inversion is $$[X]^{-1} = [\mathrm{rev}(X)]$$

That defines the mapping $$T$$. Next time I will give a right inverse to $$T$$ which turns any group into a $$\mathcal{V}$$-enriched category where $$\mathcal{V}$$ is a monoidal preorder over a poset.

Comment Source:> Define the category **DiscretePosEnrich** to be the category of \$$\mathcal{V}\$$-enriched categories where \$$\mathcal{V}\$$ is any monoidal preorder over a discrete poset. Morphisms over this category are maps \$$\phi : \mathcal{X} \to \mathcal{Y} \$$ obeying: > > \$$\mathcal{X}(a,b) \leq_{\mathcal{X}} \mathcal{X}(c,d) \implies \mathcal{Y}(\phi(a),\phi(b)) \leq_{\mathcal{Y}} \mathcal{Y}(\phi(c),\phi(d)) \$$ > > \$$\mathcal{X}(a,b) \otimes_{\mathcal{X}} \mathcal{X}(c,d) \leq_{\mathcal{X}} \mathcal{X}(e,f) \implies \mathcal{Y}(\phi(a),\phi(b)) \otimes_{\mathcal{Y}} \mathcal{Y}(\phi(c),\phi(d)) \leq_{\mathcal{Y}} \mathcal{Y}(\phi(e),\phi(f)) \$$ > > \$$\mathcal{X}(a,b) = I \implies \mathcal{Y}(\phi(a),\phi(b)) = I \$$ > > Let **Grp** be the category of [Groups](https://en.wikipedia.org/wiki/Abelian_group) with group homomorphisms as its morphisms. > > **Puzzle MD1**. Show that there is mapping \$$T : \mathbf{DiscretePosEnrich} \hookrightarrow \mathbf{Grp}\$$ which is *surjective* up to isomorphism. Lift this map into a functor. I am afraid this puzzle is a little lame. I'll try to cook up something more fun next time. However, I did think this was cute when I thought it up so I'll just go over the answer. First, consider any \$$\mathcal{V}\$$-enriched category \$$\mathcal{X}\$$ where \$$\mathcal{V}\$$ is a monoidal preorder over the discrete poset. This is an object in \$$\mathbf{DiscretePosEnrich}\$$. There is a key observation: *\$$\mathcal{X}\$$ has a natural group associated with it* To see this, first see that for any \$$x\$$ from rule (a) of enriched categories $$I \leq \mathcal{X}(x,x)$$ But since \$$\leq\$$ is the discrete poset, it must be that $$I = \mathcal{X}(x,x)$$ Next, observe that from rule (b) of enriched categories that $$\mathcal{X}(x,y) \otimes \mathcal{X}(y,x) \leq \mathcal{X}(x,x)$$ But since \$$I = \mathcal{X}(x,x)\$$ and \$$(\leq)\$$ is the same as \$$(=)\$$, we have: $$\mathcal{X}(x,y) \otimes \mathcal{X}(y,x) = I$$ We can also see that $$\mathcal{X}(y,x) \otimes \mathcal{X}(x,y) = I$$ In other words, \$$\mathcal{X}(y,x)\$$ is the *inverse* of \$$\mathcal{X}(x,y)\$$. This relation lifts further - consider any finite sequence \$$\langle (a,b), (c,d), \ldots, (y,z)\rangle \$$ of pairs of elements of \$$\mathcal{X}\$$. We can define \$$\bigotimes_{\mathcal{X}}\$$ to act on sequences as follows: $$\bigotimes_{\mathcal{X}} \langle (a,b), (c,d), \ldots, (y,z)\rangle = I \otimes \mathcal{X}(a,b) \otimes \mathcal{X}(c,d) \otimes \ldots \otimes \mathcal{X}(y,z)$$ Next, define \$$\mathrm{rev}\$$ such that: $$\mathrm{rev}(\langle (a,b), (c,d), \ldots, (y,z)\rangle ) = \langle (z,y), \ldots, (d,c), (b,a)\rangle$$ Then we have: \begin{align} \bigotimes_{\mathcal{X}} \langle(a,b), (c,d), \ldots, (y,z)\rangle \otimes \bigotimes_{\mathcal{X}} \mathrm{rev}(\langle (a,b), (c,d), \ldots, (y,z)\rangle) = I \end{align} We also have a *congruence relation* \$$\cong_{\mathcal{X}}\$$ on finite sequences, defined as: $$S_1 \cong_{\mathcal{X}} S_2 \iff \bigotimes_{\mathcal{X}} S_1 = \bigotimes_{\mathcal{X}} S_2$$ Let \$$[S]\$$ the equivalence class for \$$S\$$ under this congruence. We now have enough to define the mapping \$$T: \mathbf{DiscretePosEnrich} \to \mathbf{Grp}\$$. In particular, it maps \$$\mathcal{X} \mapsto G_\mathcal{X}\$$, where \$$G_\mathcal{X}\$$ is defined: - The *domain* of \$$G_\mathcal{X}\$$ are just the equivalence classes of finite sequences of pairs of elements of \$$\mathcal{X}\$$ [modulo](https://en.wikipedia.org/wiki/Quotient_algebra) \$$\cong_{\mathcal{X}}\$$ - The group operation \$$\odot\$$ is defined by: $$[X] \odot [Y] = \left[\bigotimes_{\mathcal{X}} X\otimes \bigotimes_{\mathcal{X}} Y\right]$$ (this can be defined other ways that make it easier to see its well defined, but they may not be so clear) - Identity is \$$[\langle\rangle]\$$, the equivalence class generated by the empty sequence - Inversion is \$$[X]^{-1} = [\mathrm{rev}(X)]\$$ --------------------------- That defines the mapping \$$T\$$. Next time I will give a *right inverse* to \$$T\$$ which turns any group into a \$$\mathcal{V}\$$-enriched category where \$$\mathcal{V}\$$ is a monoidal preorder over a poset.
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71.

Matthew, I don't understand your definition of maps in your category are. You say $$\mathcal{X}(a,b) \leq_{\mathcal{X}} \mathcal{X}(c,d)$$ but I don't know what $$\leq_{\mathcal{X}}$$ means.

My gut is saying that you should fix $$\mathcal{V}$$ and then look at the enriched categories for that $$\mathcal{V}$$ rather than sticking together all the enriched categories for all $$\mathcal{V}$$.

Anyway, it might be worth you going back a step. What structure does what you call 'a monoidal preorder over a discrete poset' have? What is another name for a monoidal preorder over a discrete poset?

Can you give an example of such a thing?

Can you give an example of an enriched category over such a thing?

Comment Source:Matthew, I don't understand your definition of maps in your category are. You say \$$\mathcal{X}(a,b) \leq_{\mathcal{X}} \mathcal{X}(c,d) \$$ but I don't know what \$$\leq_{\mathcal{X}}\$$ means. My gut is saying that you should fix \$$\mathcal{V}\$$ and then look at the enriched categories for that \$$\mathcal{V}\$$ rather than sticking together all the enriched categories for all \$$\mathcal{V}\$$. Anyway, it might be worth you going back a step. What structure does what you call 'a monoidal preorder over a discrete poset' have? What is another name for a monoidal preorder over a discrete poset? Can you give an example of such a thing? Can you give an example of an enriched category over such a thing?
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72.
edited May 2018

I am currently on the road, so please excuse my poor formatting.

This theorem does not fix $$\mathcal{V}$$. I have been reading your blog article from a few years ago and it’s references to acquaint myself with enriched categories, Simon. I see now that not fixing $$\mathcal{V}$$ is unorthodox. I am sorry, I didn’t intend to defy convention.

Consider some $$\mathcal{X}$$ enriched in some $$\mathcal{V}$$. Since I am not fixing a $$\mathcal{V}$$, I can’t talk about a particular preorder $$\le$$ when comparing $$\mathcal{X}(a,b)$$ and $$\mathcal{X}(c,d)$$. Because of this, I thought it would make since to clarify the order associated with $$\mathcal{X}$$ as $$\le_\mathcal{X}$$.

A ‘monoidal preorder over a discrete poset’ is a monoidal preorder where the underlying preorder is a discrete poset. That is not a great definition. I will attempt to clarify.

In a discrete poset, $$a \le b$$ is true if and only if $$a = b$$.

Every monoid can be made into a monoidal preorder over a discrete poset. Just construct the trivial poset over the domain of the monoid.

A simple example would be $$\mathbb{Z}/(2)$$. Since discrete posets are equivalence relations, we can represent them with their equivalence classes. In this case the equivalence class is {{0},{1}}.

I will try to go into how to construct an enriched category for the discretely ordered $$\mathbb{Z}/(2)$$ group or any other discretely ordered group later.

Comment Source:I am currently on the road, so please excuse my poor formatting. This theorem does not fix \$$\mathcal{V}\$$. I have been reading your blog article from a few years ago and it’s references to acquaint myself with enriched categories, Simon. I see now that not fixing \$$\mathcal{V}\$$ is unorthodox. I am sorry, I didn’t intend to defy convention. Consider some \$$\mathcal{X}\$$ enriched in some \$$\mathcal{V}\$$. Since I am not fixing a \$$\mathcal{V}\$$, I can’t talk about a particular preorder \$$\le\$$ when comparing \$$\mathcal{X}(a,b)\$$ and \$$\mathcal{X}(c,d)\$$. Because of this, I thought it would make since to clarify the order associated with \$$\mathcal{X}\$$ as \$$\le_\mathcal{X}\$$. A ‘monoidal preorder over a discrete poset’ is a monoidal preorder where the underlying preorder is a discrete poset. That is not a great definition. I will attempt to clarify. In a discrete poset, \$$a \le b\$$ is true if and only if \$$a = b\$$. Every monoid can be made into a monoidal preorder over a discrete poset. Just construct the trivial poset over the domain of the monoid. A simple example would be \$$\mathbb{Z}/(2)\$$. Since discrete posets are equivalence relations, we can represent them with their equivalence classes. In this case the equivalence class is {{0},{1}}. I will try to go into how to construct an enriched category for the discretely ordered \$$\mathbb{Z}/(2)\$$ group or any other discretely ordered group later.
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73.
edited May 2018

Christopher wrote:

The $$\mathbf{Bool} \times \mathbf{Bool}$$ interpretation seems to extend nicely to let one talk about $$S$$ indexed preorders as categories enriched in the space of (monotonic?) functions from $$S$$ to $$\mathbf{Bool}$$.

Yes! You didn't say what $$S$$ is, but if it's just a set we don't need "monotonic" functions from $$S$$ to $$\mathbf{Bool}$$; we can just look at all functions from $$S$$ to $$\mathbf{Bool}$$. This is called $$\mathbf{Bool}^S$$, because you can also think of it as a product of copies of $$\mathbf{Bool}$$, one for each element of $$S$$:

$$\prod_{s \in S} \mathbf{Bool} .$$ $$\mathbf{Bool}^S$$ becomes a symmetric monoidal poset in an obvious way, and an $$\mathbf{Bool}^S$$-category is the same as a set $$X$$ with a bunch of different preorders, one for each element of $$S$$.

The "obvious way": given $$f, g : S \to \mathbf{Bool}$$, we say $$f \le g$$ iff

$$f(s) \le g(s) \textrm{ for all } s \in S .$$ This makes $$\mathbf{Bool}^S$$ into a poset, and then we make it into a symmetric monoidal poset using $$\wedge$$.

But in fact, the poset $$\mathbf{Bool}^S$$ is isomorphic to $$P(S)$$, the power set of $$S$$, made into a poset using $$\subseteq$$. It's easiest to understand this by looking at the case where $$S$$ has 3 elements. Then $$P(S)$$ is a cube:

and you can see it's isomorphic to $$\mathbf{Bool}^S \cong \mathbf{Bool} \times \mathbf{Bool} \times \mathbf{Bool}$$. (Sorry, this picture calls $$S$$ "$$X$$".)

It's also interesting to think about what happens when $$S$$ is a preorder, and we look at monotone maps $$f : S \to \mathbf{Bool}$$.

Comment Source:Christopher wrote: > The \$$\mathbf{Bool} \times \mathbf{Bool} \$$ interpretation seems to extend nicely to let one talk about \$$S\$$ indexed preorders as categories enriched in the space of (monotonic?) functions from \$$S\$$ to \$$\mathbf{Bool}\$$. Yes! You didn't say what \$$S\$$ is, but if it's just a set we don't need "monotonic" functions from \$$S\$$ to \$$\mathbf{Bool}\$$; we can just look at _all_ functions from \$$S\$$ to \$$\mathbf{Bool}\$$. This is called \$$\mathbf{Bool}^S\$$, because you can also think of it as a product of copies of \$$\mathbf{Bool}\$$, one for each element of \$$S\$$: $\prod_{s \in S} \mathbf{Bool} .$ \$$\mathbf{Bool}^S\$$ becomes a symmetric monoidal poset in an obvious way, and an \$$\mathbf{Bool}^S\$$-category is the same as a set \$$X\$$ with a bunch of different preorders, one for each element of \$$S\$$. The "obvious way": given \$$f, g : S \to \mathbf{Bool}\$$, we say \$$f \le g\$$ iff $f(s) \le g(s) \textrm{ for all } s \in S .$ This makes \$$\mathbf{Bool}^S\$$ into a poset, and then we make it into a symmetric monoidal poset using \$$\wedge \$$. But in fact, the poset \$$\mathbf{Bool}^S\$$ is isomorphic to \$$P(S)\$$, the power set of \$$S\$$, made into a poset using \$$\subseteq\$$. It's easiest to understand this by looking at the case where \$$S\$$ has 3 elements. Then \$$P(S)\$$ is a cube: <center><img src = "http://math.ucr.edu/home/baez/mathematical/7_sketches/P3_hasse_diagram.png"></center> and you can see it's isomorphic to \$$\mathbf{Bool}^S \cong \mathbf{Bool} \times \mathbf{Bool} \times \mathbf{Bool} \$$. (Sorry, this picture calls \$$S\$$ "\$$X\$$".) It's also interesting to think about what happens when \$$S\$$ is a _preorder_, and we look at _monotone_ maps \$$f : S \to \mathbf{Bool}\$$.
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74.
edited May 2018

So if i am reading Wikipedia right:

If we take define p as so: $p(X) = \{ s | s \subset X \land \text{s is finite} \}$ then (contravariant) monotone functions to bool from S are isomorphic to elements of the free distributive lattice on X.

Now, these still are preorders enriched over p(X), so that tells us what the "transitivity" looks like, its pointwise relative ti p(x), and so: $\mathcal{X}(a,b)=p\, \land \, \mathcal{X}(b,c)=q\quad\to\quad \mathcal{X}(a,c)\ge\,p\land q$

So these are categories enriched in a distributive lattice using $$\land$$. (honestly the terminology [name of monoidal preorder]-enriched set would make a lot more sense. Oh well.)

And I think, Birkhoff's representation theorem (as mentioned here https://en.wikipedia.org/wiki/Antichain) means that we can construct every category enriched by a distributive latice in this way.

Now, I am not sure what this does exactly if we use it on some preorder S which is $$not$$ isomorphic to a powerset. but it should still have something to do with lattices.

Comment Source:So if i am reading Wikipedia right: If we take define p as so: \$p(X) = \\{ s | s \subset X \land \text{s is finite} \\} \$ then (contravariant) monotone functions to bool from S are isomorphic to elements of the free distributive lattice on X. Now, these still are preorders enriched over p(X), so that tells us what the "transitivity" looks like, its pointwise relative ti p(x), and so: \$\mathcal{X}(a,b)=p\, \land \, \mathcal{X}(b,c)=q\quad\to\quad \mathcal{X}(a,c)\ge\,p\land q\$ So these are categories enriched in a distributive lattice using \$$\land\$$. (honestly the terminology [name of monoidal preorder]-enriched set would make a lot more sense. Oh well.) And I think, Birkhoff's representation theorem (as mentioned here https://en.wikipedia.org/wiki/Antichain) means that we can construct every category enriched by a distributive latice in this way. Now, I am not sure what this does exactly if we use it on some preorder S which is \$$not\$$ isomorphic to a powerset. but it should still have something to do with lattices.
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75.
edited May 2018

Just one small remark while I ponder the deeper stuff:

Honestly the terminology [name of monoidal preorder]-enriched set would make a lot more sense. Oh well.

We'll soon see that

• preorders are a special case of categories
• monoidal preorders are a special case of monoidal categories, and
• categories enriched over monoidal preorders are a special case of categories enriched over monoidal categories.

So, our use of the term "enriched categories" is looking forward to this realization. The things we're calling enriched categories now are enriched categories - just a special case.

Comment Source:Just one small remark while I ponder the deeper stuff: > Honestly the terminology [name of monoidal preorder]-enriched set would make a lot more sense. Oh well. We'll soon see that * preorders are a special case of categories * monoidal preorders are a special case of monoidal categories, and * categories enriched over monoidal preorders are a special case of categories enriched over monoidal categories. So, our use of the term "enriched categories" is looking forward to this realization. The things we're calling enriched categories now _are_ enriched categories - just a special case.
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76.

Can you give an example of an enriched category over such a thing?

Alright, I am going to attempt to show how to construct an discretely enriched category from any group $$G = \langle D, I, \odot, (\cdot)^{-1} \rangle$$.

• Order is defined to be $$a \leq b \equiv a = b$$. This is a discrete preorder.
• The monoidal identity is $$I$$. The monoidal tensor $$\otimes = \odot$$. We just use the group and forget inversion.
• $$Obj(\mathcal{X}) = D$$
• Finally, $$\mathcal{X}(x,y) = x \odot y^{-1}$$

This is the right inverse of $$T$$.

Comment Source:> Can you give an example of an enriched category over such a thing? Alright, I am going to attempt to show how to construct an discretely enriched category from any group \$$G = \langle D, I, \odot, (\cdot)^{-1} \rangle \$$. - Order is defined to be \$$a \leq b \equiv a = b\$$. This is a discrete preorder. - The monoidal identity is \$$I\$$. The monoidal tensor \$$\otimes = \odot\$$. We just use the group and forget inversion. - \$$Obj(\mathcal{X}) = D\$$ - Finally, \$$\mathcal{X}(x,y) = x \odot y^{-1}\$$ This is the *right inverse* of \$$T\$$.
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77.
edited May 2018

Matthew: when you say

Alright, I am going to attempt to show how to construct an discretely enriched category...

you need to say enriched in what. It's like saying you're going to construct a module without saying what ring it's a module over. An enriched category that's not enriched in something is not a real thing. It's like a son that's not anyone's son.

The phrase "discretely enriched" doesn't mean anything to me, and I've been around the block a few times. Maybe this is supposed to answer the question "enriched in what?" For example, maybe you mean "enriched in $$\mathbf{Set}$$" or something like that (though we haven't talked about categories enriched in monoidal categories yet, and a category enriched in $$\mathbf{Set}$$ is just a category). But it's not how people would express such a thought.

Comment Source:Matthew: when you say > Alright, I am going to attempt to show how to construct an discretely enriched category... you need to say enriched _in what_. It's like saying you're going to construct a module without saying what ring it's a module over. An enriched category that's not enriched in something is not a real thing. It's like a son that's not anyone's son. The phrase "discretely enriched" doesn't mean anything to me, and I've been around the block a few times. Maybe this is supposed to answer the question "enriched in _what?_" For example, maybe you mean "enriched in \$$\mathbf{Set}\$$" or something like that (though we haven't talked about categories enriched in monoidal categories yet, and a category enriched in \$$\mathbf{Set}\$$ is just a category). But it's not how people would express such a thought.
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78.

I´ve been trying to approach it by structures..

I mean, given a structure that is related to a categorie X (if degenerated, also a preorder).. if one turn this structure more complex somehow, in some specific feature. then it was reached an enriched (on that feature) category X,is it ok? But, if is only about structure, the enriched categorie X would have some "analog" categorie Y, both sharing the same structure.. I guess this is wrong... because there is some structures that would be "reserved" only to enriched categories..I did get it this way ..Could you check this, please? best

Comment Source:I´ve been trying to approach it by structures.. I mean, given a structure that is related to a categorie X (if degenerated, also a preorder).. if one turn this structure more complex somehow, in some specific feature. then it was reached an enriched (on that feature) category X,is it ok? But, if is only about structure, the enriched categorie X would have some "analog" categorie Y, both sharing the same structure.. I guess this is wrong... because there is some structures that would be "reserved" only to enriched categories..I did get it this way ..Could you check this, please? best
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79.
edited May 2018

you need to say enriched in what. It's like saying you're going to construct a module without saying what ring it's a module over. An enriched category that's not enriched in something is not a real thing. It's like a son that's not anyone's son.

I am very sorry, I did not initially understand the terminology and common way that Mathematicians use the term "enriched category".

That being said, I do not feel what I proved is wrong. I make mistakes all the time, but this time I believe I did not make a mistake in reasoning. I made a mistake in understanding convention.

I will try to explain my reasoning from the top, in case anyone is interested.

Let $$\mathcal{V} = \langle V, I, \otimes, \leq\rangle$$ be a monoidal preorder. Furthermore, let the preorder be a discrete order.

Let $$\mathcal{X}$$ be a $$\mathcal{V}$$-enriched preorder. Then $$\mathcal{X}(a,b) \otimes \mathcal{X}(b,a) = I$$. Take the set $$S = \{\mathcal{X}(a,b)\; :\; a,b \in \mathrm{Obj}(\mathcal{X})\}$$. Consider the submonoid of $$\mathcal{V}$$ generated by $$S$$. Every element of this submonoid has an inverse. So it is a group. Call this group $$G_\mathcal{X}$$.

For any monoidal preorder $$\mathcal{V}$$ where the preorder is discrete, for any $$\mathcal{X}$$ enriched in $$\mathcal{V}$$, let $$T$$ be a map $$\mathcal{X} \mapsto G_\mathcal{X}$$.

Now consider any group $$G = \langle Dom_G, I, \odot, (\cdot)^{-1}\rangle$$. $$G$$ is a monoidal preorder under the discrete order. We can define a $$G$$-enriched category $$\mathcal{X}_G$$. Here $$\mathrm{Obj}(\mathcal{X}_G) = Dom_G$$ and $$\mathcal{X}_G(x,y) = x \odot y^{-1}$$.

Let $$U$$ be a map such that $$G \mapsto \mathcal{X}_G$$ for any group.

Moreover, we have $$(T \circ U) (G) \cong G$$. This means that $$T$$ is a left inverse of $$U$$ (up to isomorphism).

We can lift $$T$$ and $$U$$ into functors, but we have to use some nonstandard definitions of morphisms between enriched categories.

Hopefully, that makes a some sense. I am so very sorry for not being clear about this, and my misunderstanding of the conventions.

Comment Source:> you need to say enriched _in what_. It's like saying you're going to construct a module without saying what ring it's a module over. An enriched category that's not enriched in something is not a real thing. It's like a son that's not anyone's son. I am very sorry, I did not initially understand the terminology and common way that Mathematicians use the term "enriched category". That being said, I do not feel what I proved is wrong. I make mistakes all the time, but this time I believe I did not make a mistake in reasoning. I made a mistake in understanding convention. I will try to explain my reasoning from the top, in case anyone is interested. Let \$$\mathcal{V} = \langle V, I, \otimes, \leq\rangle \$$ be a monoidal preorder. Furthermore, let the preorder be a [*discrete order*](https://www.encyclopediaofmath.org/index.php/Discrete_order). Let \$$\mathcal{X}\$$ be a \$$\mathcal{V}\$$-enriched preorder. Then \$$\mathcal{X}(a,b) \otimes \mathcal{X}(b,a) = I\$$. Take the set \$$S = \\{\mathcal{X}(a,b)\; :\; a,b \in \mathrm{Obj}(\mathcal{X})\\}\$$. Consider the submonoid of \$$\mathcal{V}\$$ [generated](https://en.wikipedia.org/wiki/Monoid#Generators) by \$$S\$$. Every element of this submonoid has an inverse. So it is a group. Call this group \$$G_\mathcal{X}\$$. For any monoidal preorder \$$\mathcal{V}\$$ where the preorder is discrete, for any \$$\mathcal{X}\$$ enriched in \$$\mathcal{V}\$$, let \$$T\$$ be a map \$$\mathcal{X} \mapsto G_\mathcal{X}\$$. Now consider any group \$$G = \langle Dom_G, I, \odot, (\cdot)^{-1}\rangle \$$. \$$G\$$ is a monoidal preorder under the discrete order. We can define a \$$G\$$-enriched category \$$\mathcal{X}_G\$$. Here \$$\mathrm{Obj}(\mathcal{X}_G) = Dom_G\$$ and \$$\mathcal{X}_G(x,y) = x \odot y^{-1}\$$. Let \$$U\$$ be a map such that \$$G \mapsto \mathcal{X}_G\$$ for any group. Moreover, we have \$$(T \circ U) (G) \cong G\$$. This means that \$$T\$$ is a left inverse of \$$U\$$ (up to isomorphism). We can lift \$$T\$$ and \$$U\$$ into functors, but we have to use some nonstandard definitions of morphisms between enriched categories. Hopefully, that makes a some sense. I am so very sorry for not being clear about this, and my misunderstanding of the conventions.
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80.
edited May 2018

So if I am reading wikipedia right, if we take $$p(X) = \{ s | s \subset X \land s \text{ is finite}\}$$ then monotone functions to $$\mathbf{Bool}$$ from $$p(X)$$ are isomorphic to elements of the free distributive lattice.

Now, there's an identity on objects functor whose actions on the Hom objects is the inclusion of monotonic functions in the set of functions. So these are still preorders indexed by $$p(X)$$, but ones that respect inclusion, that is each link exists if and only if a expression built From elements of $$X$$, and the operators or and and is true, given that the variables in the indexing power set are true.

Now this is a interesting logic to work in, because it doesn't distinguish Boolean and intuitonistic logic. You need to be able to use "not" just to talk about q being false or unknown, let alone distinguish the two.

You can distinguish the partition logic though, because accumulation isn't true for dit sets on more then 2 elements.

Comment Source:So if I am reading wikipedia right, if we take \$$p(X) = \\{ s | s \subset X \land s \text{ is finite}\\} \$$ then monotone functions to \$$\mathbf{Bool}\$$ from \$$p(X)\$$ are isomorphic to elements of the free distributive lattice. Now, there's an identity on objects functor whose actions on the Hom objects is the inclusion of monotonic functions in the set of functions. So these are still preorders indexed by \$$p(X)\$$, but ones that respect inclusion, that is each link exists if and only if a expression built From elements of \$$X\$$, and the operators or and and is true, given that the variables in the indexing power set are true. Now this is a interesting logic to work in, because it doesn't distinguish Boolean and intuitonistic logic. You need to be able to use "not" just to talk about q being false or unknown, let alone distinguish the two. You can distinguish the partition logic though, because accumulation isn't true for dit sets on more then 2 elements.
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81.
edited May 2018

Matthew wrote:

I am very sorry, I did not initially understand the terminology and common way that mathematicians use the term "enriched category".

Sorry to be so fierce: I'm just playing the teacher, and one of the teacher's jobs is to get people to talk the official standard way, so they can communicate. I figured that if you, one of the best students, hadn't yet absorbed how to talk about enriched categories, I really need to emphasize how it's done.

That being said, I do not feel what I proved is wrong. I make mistakes all the time, but this time I believe I did not make a mistake in reasoning.

I bet you're doing something interesting, because you usually do. Let's see if I can understand it.

I will try to explain my reasoning from the top, in case anyone is interested.

Let $$\mathcal{V} = \langle V, I, \otimes, \leq\rangle$$ be a monoidal preorder. Furthermore, let the preorder be a discrete order.

Okay: a discrete preorder is one where $$x \le y$$ iff $$x = y$$. A discrete preorder is secretly just a way of treating a set as a preorder. So, a discrete monoidal preorder is secretly just a monoid!

We can indeed do enrichment in monoids, and this can be very interesting. For example consider any group $$G$$ regarded as a monoidal preorder as above. Then there's a cute way to get a category $$\mathcal{G}$$ that's enriched in $$G$$. The objects of $$\mathcal{G}$$ are just elements of $$G$$, and for any two objects $$g,h$$ we define

$$\mathcal{G}(g,h) = g^{-1} h \in G .$$ People say for short that "any group is a category enriched in itself".

You seem to be doing something related to this. Let's see:

Let $$\mathcal{X}$$ be a $$\mathcal{V}$$-enriched preorder. Then $$\mathcal{X}(a,b) \otimes \mathcal{X}(b,a) = I$$.

Hmm, interesting! At first I didn't see how you got this, but now it's clear to me. In any category $$\mathcal{X}$$ enriched over any monoidal preorder $$\mathcal{V}$$ we have

$$I \le \mathcal{X}(a,a)$$ and

$$\mathcal{X}(a,b) \otimes \mathcal{X}(b,a) \le \mathcal{X}(a,a)$$ but if $$\mathcal{V}$$ is discrete these inequalities are equations, so we get

$$I = \mathcal{X}(a,b) \otimes \mathcal{X}(b,a)$$ which means that the subset

$$\{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \} \subseteq \mathcal{V}$$ is actually a group contained in the monoid $$\mathcal{V}$$.

Okay, I'll read on and see what you're doing....

Comment Source:Matthew wrote: > I am very sorry, I did not initially understand the terminology and common way that mathematicians use the term "enriched category". Sorry to be so fierce: I'm just playing the teacher, and one of the teacher's jobs is to get people to talk the official standard way, so they can communicate. I figured that if _you_, one of the best students, hadn't yet absorbed how to talk about enriched categories, I really need to emphasize how it's done. > That being said, I do not feel what I proved is wrong. I make mistakes all the time, but this time I believe I did not make a mistake in reasoning. I bet you're doing something interesting, because you usually do. Let's see if I can understand it. > I will try to explain my reasoning from the top, in case anyone is interested. > Let \$$\mathcal{V} = \langle V, I, \otimes, \leq\rangle \$$ be a monoidal preorder. Furthermore, let the preorder be a [*discrete order*](https://www.encyclopediaofmath.org/index.php/Discrete_order). Okay: a discrete preorder is one where \$$x \le y\$$ iff \$$x = y\$$. A discrete preorder is secretly just a way of treating a set as a preorder. So, a discrete monoidal preorder is secretly just a monoid! We can indeed do enrichment in monoids, and this can be very interesting. For example consider any group \$$G\$$ regarded as a monoidal preorder as above. Then there's a cute way to get a category \$$\mathcal{G}\$$ that's enriched in \$$G\$$. The objects of \$$\mathcal{G}\$$ are just elements of \$$G\$$, and for any two objects \$$g,h\$$ we define $\mathcal{G}(g,h) = g^{-1} h \in G .$ People say for short that "any group is a category enriched in itself". You seem to be doing something related to this. Let's see: > Let \$$\mathcal{X}\$$ be a \$$\mathcal{V}\$$-enriched preorder. Then \$$\mathcal{X}(a,b) \otimes \mathcal{X}(b,a) = I\$$. Hmm, interesting! At first I didn't see how you got this, but now it's clear to me. In any category \$$\mathcal{X}\$$ enriched over any monoidal preorder \$$\mathcal{V}\$$ we have $I \le \mathcal{X}(a,a)$ and $\mathcal{X}(a,b) \otimes \mathcal{X}(b,a) \le \mathcal{X}(a,a)$ but if \$$\mathcal{V}\$$ is discrete these inequalities are equations, so we get $I = \mathcal{X}(a,b) \otimes \mathcal{X}(b,a)$ which means that the subset $\\{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \\} \subseteq \mathcal{V}$ is actually a group contained in the monoid \$$\mathcal{V}\$$. Okay, I'll read on and see what you're doing....
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82.
edited May 2018

Sorry to be so fierce: I'm just playing the teacher, and one of the teacher's jobs is to get people to talk the official standard way, so they can communicate. I figured that if you, one of the best students, hadn't yet absorbed how to talk about enriched categories, I really need to emphasize how it's done.

Thank you for your kind words John.

I have been learning an awful lot from you and the other members on this forum.

I believe I have a handle on enriched categories now!

If $$\mathcal{V}$$ is discrete these inequalities are equations, so we get

$$I = \mathcal{X}(a,b) \otimes \mathcal{X}(b,a)$$ which means that the subset $$\{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \} \subseteq \mathcal{V}$$ is actually a group contained in the monoid $$\mathcal{V}$$.

Not exactly.

The set $$\{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \}$$ may not be closed under $$\otimes$$.

It generates a group, however.

Comment Source:> Sorry to be so fierce: I'm just playing the teacher, and one of the teacher's jobs is to get people to talk the official standard way, so they can communicate. I figured that if you, one of the best students, hadn't yet absorbed how to talk about enriched categories, I really need to emphasize how it's done. Thank you for your kind words John. I have been learning an awful lot from you and the other members on this forum. I believe I have a handle on enriched categories now! > If \$$\mathcal{V}\$$ is discrete these inequalities are equations, so we get > > $I = \mathcal{X}(a,b) \otimes \mathcal{X}(b,a)$ > > which means that the subset > > $\{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \} \subseteq \mathcal{V}$ > > is actually a group contained in the monoid \$$\mathcal{V}\$$. Not exactly. The set \$$\\{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \\}\$$ may not be closed under \$$\otimes\$$. It *generates* a group, however.
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83.

Maybe these are very basic questions or I have neglected something, but I'm a bit confused by the two requirements in the definition of "enriched category", i.e. $$I\leq\mathcal{X}(x,x)$$ and $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$. First, why does the definition use "≤" instead of "=" or "≥"? Second, why does the relationship between $$\mathcal{X}(x,x)$$ and $$I$$ matter (such that it must be written into the definition)?

Comment Source:Maybe these are very basic questions or I have neglected something, but I'm a bit confused by the two requirements in the definition of "enriched category", i.e. \$$I\leq\mathcal{X}(x,x) \$$ and \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \$$. First, why does the definition use "≤" instead of "=" or "≥"? Second, why does the relationship between \$$\mathcal{X}(x,x) \$$ and \$$I \$$ matter (such that it must be written into the definition)?
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84.
edited June 2018

Maybe these are very basic questions or I have neglected something, but I'm a bit confused by the two requirements in the definition of "enriched category", i.e. $$I\leq\mathcal{X}(x,x)$$ and $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$. First, why does the definition use "≤" instead of "=" or "≥"? Second, why does the relationship between $$\mathcal{X}(x,x)$$ and $$I$$ matter (such that it must be written into the definition)?

In category theory, when we have two objects $$X$$ and $$Y$$ in a category we often care about the set of morphisms between them. This is the hom-set $$\mathrm{Hom}(X,Y)$$. This pays a central role in the Yoneda Lemma, for instance.

While this is a powerful generalization, it can be more general. Why do we demand the morphisms between $$X$$ and $$Y$$ be a set? Why not $$\{\mathtt{false}, \mathtt{true}\}$$, or a number, or a member of a group?

Enriched categories substitute the sets in hom-sets with an arbitrary monoidal category. Regular categories can be see as Set-enriched categories, where Set is the category of sets.

In order to make this happen, you need two things: composition of arrows and identity morphisms.

$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$ ensures morphisms are closed under composition.

$$I\leq\mathcal{X}(x,x)$$ effectively ensures the presence of identity morphisms.

Comment Source:> Maybe these are very basic questions or I have neglected something, but I'm a bit confused by the two requirements in the definition of "enriched category", i.e. \$$I\leq\mathcal{X}(x,x) \$$ and \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \$$. First, why does the definition use "≤" instead of "=" or "≥"? Second, why does the relationship between \$$\mathcal{X}(x,x) \$$ and \$$I \$$ matter (such that it must be written into the definition)? In category theory, when we have two objects \$$X\$$ and \$$Y\$$ in a category we often care about the set of morphisms between them. This is the [*hom-set*](https://en.wikipedia.org/wiki/Morphism#Hom-set) \$$\mathrm{Hom}(X,Y)\$$. This pays a central role in the [Yoneda Lemma](https://en.wikipedia.org/wiki/Yoneda_lemma), for instance. While this is a powerful generalization, it can be more general. Why do we demand the morphisms between \$$X\$$ and \$$Y\$$ be a set? Why not \$$\\{\mathtt{false}, \mathtt{true}\\}\$$, or a number, or a member of a group? Enriched categories substitute the sets in hom-sets with an arbitrary monoidal category. Regular categories can be see as **Set**-enriched categories, where **Set** is the category of sets. In order to make this happen, you need two things: composition of arrows and identity morphisms. \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \$$ ensures morphisms are closed under composition. \$$I\leq\mathcal{X}(x,x) \$$ effectively ensures the presence of identity morphisms.
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85.
edited May 2018

$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$ ensures morphisms are closed under composition.

One way to read this is that, while $$\mathcal{X}(x, z)$$ can be many things, it must be at least be the composition of paths from $$x$$ to $$z$$, where "at least" is defined however you like (i.e. by giving a preorder to enrich over). We're enforcing that compositions exist in just about as abstract a way as possible, substituting "be at least" in where we would otherwise say "contain" or "include".

This angle also explains $$I\leq\mathcal{X}(x,x)$$: whatever $$I$$ is, $$\mathcal{X}(x, x)$$ better be at least that much. It could have more (as is the case with monoid objects in general), but it should be at least as much as $$I$$.

Comment Source:[Matthew Doty wrote](https://forum.azimuthproject.org/discussion/comment/18702/#Comment_18702): > \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \$$ ensures morphisms are closed under composition. One way to read this is that, while \$$\mathcal{X}(x, z)\$$ can be many things, it must be at least be the composition of paths from \$$x\$$ to \$$z\$$, where "at least" is defined however you like (i.e. by giving a preorder to enrich over). We're enforcing that compositions exist in just about as abstract a way as possible, substituting "be at least" in where we would otherwise say "contain" or "include". This angle also explains \$$I\leq\mathcal{X}(x,x)\$$: whatever \$$I\$$ is, \$$\mathcal{X}(x, x)\$$ better be _at least_ that much. It could have more (as is the case with monoid objects in general), but it should be at least as much as \$$I\$$.
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86.

Matthew, so John said what you probably knew, but hadn't said explicitly, and I was trying to get you to say (!), a monoidal preorder over a discrete poset is just a monoid (thought of in a specific way).

You said

I see now that not fixing $$\mathcal{V}$$ is unorthodox. I am sorry, I didn’t intend to defy convention.

I don't care too much about you defying convention, I care a lot more about you being right!

You are trying to define a functor from a category of categories enriched over (all) monoids to the category of groups. I'm not convinced that you have yet defined what the morphisms in your category of enriched categories are. Your functor to the category of groups needs to take morphisms to group homomorphisms.

Comment Source:Matthew, so John said what you probably knew, but hadn't said explicitly, and I was trying to get you to say (!), a monoidal preorder over a discrete poset is just a monoid (thought of in a specific way). You said > I see now that not fixing \$$\mathcal{V}\$$ is unorthodox. I am sorry, I didn’t intend to defy convention. I don't care too much about you defying convention, I care a lot more about you being right! You are trying to define a functor from a category of categories enriched over (all) monoids to the category of groups. I'm not convinced that you have yet defined what the morphisms in your category of enriched categories are. Your functor to the category of groups needs to take morphisms to group homomorphisms.
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87.
edited May 2018

Hey Simon,

I was trying to get you to say (!), a monoidal preorder over a discrete poset is just a monoid (thought of in a specific way).

Oh, I am sorry for not saying that.

I'm not convinced that you have yet defined what the morphisms in your category of enriched categories are.

Let $$\mathcal{X}$$ be enriched in $$\mathcal{V}$$ and $$\mathcal{Y}$$ be enriched in $$\mathcal{U}$$. Morphisms are maps $$\phi : \mathcal{X} \to \mathcal{Y}$$ obeying:

$$\mathcal{X}(a,b) \leq_{\mathcal{V}} \mathcal{X}(c,d) \implies \mathcal{Y}(\phi(a),\phi(b)) \leq_{\mathcal{U}} \mathcal{Y}(\phi(c),\phi(d))$$

$$\mathcal{X}(a,b) \otimes_{\mathcal{V}} \mathcal{X}(c,d) \leq_{\mathcal{V}} \mathcal{X}(e,f) \implies \mathcal{Y}(\phi(a),\phi(b)) \otimes_{\mathcal{U}} \mathcal{Y}(\phi(c),\phi(d)) \leq_{\mathcal{U}} \mathcal{Y}(\phi(e),\phi(f))$$

$$\mathcal{X}(a,b) = I_{\mathcal{V}} \implies \mathcal{Y}(\phi(a),\phi(b)) = I_{\mathcal{U}}$$

My idea is that $$\mathcal{X}$$ would be enriched in one monoid and $$\mathcal{Y}$$ would be enriched in another, maybe different monoid.

Since $$\mathcal{X}$$ and $$\mathcal{Y}$$ may have different monoids associated with them, I indexed their operations.

Now I look closely as this, I think these axioms need to be strengthened.

Here is just one axiom that strengthens all three. I believe this works. It is rather long...

$$I_{\mathcal{V}} \otimes_{\mathcal{V}} \mathcal{X}(a,b) \otimes_{\mathcal{V}} \mathcal{X}(c,d) \otimes_{\mathcal{V}} \cdots \otimes_{\mathcal{V}} \mathcal{X}(y,z) \leq_{\mathcal{V}} I_{\mathcal{V}} \otimes_{\mathcal{V}} \mathcal{X}(p,q) \otimes_{\mathcal{V}} \mathcal{X}(r,s) \otimes_{\mathcal{V}} \cdots \otimes_{\mathcal{V}} \mathcal{X}(w,x)$$

$$\implies I_{\mathcal{U}} \otimes_{\mathcal{U}} \mathcal{Y}(\phi(a),\phi(b)) \otimes_{\mathcal{U}} \mathcal{Y}(\phi(c),\phi(d)) \otimes_{\mathcal{U}} \cdots \otimes_{\mathcal{U}} \mathcal{Y}(\phi(y),\phi(z))$$

$$\ \ \ \ \ \ \ \ \ \ \leq_{\mathcal{U}} I_{\mathcal{U}} \otimes_{\mathcal{U}} \mathcal{Y}(\phi(p),\phi(q)) \otimes_{\mathcal{U}} \mathcal{Y}(\phi(r),\phi(s)) \otimes_{\mathcal{U}} \cdots \otimes_{\mathcal{U}} \mathcal{Y}(\phi(w),\phi(x))$$

I can prove this turns into a group homomorphism for the generated groups John mentions in #81. Just tell me if you would like to see me do the proof.

Similarly John also mentions how to take a group $$G$$ to a category $$\mathcal{G}$$ enriched in $$G$$ in #81 (I mention another way in #79 and #76). This operation can be lifted to make enriched-category morphisms that obey the monster axiom. Again, I can attempt to prove this if you want.

Comment Source:Hey Simon, > I was trying to get you to say (!), a monoidal preorder over a discrete poset is just a monoid (thought of in a specific way). Oh, I am sorry for not saying that. > I'm not convinced that you have yet defined what the morphisms in your category of enriched categories are. I had written in [#63](https://forum.azimuthproject.org/discussion/comment/18495/#Comment_18495) > Let \$$\mathcal{X}\$$ be enriched in \$$\mathcal{V}\$$ and \$$\mathcal{Y}\$$ be enriched in \$$\mathcal{U}\$$. Morphisms are maps \$$\phi : \mathcal{X} \to \mathcal{Y} \$$ obeying: > > \$$\mathcal{X}(a,b) \leq_{\mathcal{V}} \mathcal{X}(c,d) \implies \mathcal{Y}(\phi(a),\phi(b)) \leq_{\mathcal{U}} \mathcal{Y}(\phi(c),\phi(d)) \$$ > > \$$\mathcal{X}(a,b) \otimes_{\mathcal{V}} \mathcal{X}(c,d) \leq_{\mathcal{V}} \mathcal{X}(e,f) \implies \mathcal{Y}(\phi(a),\phi(b)) \otimes_{\mathcal{U}} \mathcal{Y}(\phi(c),\phi(d)) \leq_{\mathcal{U}} \mathcal{Y}(\phi(e),\phi(f)) \$$ > > \$$\mathcal{X}(a,b) = I_{\mathcal{V}} \implies \mathcal{Y}(\phi(a),\phi(b)) = I_{\mathcal{U}} \$$ My idea is that \$$\mathcal{X}\$$ would be enriched in one monoid and \$$\mathcal{Y}\$$ would be enriched in another, maybe different monoid. Since \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$ may have different monoids associated with them, I indexed their operations. Now I look closely as this, I think these axioms need to be strengthened. Here is just one axiom that strengthens all three. I believe this works. It is rather long... \$$I_{\mathcal{V}} \otimes_{\mathcal{V}} \mathcal{X}(a,b) \otimes_{\mathcal{V}} \mathcal{X}(c,d) \otimes_{\mathcal{V}} \cdots \otimes_{\mathcal{V}} \mathcal{X}(y,z) \leq_{\mathcal{V}} I_{\mathcal{V}} \otimes_{\mathcal{V}} \mathcal{X}(p,q) \otimes_{\mathcal{V}} \mathcal{X}(r,s) \otimes_{\mathcal{V}} \cdots \otimes_{\mathcal{V}} \mathcal{X}(w,x) \$$ \$$\implies I_{\mathcal{U}} \otimes_{\mathcal{U}} \mathcal{Y}(\phi(a),\phi(b)) \otimes_{\mathcal{U}} \mathcal{Y}(\phi(c),\phi(d)) \otimes_{\mathcal{U}} \cdots \otimes_{\mathcal{U}} \mathcal{Y}(\phi(y),\phi(z))\$$ \$$\ \ \ \ \ \ \ \ \ \ \leq_{\mathcal{U}} I_{\mathcal{U}} \otimes_{\mathcal{U}} \mathcal{Y}(\phi(p),\phi(q)) \otimes_{\mathcal{U}} \mathcal{Y}(\phi(r),\phi(s)) \otimes_{\mathcal{U}} \cdots \otimes_{\mathcal{U}} \mathcal{Y}(\phi(w),\phi(x)) \$$ I can prove this turns into a group homomorphism for the generated groups John mentions in [#81](https://forum.azimuthproject.org/discussion/comment/18687/#Comment_18687). Just tell me if you would like to see me do the proof. Similarly John also mentions how to take a group \$$G\$$ to a category \$$\mathcal{G}\$$ enriched in \$$G\$$ in [#81](https://forum.azimuthproject.org/discussion/comment/18687/#Comment_18687) (I mention another way in [#79](https://forum.azimuthproject.org/discussion/comment/18675/#Comment_18675) and [#76](https://forum.azimuthproject.org/discussion/comment/18647/#Comment_18647)). This operation can be lifted to make enriched-category morphisms that obey the monster axiom. Again, I can attempt to prove this if you want.
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88.
edited May 2018

Thanks a lot, Matthew and Jonathan! That makes a lot of sense to me. Two follow-up questions:

First, Matthew said

Enriched categories substitute the sets in hom-sets with an arbitrary category. Regular categories can be see as Set-enriched categories, where Set is the category of sets. In order to make this happen, you need two thing: composition of arrows and identity morphisms.

Does this mean the reason why we need the two requirements when substituting the hom-set with an arbitrary category is because we need to make such extra effort to guarantee what we end up with after the substitution is still a category (I mean, still an "enriched category")? If so, what would the substitution yield if e.g. the arbitrary category we use is nothing (such that we effectively leave no enrichment base at all, not even Set)?

Second, Jonathan said

This angle also explains $$I\leq\mathcal{X}(x,x)$$: whatever $$I$$ is, $$\mathcal{X}(x, x)$$ better be at least that much.

I managed to understand the adoption of "≤" in $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$ via this angle, but am still not sure I fully grasp its necessity in $$I\leq\mathcal{X}(x,x)$$ - I (think I) see that $$\mathcal{X}(x,x)$$ need not be exactly $$I$$, but why must it be at least that much? What would happen if $$I>\mathcal{X}(x,x)$$?

Comment Source:Thanks a lot, [Matthew](https://forum.azimuthproject.org/discussion/comment/18702/#Comment_18702) and [Jonathan](https://forum.azimuthproject.org/discussion/comment/18703/#Comment_18703)! That makes a lot of sense to me. Two follow-up questions: First, Matthew said > Enriched categories substitute the sets in hom-sets with an arbitrary category. Regular categories can be see as Set-enriched categories, where Set is the category of sets. > In order to make this happen, you need two thing: composition of arrows and identity morphisms. Does this mean the reason why we need the two requirements when substituting the hom-set with an arbitrary category is because we need to make such extra effort to guarantee what we end up with after the substitution is still a category (I mean, still an "enriched category")? If so, what would the substitution yield if e.g. the arbitrary category we use is _nothing_ (such that we effectively leave no enrichment base at all, not even **Set**)? Second, Jonathan said >This angle also explains \$$I\leq\mathcal{X}(x,x)\$$: whatever \$$I\$$ is, \$$\mathcal{X}(x, x)\$$ better be _at least_ that much. I managed to understand the adoption of "≤" in \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \$$ via this angle, but am still not sure I fully grasp its necessity in \$$I\leq\mathcal{X}(x,x)\$$ - I (think I) see that \$$\mathcal{X}(x,x)\$$ need not be exactly \$$I\$$, but why must it be _at least_ that much? What would happen if \$$I>\mathcal{X}(x,x)\$$?
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89.
edited June 2018

Well, I think it comes down $$\le$$ being easier to work with and generalize then $$\lt$$.

So you want reflexivity.

Comment Source:Well, I think it comes down \$$\le\$$ being easier to work with and generalize then \$$\lt\$$. So you want reflexivity.
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90.

Julio said

Maybe these are very basic questions or I have neglected something, but I'm a bit confused by the two requirements in the definition of "enriched category", i.e. $$I\leq\mathcal{X}(x,x)$$ and $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$. First, why does the definition use "≤" instead of "=" or "≥"? Second, why does the relationship between $$\mathcal{X}(x,x)$$ and $$I$$ matter (such that it must be written into the definition)?

Why do you ever have requirements in a definition? What makes the axiom of a group right? Usually a mathematical definition is an abstraction of lots of examples. (Or occassionally it is something with just the right structure to make nice theorems hold, and then interesting examples come along afterwards.)

In this case, one way of thinking about this is that categories are a fundamental and useful structure in mathematics. Lots of things form categories. It was then noticed that one can encode extra structure that is often seen in categories via the notion of enriched category, and in fact this introduced new examples (like metric spaces) which hadn't been thought of in terms of categories before. Specialising to a particular type of enrichment (by using a monoidal preorder) gives the notion under consideration here. That's why we have these axioms. John explains this in #56, and as he says, you'll see more of where these come from as we progress.

There are many interesting things satisfying these axioms. To understand the axioms you should play with them. That means working out examples, or working out non-examples, or trying to prove things using the axioms, or changing the axioms to see what goes wrong. There may well be other interesting things satisfying similar axioms. (Groups and monoids have similar axiom sets, and both are interesting and useful notions.)

You suggest some alternatives to the axioms. Do the previous examples still work? What examples can you find of the new structure? Can you prove any theorems?

Comment Source:Julio [said](https://forum.azimuthproject.org/discussion/comment/18701/#Comment_18701) > Maybe these are very basic questions or I have neglected something, but I'm a bit confused by the two requirements in the definition of "enriched category", i.e. \$$I\leq\mathcal{X}(x,x) \$$ and \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \$$. First, why does the definition use "≤" instead of "=" or "≥"? Second, why does the relationship between \$$\mathcal{X}(x,x) \$$ and \$$I \$$ matter (such that it must be written into the definition)? Why do you ever have requirements in a definition? What makes the axiom of a group right? Usually a mathematical definition is an abstraction of lots of examples. (Or occassionally it is something with just the right structure to make nice theorems hold, and then interesting examples come along afterwards.) In this case, one way of thinking about this is that categories are a fundamental and useful structure in mathematics. Lots of things form categories. It was then noticed that one can encode extra structure that is often seen in categories via the notion of enriched category, and in fact this introduced new examples (like metric spaces) which hadn't been thought of in terms of categories before. Specialising to a particular type of enrichment (by using a monoidal preorder) gives the notion under consideration here. That's why we have these axioms. John explains this in [#56](https://forum.azimuthproject.org/discussion/comment/18474/#Comment_18474), and as he says, you'll see more of where these come from as we progress. There are many interesting things satisfying these axioms. To understand the axioms you should play with them. That means working out examples, or working out non-examples, or trying to prove things using the axioms, or changing the axioms to see what goes wrong. There may well be other interesting things satisfying similar axioms. (Groups and monoids have similar axiom sets, and both are interesting and useful notions.) You suggest some alternatives to the axioms. Do the previous examples still work? What examples can you find of the new structure? Can you prove any theorems?
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91.
edited May 2018

Matthew, you say in #87

I had written in #63 [something involving $$\le_{\mathcal{V}}$$]

No you hadn't :-) you wrote something involving $$\le_{\mathcal{X}}$$ which I didn't understand and which I questioned. Anyway, now that you've established that $$\mathcal{V}$$ is just a monoid, you can get rid of $$\le_{\mathcal{V}}$$ and just write $$=$$ instead everywhere. That will simplify your (initial three) axioms somewhat. Hopefully you'll be able to see a little more clearly what the morphisms are, and be able to say in a few words what structure you are actually seeing there.

Also, you need to be a bit more careful and say precisely what you mean by

maps $$\phi : \mathcal{X} \to \mathcal{Y}$$

Comment Source:Matthew, you say in [#87](https://forum.azimuthproject.org/discussion/comment/18707/#Comment_18707) > I had written in #63 [something involving \$$\le_{\mathcal{V}}\$$] No you hadn't :-) you wrote something involving \$$\le_{\mathcal{X}}\$$ which I didn't understand and which I questioned. Anyway, now that you've established that \$$\mathcal{V}\$$ is just a monoid, you can get rid of \$$\le_{\mathcal{V}}\$$ and just write \$$=\$$ instead everywhere. That will simplify your (initial three) axioms somewhat. Hopefully you'll be able to see a little more clearly what the morphisms are, and be able to say in a few words what structure you are actually seeing there. Also, you need to be a bit more careful and say precisely what you mean by > maps \$$\phi : \mathcal{X} \to \mathcal{Y} \$$
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92.

No you hadn't you wrote something involving ≤X which I didn't understand and which I questioned. Now that you've established that V is just a monoid, you can get rid of ≤V and just write = instead everywhere.

Huh? I'm confused.

Comment Source:> No you hadn't you wrote something involving ≤X which I didn't understand and which I questioned. Now that you've established that V is just a monoid, you can get rid of ≤V and just write = instead everywhere. Huh? I'm confused.
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93.

Christopher, help me by saying which bit is confusing you. The first sentence or the second?

Comment Source:Christopher, help me by saying which bit is confusing you. The first sentence or the second?
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94.
edited May 2018

Hi Julio,

I think the choice of $$\leq$$ is to comply with how things are done in the book, that goes from more concrete to more abstract ideas, to scare people per degrees.

Note that here we have:

We'll start by picking a monoidal preorder $$(\mathcal{V},\leq,\otimes,I)$$, like $$\mathbf{Bool}$$ or $$\mathbf{Cost}$$. Then:

Definition. A $$\mathcal{V}$$-enriched category $$\mathcal{X}$$ consists of two parts, satisfying two properties...

But your question cries for for the more general and standard notion of enriched category, though that requires the definition of category and of monoidal category first, to say in what you are enriching. If your $$\mathcal{V}$$ has to be a monoidal preorder, your morphims come from the order and with $$\otimes$$ that gives a monoidal category to enrich in. But since the category of sets Set is not a monoidal preorder, you couldn't even say that a category is, trivially, a category enriched in Set as indeed is. You need the more general definition.

If one has already peeked at the definition of category (think it's in chapter 3), this slides from Simon Willerton (thanks!) give an animated springboard, starting in slide 9, to view the conditions of "enriched category" as they emerge from those of ordinary category.

Take into account also that, for $$v_1,v_2 \in \mathcal{V}$$, $$v_1 \leq v_2$$ is a morphism, just as $$f:A \to B$$ is in the category of sets.

Comment Source:Hi Julio, About your [#83](https://forum.azimuthproject.org/discussion/comment/18701/#Comment_18701), I think the choice of \$$\leq\$$ is to comply with how things are done in the book, that goes from more concrete to more abstract ideas, to scare people per degrees. Note that here we have: > We'll start by picking a monoidal _preorder_ \$$(\mathcal{V},\leq,\otimes,I)\$$, like \$$\mathbf{Bool}\$$ or \$$\mathbf{Cost}\$$. Then: > **Definition.** A **\$$\mathcal{V}\$$-enriched category** \$$\mathcal{X}\$$ consists of two parts, satisfying two properties... But your question cries for for the more general and standard notion of enriched category, though that requires the definition of category and of monoidal category first, to say in *what* you are enriching. If your \$$\mathcal{V}\$$ has to be a monoidal preorder, your morphims come from the order and with \$$\otimes\$$ that gives a monoidal category to enrich in. But since the category of sets **Set** is not a monoidal preorder, you couldn't even say that a category is, trivially, a category enriched in **Set** as indeed is. You need the more general definition. If one has already peeked at the definition of category (think it's in chapter 3), this [slides](http://www.simonwillerton.staff.shef.ac.uk/ftp/AberdeenMetric.pdf) from Simon Willerton (thanks!) give an animated springboard, starting in slide 9, to view the conditions of "enriched category" as they emerge from those of ordinary category. Take into account also that, for \$$v_1,v_2 \in \mathcal{V}\$$, \$$v_1 \leq v_2\$$ **is** a morphism, just as \$$f:A \to B\$$ is in the category of sets.
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95.
edited May 2018

Hey Simon,

Matthew, you say in #87

I had written in #63 [something involving $$\le_{\mathcal{V}}$$]

No you hadn't :-) you wrote something involving $$\le_{\mathcal{X}}$$ which I didn't understand and which I questioned.

I admittedly changed my notation because you didn't understand $$\le_{\mathcal{X}}$$ and I thought this would be clearer.

I am trying to change my notation in order to clarify myself.

I am sorry for the confusion.

Anyway, now that you've established that $$\mathcal{V}$$ is just a monoid, you can get rid of $$\le_{\mathcal{V}}$$ and just write $$=$$ instead everywhere.

We still need to keep $$\otimes_\mathcal{V}$$ and $$\otimes_\mathcal{U}$$ around, since there are different monoids.

Below, I started to use $$\otimes$$ and $$\odot$$ instead.

Hopefully this clears up some clutter.

That will simplify your (initial three) axioms somewhat.

Unfortunately those axioms do not work.

I believe we have to use the big axiom in #87.

Also, you need to be a bit more careful and say precisely what you mean by

maps $$\phi : \mathcal{X} \to \mathcal{Y}$$

Okay.

I am assuming $$\mathcal{X}$$ is enriched in one monoid $$\mathcal{V}$$ and $$\mathcal{Y}$$ is enriched in $$\mathcal{U}$$.

In general, $$\mathcal{V} \neq \mathcal{U}$$.

I need some way of doing the book-keeping.

Below is my attempt at a careful formulation.

Define a new category $$\mathbb{M}\mathbb{E}$$. Here $$\mathbb{M}\mathbb{E}$$ stands for monoid enriched. Previously I called this DiscretePosEnrich, but that name is confusing.

• Objects

Objects are $$\mathcal{V}$$-enriched categories $$\mathcal{X}$$ for various monoids.

For instance there are $$\mathbb{Z}/2\mathbb{Z}$$-enriched categories in $$\mathbb{M}\mathbb{E}$$.

There are also $$D_8$$-enriched categories in $$\mathbb{M}\mathbb{E}$$.

For any semilattice $$\mathbb{L}$$ then $$\mathbb{L}$$-enriched categories are members of $$\mathbb{M}\mathbb{E}$$.

John mentioned how to take any group $$G$$ to a $$G$$-enriched category $$\mathcal{G}$$ in #81. These are members of $$\mathbb{M}\mathbb{E}$$.

• Morphisms

Let $$\mathcal{V} = \langle V, I_{\mathcal{V}}, \otimes \rangle$$ and $$\mathcal{U} = \langle U, I_{\mathcal{U}}, \odot \rangle$$ be arbitrary monoids.

Let $$\mathcal{X}$$ be $$\mathcal{V}$$-enriched.

Let $$\mathcal{Y}$$ be $$\mathcal{U}$$-enriched.

A morphism in $$\mathbb{M}\mathbb{E}$$ is a function $$\phi_{\mathcal{V},\mathcal{U}} : \mathrm{Obj}(\mathcal{X}) \to \mathrm{Obj}(\mathcal{Y})$$ which obeys the following law:

$\text{if }$

$I_{\mathcal{V}} \otimes \mathcal{X}(a,b) \otimes \mathcal{X}(c,d) \otimes \cdots \otimes \mathcal{X}(y,z) = I_{\mathcal{V}} \otimes \mathcal{X}(p,q) \otimes \mathcal{X}(r,s) \otimes \cdots \otimes \mathcal{X}(w,x)$

$\text{then}$

$I_{\mathcal{U}} \odot \mathcal{Y}(\phi(a),\phi(b)) \odot \mathcal{Y}(\phi(c),\phi(d)) \odot \cdots \odot \mathcal{Y}(\phi(y),\phi(z)) = I_{\mathcal{U}} \odot \mathcal{Y}(\phi(p),\phi(q)) \odot \mathcal{Y}(\phi(r),\phi(s)) \odot \cdots \odot \mathcal{Y}(\phi(w),\phi(x))$

I know your gut says you want to use just one category for enrichment, Simon. I am sorry I am not doing that.

As I mentioned, I have changed my notation around a bit above in order to clarify what I have in mind.

John already showed how to lift any group $$G$$ into $$\mathbb{M}\mathbb{E}$$. If you want, I can show how a group homomorphism gives rise to an $$\mathbb{M E}$$ morphism.

Thank you for your patience as I attempt to clarify myself and work out the kinks of this idea.

Comment Source:Hey [Simon](https://forum.azimuthproject.org/profile/1689/Simon%20Willerton), > Matthew, you say in [#87](https://forum.azimuthproject.org/discussion/comment/18707/#Comment_18707) > > I had written in #63 [something involving \$$\le_{\mathcal{V}}\$$] > > No you hadn't :-) you wrote something involving \$$\le_{\mathcal{X}}\$$ which I didn't understand and which I questioned. I admittedly changed my notation because you didn't understand \$$\le_{\mathcal{X}}\$$ and I thought this would be clearer. I am trying to change my notation in order to clarify myself. I am sorry for the confusion. > Anyway, now that you've established that \$$\mathcal{V}\$$ is just a monoid, you can get rid of \$$\le_{\mathcal{V}}\$$ and just write \$$=\$$ instead everywhere. We still need to keep \$$\otimes_\mathcal{V}\$$ and \$$\otimes_\mathcal{U}\$$ around, since there are different monoids. Below, I started to use \$$\otimes\$$ and \$$\odot\$$ instead. Hopefully this clears up some clutter. > That will simplify your (initial three) axioms somewhat. Unfortunately those axioms do not work. I believe we have to use the big axiom in [#87](https://forum.azimuthproject.org/discussion/comment/18707/#Comment_18707). > Also, you need to be a bit more careful and say precisely what you mean by > > maps \$$\phi : \mathcal{X} \to \mathcal{Y} \$$ Okay. I am assuming \$$\mathcal{X}\$$ is enriched in one monoid \$$\mathcal{V}\$$ and \$$\mathcal{Y}\$$ is enriched in \$$\mathcal{U}\$$. In general, \$$\mathcal{V} \neq \mathcal{U}\$$. I need some way of doing the book-keeping. Below is my attempt at a careful formulation. --------------------------------- Define a new category \$$\mathbb{M}\mathbb{E}\$$. Here \$$\mathbb{M}\mathbb{E}\$$ stands for *monoid enriched*. Previously I called this **DiscretePosEnrich**, but that name is confusing. - **Objects** Objects are \$$\mathcal{V}\$$-enriched categories \$$\mathcal{X}\$$ for various monoids. For instance there are [\$$\mathbb{Z}/2\mathbb{Z}\$$](https://en.wikipedia.org/wiki/Cyclic_group)-enriched categories in \$$\mathbb{M}\mathbb{E}\$$. There are also [\$$D_8\$$](https://en.wikipedia.org/wiki/Dihedral_group)-enriched categories in \$$\mathbb{M}\mathbb{E}\$$. For any [semilattice](https://en.wikipedia.org/wiki/Semilattice) \$$\mathbb{L}\$$ then \$$\mathbb{L}\$$-enriched categories are members of \$$\mathbb{M}\mathbb{E}\$$. John mentioned how to take any group \$$G\$$ to a \$$G\$$-enriched category \$$\mathcal{G}\$$ in [#81](https://forum.azimuthproject.org/discussion/comment/18687/#Comment_18687). These are members of \$$\mathbb{M}\mathbb{E}\$$. - **Morphisms** Let \$$\mathcal{V} = \langle V, I_{\mathcal{V}}, \otimes \rangle\$$ and \$$\mathcal{U} = \langle U, I_{\mathcal{U}}, \odot \rangle\$$ be arbitrary monoids. Let \$$\mathcal{X}\$$ be \$$\mathcal{V}\$$-enriched. Let \$$\mathcal{Y}\$$ be \$$\mathcal{U}\$$-enriched. A morphism in \$$\mathbb{M}\mathbb{E}\$$ is a function \$$\phi_{\mathcal{V},\mathcal{U}} : \mathrm{Obj}(\mathcal{X}) \to \mathrm{Obj}(\mathcal{Y}) \$$ which obeys the following law: \$\text{if } \$ \$I_{\mathcal{V}} \otimes \mathcal{X}(a,b) \otimes \mathcal{X}(c,d) \otimes \cdots \otimes \mathcal{X}(y,z) = I_{\mathcal{V}} \otimes \mathcal{X}(p,q) \otimes \mathcal{X}(r,s) \otimes \cdots \otimes \mathcal{X}(w,x) \$ \$\text{then} \$ \$I_{\mathcal{U}} \odot \mathcal{Y}(\phi(a),\phi(b)) \odot \mathcal{Y}(\phi(c),\phi(d)) \odot \cdots \odot \mathcal{Y}(\phi(y),\phi(z)) = I_{\mathcal{U}} \odot \mathcal{Y}(\phi(p),\phi(q)) \odot \mathcal{Y}(\phi(r),\phi(s)) \odot \cdots \odot \mathcal{Y}(\phi(w),\phi(x)) \$ --------------------------------- I know your gut says you want to use just one category for enrichment, Simon. I am sorry I am not doing that. As I mentioned, I have changed my notation around a bit above in order to clarify what I have in mind. John already showed how to lift any group \$$G\$$ into \$$\mathbb{M}\mathbb{E}\$$. If you want, I can show how a group homomorphism gives rise to an \$$\mathbb{M E}\$$ morphism. Thank you for your patience as I attempt to clarify myself and work out the kinks of this idea.
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96.
edited May 2018

I managed to understand the adoption of "≤" in $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$ via this angle, but am still not sure I fully grasp its necessity in $$I\leq\mathcal{X}(x,x)$$ - I (think I) see that $$\mathcal{X}(x,x)$$ need not be exactly $$I$$, but why must it be at least that much? What would happen if $$I>\mathcal{X}(x,x)$$?

In a preorder, we always want the "empty path", the reflexive relation $$x \le x$$, to be present for all $$x$$. When we start working with categories, this translates to having an identity morphism $$Id_x \in \mathrm{Hom}_\mathcal{X}(x, x)$$ that embodies the idea of "doing nothing". In an enriched category, where we substitute the traditional hom-set for a value in an arbitrary monoidal preorder, we want to translate this same idea of guaranteeing the presence of "do nothing". Preorders have as a fundamental notion the idea of "at least", in the form of its relation $$\le$$, and monoids have as fundamental the notion of a thing that "does nothing": its identity element. So, purely following the analogy, we'd like to guarantee that whatever $$\mathcal{X}(x, x)$$ is, it should "at least do nothing": $$I \le \mathcal{X}(x, x)$$.

Comment Source:[Julio wrote](https://forum.azimuthproject.org/discussion/comment/18708/#Comment_18708): > I managed to understand the adoption of "≤" in \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \$$ via this angle, but am still not sure I fully grasp its necessity in \$$I\leq\mathcal{X}(x,x)\$$ - I (think I) see that \$$\mathcal{X}(x,x)\$$ need not be exactly \$$I\$$, but why must it be _at least_ that much? What would happen if \$$I>\mathcal{X}(x,x)\$$? In a preorder, we always want the "empty path", the reflexive relation \$$x \le x\$$, to be present for all \$$x\$$. When we start working with categories, this translates to having an identity morphism \$$Id\_x \in \mathrm{Hom}\_\mathcal{X}(x, x)\$$ that embodies the idea of "doing nothing". In an _enriched_ category, where we substitute the traditional hom-set for a value in an arbitrary monoidal preorder, we want to translate this same idea of guaranteeing the presence of "do nothing". Preorders have as a fundamental notion the idea of "at least", in the form of its relation \$$\le\$$, and monoids have as fundamental the notion of a thing that "does nothing": its identity element. So, purely following the analogy, we'd like to guarantee that whatever \$$\mathcal{X}(x, x)\$$ is, it should "at least do nothing": \$$I \le \mathcal{X}(x, x)\$$.
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97.
edited May 2018

Thanks for the background explanation, Simon! That really helps me see the bigger picture. And no, I have nothing against the standard axioms - I was just unclear about the methodology/epistemology in mathematics (had naively assumed it relied little on empirical observations). :-)

Great, Jesus - you literally spelled out my thoughts (in a much more explicit way than I could have done)! Thanks a lot for pointing me to the fantastic slides! I always find it easier to grasp notions from slides than from books.

And thanks, Jonathan, for the cool phrase "at least do nothing" - that perspective immediately dissolves my confusion. :-bd

Comment Source:Thanks for the background explanation, [Simon](https://forum.azimuthproject.org/discussion/comment/18711/#Comment_18711)! That really helps me see the bigger picture. And no, I have nothing against the standard axioms - I was just unclear about the methodology/epistemology in mathematics (had naively assumed it relied little on empirical observations). :-) Great, [Jesus](https://forum.azimuthproject.org/discussion/comment/18723/#Comment_18723) - you literally spelled out my thoughts (in a much more explicit way than I could have done)! Thanks a lot for pointing me to the fantastic slides! I always find it easier to grasp notions from slides than from books. And thanks, [Jonathan](https://forum.azimuthproject.org/discussion/comment/18733/#Comment_18733), for the **cool** phrase "at least do nothing" - that perspective immediately dissolves my confusion. :-bd
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98.
edited May 2018

Matthew, don't apologise for not listening to my gut. Follow your gut and learn in the process whether my gut was right or wrong!

About what you've written down, you've got some "multiply by identity element"s, as you're in a monoid you can safely remove those.

Now you've got a subset of $$\mathcal{V}$$ that we can call $$\mathcal{V}_X$$ which is $$\{\mathcal{X}(a,b) \,|\, a,b\in Ob(\mathcal{X})\}$$. Given any element of $$\mathcal{V}$$ which is a product of elements of $$\mathcal{V}_X$$ you associate an element of $$\mathcal{U}$$ which is independent of how it is expressed as a product. There's a more 'high-level' way of saying this, which will make what you've written easier to relate to other ideas about functors between enriched categories. I suspect you know this higher level description and use it when you bring groups in to the picture.

Comment Source:Matthew, don't apologise for not listening to my gut. Follow your gut and learn in the process whether my gut was right or wrong! About what you've written down, you've got some "multiply by identity element"s, as you're in a monoid you can safely remove those. Now you've got a subset of \$$\mathcal{V}\$$ that we can call \$$\mathcal{V}_X\$$ which is \$$\\{\mathcal{X}(a,b) \,|\, a,b\in Ob(\mathcal{X})\\}\$$. Given any element of \$$\mathcal{V}\$$ which is a product of elements of \$$\mathcal{V}_X\$$ you associate an element of \$$\mathcal{U}\$$ which is independent of how it is expressed as a product. There's a more 'high-level' way of saying this, which will make what you've written easier to relate to other ideas about functors between enriched categories. I suspect you know this higher level description and use it when you bring groups in to the picture.
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99.

Now you've got a subset of $$\mathcal{V}$$ that we can call $$\mathcal{V}_X$$ which is $$\{\mathcal{X}(a,b) \,|\, a,b\in Ob(\mathcal{X})\}$$. Given any element of $$\mathcal{V}$$ which is a product of elements of $$\mathcal{V}_X$$ you associate an element of $$\mathcal{U}$$ which is independent of how it is expressed as a product. There's a more 'high-level' way of saying this, which will make what you've written easier to relate to other ideas about functors between enriched categories. I suspect you know this higher level description and use it when you bring groups in to the picture.

Hmm... I am not sure what word you are looking for.

It's charitable to assume I know things. I sort of have the "jack of all trades, master of none" going for me, only it's more like "jack of measure-zero trades, master of $$-\infty$$" where $$-\infty < 0$$ is the thing Chris Upshaw invented in the other thread.

Comment Source:> Now you've got a subset of \$$\mathcal{V}\$$ that we can call \$$\mathcal{V}_X\$$ which is \$$\\{\mathcal{X}(a,b) \,|\, a,b\in Ob(\mathcal{X})\\}\$$. Given any element of \$$\mathcal{V}\$$ which is a product of elements of \$$\mathcal{V}_X\$$ you associate an element of \$$\mathcal{U}\$$ which is independent of how it is expressed as a product. There's a more 'high-level' way of saying this, which will make what you've written easier to relate to other ideas about functors between enriched categories. I suspect you know this higher level description and use it when you bring groups in to the picture. Hmm... I am not sure what word you are looking for. It's charitable to assume I know things. I sort of have the "jack of all trades, master of none" going for me, only it's more like "jack of measure-zero trades, master of \$$-\infty\$$" where \$$-\infty < 0\$$ is the thing Chris Upshaw invented in [the other thread](https://forum.azimuthproject.org/discussion/comment/18568/#Comment_18568).
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100.
edited June 2018

The word I was looking for was 'function'. Sorry.

You have defined a function on a certain subset of $$\mathcal{V}$$, specifically the subset $$\langle\mathcal{V}_X \rangle\subset \mathcal{V}$$ defined to be the submonoid generated by the elements of $$\mathcal{V}_X :=\{\mathcal{X}(a,b) \,|\, a,b\in Ob(\mathcal{X})\}$$.

Now, $$\langle \mathcal{V}_X \rangle$$ is a monoid (by construction) and is the smallest monoid that $$\mathcal{X}$$ can be considered to be an enriched category over. In fact, as you've probably pointed out somewhere, $$\langle \mathcal{V}_X \rangle$$ is a group.

So what you've constructed (I think!) as a morphism $$\mathcal{X}\to\mathcal{Y}$$ in ME is a function $$\phi\colon\text{Ob}(\mathcal{X}) \to \text{Ob}(\mathcal{Y})$$ and a monoid/group homomorphism $$\hat\phi\colon \langle \mathcal{V}_X \rangle\to \langle \mathcal{U}_Y \rangle$$, such that $$\hat\phi(\mathcal{X}(a,b)) = \mathcal{Y}(\phi(a),\phi(b))$$ .

Comment Source:The word I was looking for was 'function'. Sorry. You have defined a function on a certain subset of \$$\mathcal{V}\$$, specifically the subset \$$\langle\mathcal{V}_X \rangle\subset \mathcal{V}\$$ defined to be the submonoid generated by the elements of \$$\mathcal{V}_X :=\\{\mathcal{X}(a,b) \,|\, a,b\in Ob(\mathcal{X})\\}\$$. Now, \$$\langle \mathcal{V}_X \rangle\$$ is a monoid (by construction) and is the smallest monoid that \$$\mathcal{X}\$$ can be considered to be an enriched category over. In fact, as you've probably pointed out somewhere, \$$\langle \mathcal{V}_X \rangle\$$ is a group. So what you've constructed (I think!) as a morphism \$$\mathcal{X}\to\mathcal{Y}\$$ in ME is a function \$$\phi\colon\text{Ob}(\\mathcal{X}) \to \text{Ob}(\\mathcal{Y})\$$ and a monoid/group homomorphism \$$\hat\phi\colon \langle \mathcal{V}_X \rangle\to \langle \mathcal{U}_Y \rangle\$$, such that \$$\hat\phi(\mathcal{X}(a,b)) = \mathcal{Y}(\phi(a),\phi(b))\$$ .