It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.2K
- Applied Category Theory Course 348
- Applied Category Theory Seminar 2
- Exercises 149
- Discussion Groups 48
- How to Use MathJax 15
- Chat 475
- Azimuth Code Project 108
- News and Information 145
- Azimuth Blog 148
- Azimuth Forum 29
- Azimuth Project 190
- - Strategy 109
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 708
- - Latest Changes 700
- - - Action 14
- - - Biodiversity 8
- - - Books 2
- - - Carbon 9
- - - Computational methods 38
- - - Climate 53
- - - Earth science 23
- - - Ecology 43
- - - Energy 29
- - - Experiments 30
- - - Geoengineering 0
- - - Mathematical methods 69
- - - Meta 9
- - - Methodology 16
- - - Natural resources 7
- - - Oceans 4
- - - Organizations 34
- - - People 6
- - - Publishing 4
- - - Reports 3
- - - Software 20
- - - Statistical methods 2
- - - Sustainability 4
- - - Things to do 2
- - - Visualisation 1
- General 39

Options

One of the main lessons of category theory is that whenever you think about some kind of mathematical gadget, you should also think about maps *between* gadgets of this kind. For example, when you think about sets you should also think about functions. When you think about vector spaces you should also think about linear maps. And so on.

We've been talking about various kinds of monoidal preorders. So, let's think about maps *between* monoidal preorders.

As I explained in Lecture 22, a monoidal preorder is a crossbreed or hybrid of a *preorder* and a *monoid*. So let's think about maps between preorders, and maps between monoids, and try to hybridize those.

We've already seen maps between preorders: they're called monotone functions:

**Definition.** A **monotone function** from a preorder \((X,\le_X)\) to \((Y,\le_Y)\) is a function \(f : X \to Y\) such that

$$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ for all elements \(x,x' \in X\).

So, these functions preserve what a preorder has, namely the relation \(\le\). A monoid, on the other hand, has an associative operation \(\otimes\) and a unit element \(I\). So, a map between monoids should preserve th0se! That's how this game works.

Just to scare people, mathematicians call these maps "homomorphisms":

**Definition.** A **homomorphism** from a monoid \( (X,\otimes_X,I_X) \) to a monoid \( (Y,\otimes_Y,I_Y) \) is a function \(f : X \to Y\) such that:

$$ f(x \otimes_X x') = f(x) \otimes_Y f(x') $$ for all elements \(x,x' \in X\), and

$$ f(I_X) = I_Y .$$ You've probably seen a lot of homomorphisms between monoids. Some of them you barely noticed. For example, the set of integers \(\mathbb{Z}\) is a monoid with addition as \(\otimes\) and the number \(0\) as \(I\). So is the set \(\mathbb{R}\) of real numbers! There's a function that turns each integer into a real number:

$$ i: \mathbb{Z} \to \mathbb{R} . $$
It's such a bland function you may never have thought about it: it sends each integer to *itself*, but *regarded as a real number*. And this function is a homomorphism!

What does that mean? Look at the definition. It means you can either add two natural numbers and then regard the result as a real number... or first regard each of them as a real number and then add them... and you get the same answer either way. It also says that integer \(0\), regarded as a real number, is the real number we call \(0\).

Boring facts! But utterly crucial facts. Computer scientists need to worry about these things, because for them integers and real numbers (or floating-point numbers) are different data types, and \(i\) is doing "type conversion".

You've also seen a lot of other, more interesting homomorphisms between monoids.

For example, the whole point of the logarithm function is that it's a homomorphism. It carries multiplication to addition:

$$ \log(x \cdot x') = \log(x) + \log(x') $$ and it carries the identity for multiplication to the identity for addition:

$$ \log(1) = 0. $$ People invented tables of logarithms, and later slide rules, precisely for this reason! They wanted to convert multiplication problems into easier addition problems.

You may also have seen linear maps between vector spaces. A vector space gives a monoid with addition as \(\otimes\) and the zero vector as \(I\); any linear map between vector spaces then gives a homomorphism.

**Puzzle 80.** Tell me a few more homomorphisms between monoids that you routinely use, or at least know.

I hope I've convinced you: monotone functions between preorders are important, and so are homomorphisms between monoids. Thus, if we hybridize these concepts, we'll get a concept that's likely to be important.

It turns out there are a few different ways! The most obvious way is simply to combine all the conditions. There are other ways, so this way is called "strict":

**Definition.** A **strict monoidal monotone** from a monoidal preorder \( (X,\le_X,\otimes_X,I_X) \) to a monoidal preorder \( (Y,\le_Y,\otimes_Y,I_Y) \) is a function \(f : X \to Y\) such that:

$$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and

$$ f(x) \otimes_Y f(x') = f(x \otimes_X x') $$ for all elements \(x,x' \in X\), and also

$$ I_Y = f(I_X) . $$ For example, the homomorphism

$$ i : \mathbb{Z} \to \mathbb{R} ,$$ is a strict monoidal monotone: if one integer is \(\le\) another, then that's still true when we regard them as real numbers. So is the logarithm function.

What other definition could we possibly use, and why would we care? It turns out sometimes we want to replace some of the equations in the above definition by inequalities!

**Definition.** A **lax monoidal monotone** from a monoidal preorder \((X,\le_X,\otimes_X,I_X)\) to a monoidal preorder \((Y,\le_Y,\otimes_Y,I_Y)\) is a function \(f : X \to Y\) such that:

$$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and

$$ f(x) \otimes_Y f(x') \le_Y f(x \otimes_X x') $$ for all elements \(x,x' \in X\), and also

$$ I_Y \le_Y f(I_X). $$
Fong and Spivak call this simply a **monoidal monotone**, since it's their favorite kind. But I will warn you that others call it "lax".

We could also turn around those last two inequalities:

**Definition.** An **oplax monoidal monotone** from a monoidal preorder \((X,\le_X,\otimes_X,I_X)\) to a monoidal preorder \((Y,\le_Y,\otimes_Y,I_Y)\) is a function \(f : X \to Y\) such that:

$$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and

$$ f(x) \otimes_Y f(x') \ge_Y f(x \otimes_X x') $$ for all elements \(x,x' \in X\), and also

$$ I_Y \ge_Y f(I_X). $$ You are probably drowning in definitions now, so let me give some examples to show that they're justified. The monotone function

$$ i : \mathbb{Z} \to \mathbb{R} $$ has a right adjoint

$$ \lfloor \cdot \rfloor : \mathbb{R} \to \mathbb{Z} $$
which provides the approximation *from below* to the nonexistent inverse of \(i\): that is, \( \lfloor x \rfloor \) is the greatest integer that's \(\le x\). It also has a left adjoint

$$ \lceil \cdot \rceil : \mathbb{R} \to \mathbb{Z} $$
which is the best approximation *from above* to the nonexistent inverse of \(i\): that is, \( \lceil x \rceil \) is the least integer that's \(\ge x\).

**Puzzle 81.** Show that one of the functions \( \lfloor \cdot \rfloor : \mathbb{R} \to \mathbb{Z} \), \( \lceil \cdot \rceil : \mathbb{R} \to \mathbb{Z} \) is a lax monoidal monotone and the other is an oplax monoidal monotone, where we make the integers and reals into monoids using addition.

So, you should be sensing some relation between left and right adjoints, and lax and oplax monoidal monotones. We'll talk about this more! And we'll see why all this stuff is important for resource theories.

Finally, for the bravest among you:

**Puzzle 82.** Find a function between monoidal preorders that is both lax and oplax monoidal monotone but not strict monoidal monotone.

In case you haven't had enough jargon for today: a function between monoidal preorders that's both lax and oplax monoidal monotone is called **strong** monoidal monotone.

## Comments

John, for each statement like this, I think you have one too many primes:

$$ x \le_X x' \textrm{ implies } f(x') \le_Y f(x') $$

`John, for each statement like this, I think you have one too many primes: \[ x \le_X x' \textrm{ implies } f(x') \le_Y f(x') \]`

I want to test some \( \LaTeX \) out.

Definition.Amonotone functionfrom a preorder \((X,\le_X)\) to \((Y,\le_Y)\) is a function \(f : X \to Y\) such that the following diagram commutes,$$ \large\begin{matrix} x & \overset{\le_X}{\longrightarrow} & x' \\ f\downarrow & & \downarrow f \\ f(x) & \overset{\le_Y}{\longrightarrow} & f(x') \end{matrix} $$ for all elements \(x,x' \in X\).

`I want to test some \\( \LaTeX \\) out. **Definition.** A **monotone function** from a preorder \\((X,\le_X)\\) to \\((Y,\le_Y)\\) is a function \\(f : X \to Y\\) such that the following diagram commutes, $$ \large\begin{matrix} x & \overset{\le_X}{\longrightarrow} & x' \\\\ f\downarrow & & \downarrow f \\\\ f(x) & \overset{\le_Y}{\longrightarrow} & f(x') \end{matrix} $$ for all elements \\(x,x' \in X\\).`

Jonathan - thanks, fixed!

Keith - you asked a while ago if we could draw fancy diagrams here, and the answer is basically no. But yes, one can draw commutative squares! (Anyone who wants to learn how, click on the little gear to the upper right of his comment.)

By the way, the square you wrote down is not really a commutative square in any category, because \(f\) is not a morphism from \(x\) to \(f(x)\); it's a morphism from \(X\) to \(Y\). One should not write \(x \stackrel{f}{\to} f(x)\); one writes \(f: x \mapsto f(x) \) to indicate that the function \(f\) sends \(x\) to \(f(x)\).

The "mapsto" arrow, \(\mapsto\), has a very different meaning than the "to" arrow, \(\to\). For more see:

`Jonathan - thanks, fixed! Keith - you asked a while ago if we could draw fancy diagrams here, and the answer is basically no. But yes, one can draw commutative squares! (Anyone who wants to learn how, click on the little gear to the upper right of his comment.) By the way, the square you wrote down is not really a commutative square in any category, because \\(f\\) is not a morphism from \\(x\\) to \\(f(x)\\); it's a morphism from \\(X\\) to \\(Y\\). One should not write \\(x \stackrel{f}{\to} f(x)\\); one writes \\(f: x \mapsto f(x) \\) to indicate that the function \\(f\\) sends \\(x\\) to \\(f(x)\\). The "mapsto" arrow, \\(\mapsto\\), has a very different meaning than the "to" arrow, \\(\to\\). For more see: * Math Stackexchange, [Difference of mapsto and arrow.](https://math.stackexchange.com/questions/473247/difference-of-mapsto-and-right-arrow)`

Puzzle 80Let \(X\) be the monoidal preorder described by Anindya in Comment 9 of Lecture 22 . The elements of \(X\) are finite length words, the relation \(\leq_X\) is defined by word length, and \(\otimes_X\) is word concatenation. The map that takes words to their length (e.g. CAT\(\mapsto 3\)) is a homomorphism between \(X\) and \(\mathbb N\). It is a homomorphism because the lengths of two words added together is the same as concatenating the words and then taking the length of the result!`**Puzzle 80** Let \\(X\\) be the monoidal preorder described by Anindya in <a href = "https://forum.azimuthproject.org/discussion/comment/17974/#Comment_17974"> Comment 9 of Lecture 22 </a>. The elements of \\(X\\) are finite length words, the relation \\(\leq_X\\) is defined by word length, and \\(\otimes_X\\) is word concatenation. The map that takes words to their length (e.g. CAT\\(\mapsto 3\\)) is a homomorphism between \\(X\\) and \\(\mathbb N\\). It is a homomorphism because the lengths of two words added together is the same as concatenating the words and then taking the length of the result!`

To remember the condition for lax vs. oplax monoid monotone functions, I'm testing out this mnemonic:

Lax - doing things separately is cheaper than doing them together Oplax - doing things together is cheaper than doing them separately

Lax functions are like re

laxed people (i.e. people with time during the week) who find it easier to clean 10 minutes every day instead of cleaning for 1 hour once a week.`To remember the condition for lax vs. oplax monoid monotone functions, I'm testing out this mnemonic: Lax - doing things separately is cheaper than doing them together Oplax - doing things together is cheaper than doing them separately Lax functions are like re<b>lax</b>ed people (i.e. people with time during the week) who find it easier to clean 10 minutes every day instead of cleaning for 1 hour once a week.`

Puzzle 82Again let \(X\) be the monoidal preorder whose elements are words of finite length. \(s \leq_X s'\) iff the length of word \(s\) is smaller than the length of word \(s'\). \(\otimes_X\) is string concatenation and \(1_X\) is the empty string.Define a map \(f: X \to X\) by word reversal. For example, \[f(\textrm{CAT}) = \textrm{TAC}.\] First note that reversing a word doesn't change its length so \[s \leq_X s' \implies f(s) \leq_X f(s').\] Also reversing the empty string is still the empty string, so \(f(1_X) = 1_X\). The reflexive property then gives us \[ f(1_X) \leq 1_X \textrm{ and } 1_X \leq f(1_X) .\] Lastly, if I reverse two words and concatenate them, the resulting string has the same length as the string I get when I first concatenate them and then reverse them. So \[f(s) \otimes f(s') \leq f(s \otimes s') \textrm{ and } f(s \otimes s') \leq f(s) \otimes f(s').\]

Therefore \(f\) is both lax and oplax! However, \[f(\textrm{CAT}) \otimes f(\textrm{DOG}) = \textrm{TAC} \otimes \textrm{GOD} = \textrm{TACGOD}\] while \[f(\textrm{CAT} \otimes \textrm{DOG}) = f(\textrm{CATDOG}) = \textrm{GODTAC}.\] Since \[f(\textrm{CAT}) \otimes f(\textrm{DOG}) \neq f(\textrm{CAT} \otimes \textrm{DOG})\] \(f\) is not strict monoidal monotone.

`**Puzzle 82** Again let \\(X\\) be the monoidal preorder whose elements are words of finite length. \\(s \leq_X s'\\) iff the length of word \\(s\\) is smaller than the length of word \\(s'\\). \\(\otimes_X\\) is string concatenation and \\(1_X\\) is the empty string. Define a map \\(f: X \to X\\) by word reversal. For example, \\[f(\textrm{CAT}) = \textrm{TAC}.\\] First note that reversing a word doesn't change its length so \\[s \leq_X s' \implies f(s) \leq_X f(s').\\] Also reversing the empty string is still the empty string, so \\(f(1_X) = 1_X\\). The reflexive property then gives us \\[ f(1_X) \leq 1_X \textrm{ and } 1_X \leq f(1_X) .\\] Lastly, if I reverse two words and concatenate them, the resulting string has the same length as the string I get when I first concatenate them and then reverse them. So \\[f(s) \otimes f(s') \leq f(s \otimes s') \textrm{ and } f(s \otimes s') \leq f(s) \otimes f(s').\\] Therefore \\(f\\) is both lax and oplax! However, \\[f(\textrm{CAT}) \otimes f(\textrm{DOG}) = \textrm{TAC} \otimes \textrm{GOD} = \textrm{TACGOD}\\] while \\[f(\textrm{CAT} \otimes \textrm{DOG}) = f(\textrm{CATDOG}) = \textrm{GODTAC}.\\] Since \\[f(\textrm{CAT}) \otimes f(\textrm{DOG}) \neq f(\textrm{CAT} \otimes \textrm{DOG})\\] \\(f\\) is not strict monoidal monotone.`

Oh,

reversal! I was trying so many different things, and I was even trying endomorphisms on that exact monoidal preorder, but I never came up with reversal! That's a fantastic solution, Sophie.(I think that's supposed to be Puzzle 82, though, not Puzzle 81!)

`Oh, _reversal_! I was trying so many different things, and I was even trying endomorphisms on that exact monoidal preorder, but I never came up with reversal! That's a fantastic solution, Sophie. (I think that's supposed to be Puzzle 82, though, not Puzzle 81!)`

Thanks Jonathan! I edited the puzzle number :)

`Thanks Jonathan! I edited the puzzle number :)`

that's v neat @Sophie – one slight typo: the line \(f(s) \otimes f(s') = f(s \otimes s')\) should be \(f(s) \otimes f(s') \cong f(s \otimes s')\)

`that's v neat @Sophie – one slight typo: the line \\(f(s) \otimes f(s') = f(s \otimes s')\\) should be \\(f(s) \otimes f(s') \cong f(s \otimes s')\\)`

I was wondering if we needed both rules – "\(f\) preserves \(\otimes\)" and "\(f\) preserves \(I\)" – in the definition of a monoid homomorphism.

Or is "\(f\) preserves \(\otimes\)" enough on its own?

If we can pick an \(x\) such that \(f(x) = I_Y\) and then we have \[f(I_X) = f(I_X) \otimes I_Y = f(I_X) \otimes f(x) = f(I_X \otimes x) = f(x) = I_Y\]

But what if \(I_Y\) is

notin the image of \(f\)?I suspect we might be able to get a counterexample in this case – but I'm struggling to come up with one.

`I was wondering if we needed both rules – "\\(f\\) preserves \\(\otimes\\)" and "\\(f\\) preserves \\(I\\)" – in the definition of a monoid homomorphism. Or is "\\(f\\) preserves \\(\otimes\\)" enough on its own? If we can pick an \\(x\\) such that \\(f(x) = I_Y\\) and then we have \\[f(I_X) = f(I_X) \otimes I_Y = f(I_X) \otimes f(x) = f(I_X \otimes x) = f(x) = I_Y\\] But what if \\(I_Y\\) is *not* in the image of \\(f\\)? I suspect we might be able to get a counterexample in this case – but I'm struggling to come up with one.`

@Anindya: If we have a monoid homomorphism \(f : X \to Y\), then we have \(f(1_X) \otimes f(x) = f(1_X \otimes x) = f(x) = f(x \otimes_X 1_X) = f(x) \otimes_Y f(1_X)\). So if nothing else, \(f(1_X)\) is a unit for the image \(f(X) \subseteq Y\). And as you say, if \(1_Y\) is mapped to, then \(1_X\) must be one of the things that map to it. (Nice proof!)

Unfortunately, it's entirely possible for something that isn't the unit of the larger monoid to nonetheless behave like a unit for a subset of the monoid. Consider the monoid \(\langle \mathcal{P}(S), \cup, \emptyset \rangle \) of subsets of a set \(S\), with union as product and the empty set as unit. Let \(x \in S\), and consider the upset \(\operatorname{\uparrow}\{x\}\) consisting of the collection of all sets containing \(x\). Then \(\{x\}\) acts as a unit for this collection, because every subset in this collection already contains \(x\).

One reason to require that a homomorphism preserve the unit is because this lets us classify

submonoids(and subalgebras in general). A submonoid is a subset of a monoid where the unit and multiplication are the same as the containing monoid. With monoid homomorphisms as defined, every submonoid is just the image of a homomorphism into the containing monoid. (This can be as simple as an injection, like \(i : \mathbb{N} \to \mathbb{Z}\) defined by \(i(x) = x\).)`@[Anindya](https://forum.azimuthproject.org/discussion/comment/18231/#Comment_18231): If we have a monoid homomorphism \\(f : X \to Y\\), then we have \\(f(1_X) \otimes f(x) = f(1_X \otimes x) = f(x) = f(x \otimes_X 1_X) = f(x) \otimes_Y f(1_X)\\). So if nothing else, \\(f(1_X)\\) is a unit for the image \\(f(X) \subseteq Y\\). And as you say, if \\(1_Y\\) is mapped to, then \\(1_X\\) must be one of the things that map to it. (Nice proof!) Unfortunately, it's entirely possible for something that isn't the unit of the larger monoid to nonetheless behave like a unit for a subset of the monoid. Consider the monoid \\(\langle \mathcal{P}(S), \cup, \emptyset \rangle \\) of subsets of a set \\(S\\), with union as product and the empty set as unit. Let \\(x \in S\\), and consider the upset \\(\operatorname{\uparrow}\\{x\\}\\) consisting of the collection of all sets containing \\(x\\). Then \\(\\{x\\}\\) acts as a unit for this collection, because every subset in this collection already contains \\(x\\). One reason to require that a homomorphism preserve the unit is because this lets us classify _submonoids_ (and [subalgebras](http://planetmath.org/homomorphismbetweenalgebraicsystems) in general). A submonoid is a subset of a monoid where the unit and multiplication are the same as the containing monoid. With monoid homomorphisms as defined, every submonoid is just the image of a homomorphism into the containing monoid. (This can be as simple as an injection, like \\(i : \mathbb{N} \to \mathbb{Z}\\) defined by \\(i(x) = x\\).)`

@Jonathan – Ah okay! and you can simplify this to give the counterexample I'm looking for:

Take the trivial monoid \(\mathbf{1}\) and the 2-element join semilattice \((\mathbf{2}, \vee, \bot)\)

Let \(f : \mathbf{1} \rightarrow \mathbf{2}\) be the map that picks out \(\top\)

Then \(f(\bullet) \otimes f(\bullet) = f(\bullet) \vee f(\bullet) = \top \vee \top = \top = f(\bullet \otimes \bullet)\)

So \(f\) preserves \(\otimes\). But \(f(\bullet)\neq\bot\), so \(f\) does not preserve identity.

`@Jonathan – Ah okay! and you can simplify this to give the counterexample I'm looking for: Take the trivial monoid \\(\mathbf{1}\\) and the 2-element join semilattice \\((\mathbf{2}, \vee, \bot)\\) Let \\(f : \mathbf{1} \rightarrow \mathbf{2}\\) be the map that picks out \\(\top\\) Then \\(f(\bullet) \otimes f(\bullet) = f(\bullet) \vee f(\bullet) = \top \vee \top = \top = f(\bullet \otimes \bullet)\\) So \\(f\\) preserves \\(\otimes\\). But \\(f(\bullet)\neq\bot\\), so \\(f\\) does not preserve identity.`

Thanks @Anindya, I fixed the error!

`Thanks @Anindya, I fixed the error!`

I am having a hard time understanding strong monoidal monotone functions. This seems to be a paradox for me (like this water is both hot and cold). Does anyone have an example of one that can clarify how this can happen?

`I am having a hard time understanding strong monoidal monotone functions. This seems to be a paradox for me (like this water is both hot and cold). Does anyone have an example of one that can clarify how this can happen?`

Michael, can you clarify where you find Sophie's construction of a strong monoidal monotone unclear? (Also, every

strictmonoidal monotone is alsostrong; but I suspect it's the strong-but-not-strict ones that are confusing.)`Michael, can you clarify where you find [Sophie's construction](https://forum.azimuthproject.org/discussion/comment/18225/#Comment_18225) of a strong monoidal monotone unclear? (Also, every _strict_ monoidal monotone is also _strong_; but I suspect it's the strong-but-not-strict ones that are confusing.)`

Jonathan

I missed Sophie's answer to Puzzle 82. Thats pretty clear now. These creatures are some sneaky ones eh? LOL

I was also confused about this but now is clear. Thanks a lot!

`Jonathan I missed Sophie's answer to Puzzle 82. Thats pretty clear now. These creatures are some sneaky ones eh? LOL >Also, every strict monoidal monotone is also strong; but I suspect it's the strong-but-not-strict ones that are confusing. I was also confused about this but now is clear. Thanks a lot!`

John I was having a hard time proving puzzle 83 and I think it's because the unit conditions are the other way around for the lax and oplax monotones, respectively:

For the lax monotone we should require: \( I_Y \le f(I_X) . \)

For the oplax monotone we should require: \( f(I_X) \le_Y I_Y . \)

And a nitpick: one of these equations was using \(1\) rather than \(I\) to denote the unit; I've noticed this notation is also mixed at the end of the next lecture.

`[John](https://forum.azimuthproject.org/profile/17/John%20Baez) I was having a hard time proving [puzzle 83](https://forum.azimuthproject.org/discussion/2098/lecture-27-chapter-2-adjoints-of-monoidal-monotones#latest) and I think it's because the unit conditions are the other way around for the lax and oplax monotones, respectively: - For the lax monotone we should require: \\( I_Y \le f(I_X) . \\) - For the oplax monotone we should require: \\( f(I_X) \le_Y I_Y . \\) And a nitpick: one of these equations was using \\(1\\) rather than \\(I\\) to denote the unit; I've noticed this notation is also mixed at the end of the [next lecture](https://forum.azimuthproject.org/discussion/2098/lecture-27-chapter-2-adjoints-of-monoidal-monotones#latest).`

Dan, I reposted your comment in Lecture 27 so others working on the puzzle will see it there as well. Hope that's okay with you!

`Dan, I reposted your comment in Lecture 27 so others working on the puzzle will see it there as well. Hope that's okay with you!`

Sure Sophie! And thank you for the interesting points you raised there!

`> Dan, I reposted your comment in Lecture 27 so others working on the puzzle will see it there as well. Hope that's okay with you! Sure [Sophie](https://forum.azimuthproject.org/profile/2225/Sophie%20Libkind)! And thank you for the interesting points you raised there!`

Dan wrote:

You're right on all of these! I'm fixing these mistakes now. It's important to get those inequalities pointing the right way. Thanks!

`Dan wrote: > [John](https://forum.azimuthproject.org/profile/17/John%20Baez) I was having a hard time proving [puzzle 83](https://forum.azimuthproject.org/discussion/2098/lecture-27-chapter-2-adjoints-of-monoidal-monotones#latest) and I think it's because the unit conditions are the other way around for the lax and oplax monotones, respectively: > - For the lax monotone we should require: \\( I_Y \le f(I_X) . \\) > - For the oplax monotone we should require: \\( f(I_X) \le_Y I_Y . \\) > And a nitpick: one of these equations was using \\(1\\) rather than \\(I\\) to denote the unit; I've noticed this notation is also mixed at the end of the [next lecture](https://forum.azimuthproject.org/discussion/2098/lecture-27-chapter-2-adjoints-of-monoidal-monotones#latest). You're right on all of these! I'm fixing these mistakes now. It's important to get those inequalities pointing the right way. Thanks!`

Anindya wrote:

That's a good question; for group homomorphisms we don't need the second condition. But for monoid homomorphisms we do. Jonathan and you have given some nice examples from logic; here's a typical example from analysis:

Let \(X\) be the set of functions \(f : [0,1] \to \mathbb{R}\). Make this into a monoid with pointwise multiplication of functions as its multiplication and the constant function \(1\) as the unit. Let

$$ F : X \to X $$ be the map that multiplies any function \(f\) by the characteristic function of the interval \( [0,1/2] \). Then

$$ F(fg) = F(f) F(g) $$ for all \(f,g \in X\) but

$$ F(1) \ne 1 .$$ We can generalize this as follows. Suppose \(X\) is any monoid and \(p \in M\) is a

central idempotent: an element that commutes with everything in \(X\) and has \(p^2 = p \). (In the previous example, \(p\) is the characteristic function of the interval \( [0,1/2] \).) Let$$ F : X \to X $$ be the map that multiplies any element of \(X\) by \(p\). Then

$$ F(fg) = p f g = p^2 f g = p f p g = F(f) F(g) $$ but

$$ F(1) = p \ne 1 $$ unless \(p\) = 1.

Puzzle.Suppose \(X\) is any monoid and \(F : X \to X\) is any map with \(F(fg) = F(f) F(g) \) for all \(f,g \in X\). Is \(F(1)\) a central idempotent?`Anindya wrote: > I was wondering if we needed both rules – "f preserves ⊗" and "f preserves I" – in the definition of a monoid homomorphism. That's a good question; for group homomorphisms we don't need the second condition. But for monoid homomorphisms we do. Jonathan and you have given some nice examples from logic; here's a typical example from analysis: Let \\(X\\) be the set of functions \\(f : [0,1] \to \mathbb{R}\\). Make this into a monoid with pointwise multiplication of functions as its multiplication and the constant function \\(1\\) as the unit. Let \[ F : X \to X \] be the map that multiplies any function \\(f\\) by the characteristic function of the interval \\( [0,1/2] \\). Then \[ F(fg) = F(f) F(g) \] for all \\(f,g \in X\\) but \[ F(1) \ne 1 .\] We can generalize this as follows. Suppose \\(X\\) is any monoid and \\(p \in M\\) is a **central idempotent**: an element that commutes with everything in \\(X\\) and has \\(p^2 = p \\). (In the previous example, \\(p\\) is the characteristic function of the interval \\( [0,1/2] \\).) Let \[ F : X \to X \] be the map that multiplies any element of \\(X\\) by \\(p\\). Then \[ F(fg) = p f g = p^2 f g = p f p g = F(f) F(g) \] but \[ F(1) = p \ne 1 \] unless \\(p\\) = 1. **Puzzle.** Suppose \\(X\\) is any monoid and \\(F : X \to X\\) is any map with \\(F(fg) = F(f) F(g) \\) for all \\(f,g \in X\\). Is \\(F(1)\\) a central idempotent?`

Re this puzzle:

It's taken me a bit of fiddling to come up with a counterexample to that one, but I think I've got one...

First let's note that \(F(1)F(1) = F(1.1) = F(1)\), so \(F(1)\) is certainly idempotent. And \(F(1)F(g) = F(1.g) = F(g.1) = F(g)F(1)\), so \(F(1)\) commutes with anything in the image of \(F\). But \(F(1)\) need not be central (ie commute with anything in \(X\)), as the following example shows.

Let \(2 = \{0, 1\}\) be a two-element set, and let \(X\) be the set of maps \(2\rightarrow 2\).

So \(X\) has four elements:

— the identity map

— the map switching round \(0\) and \(1\)

— the constant map \(c_0\) sending everything to \(0\)

— the constant map \(c_1\) sending everything to \(1\)

\(X\) is a monoid under composition, but it is not commutative: \(c_0\circ c_1 \neq c_1 \circ c_0\)

Now let \(F : X \rightarrow X\) be the constant map sending everything in \(X\) to \(c_0\).

Then we certainly have \(F(x\circ y) = F(x)\circ F(y)\), since both sides are always \(c_0\).

But as noted above, \(c_0\) is not central because it does not commute with \(c_1\).

`Re this puzzle: > Suppose \\(X\\) is any monoid and \\(F : X \to X\\) is any map with \\(F(fg) = F(f) F(g) \\) for all \\(f,g \in X\\). Is \\(F(1)\\) a central idempotent? It's taken me a bit of fiddling to come up with a counterexample to that one, but I think I've got one... First let's note that \\(F(1)F(1) = F(1.1) = F(1)\\), so \\(F(1)\\) is certainly idempotent. And \\(F(1)F(g) = F(1.g) = F(g.1) = F(g)F(1)\\), so \\(F(1)\\) commutes with anything in the image of \\(F\\). But \\(F(1)\\) need not be central (ie commute with anything in \\(X\\)), as the following example shows. Let \\(2 = \\{0, 1\\}\\) be a two-element set, and let \\(X\\) be the set of maps \\(2\rightarrow 2\\). So \\(X\\) has four elements: — the identity map — the map switching round \\(0\\) and \\(1\\) — the constant map \\(c_0\\) sending everything to \\(0\\) — the constant map \\(c_1\\) sending everything to \\(1\\) \\(X\\) is a monoid under composition, but it is not commutative: \\(c_0\circ c_1 \neq c_1 \circ c_0\\) Now let \\(F : X \rightarrow X\\) be the constant map sending everything in \\(X\\) to \\(c_0\\). Then we certainly have \\(F(x\circ y) = F(x)\circ F(y)\\), since both sides are always \\(c_0\\). But as noted above, \\(c_0\\) is not central because it does not commute with \\(c_1\\).`

(actually just noticed we can delete the "switch" map from X above to get an even smaller counterexample – a monoid with just three elements. since any two-element monoid is commutative, this is the smallest counterexample. more generally, consider the "words on an alphabet" example where we identify any two words that begin with the same letter.)

`(actually just noticed we can delete the "switch" map from X above to get an even smaller counterexample – a monoid with just three elements. since any two-element monoid is commutative, this is the smallest counterexample. more generally, consider the "words on an alphabet" example where we identify any two words that begin with the same letter.)`

Good, Anindya! Right, saying that a map from a monoid to itself \(F: X \to X\) preserves multiplication quickly implies that \(F(1)\) is an idempotent that commute with everything

in the rangeof \(F\), but no more... so on the principle that "you don't get something for nothing", there should be examples where \(F(1)\) doesn't commute with everything in \(X\). Then the challenge is to find a counterexample... a challenge you met.`Good, Anindya! Right, saying that a map from a monoid to itself \\(F: X \to X\\) preserves multiplication quickly implies that \\(F(1)\\) is an idempotent that commute with everything _in the range_ of \\(F\\), but no more... so on the principle that "you don't get something for nothing", there should be examples where \\(F(1)\\) doesn't commute with everything in \\(X\\). Then the challenge is to find a counterexample... a challenge you met.`

Michael wrote:

I'm guessing what seemed "paradoxical" was having

$$ f(x) \otimes f(x) \le f(x \otimes x') \textrm{ and } I \le f(I) $$ and also

$$ f(x) \otimes f(x) \ge f(x \otimes x') \textrm{ and } I \ge f(I) $$ yet still not having

$$ f(x) \otimes f(x) = f(x \otimes x') \textrm{ and } I = f(I). $$ Presumably this was because you're used to

posets, where \(x \le y \) and \(y \le x\) imply \(x = y\). But in apreorderthis needn't be true.So, you need to understand preorders that aren't posets. Here is the easiest example: take a set \(X\) and decree that

everythingin this set is less than or equal toeverything else. Then the laws of a preorder hold: check them in your mind, and if you don't instantly remember what these laws are, go to jail and stay there until you do! But the law that makes a preorder a poset:$$ \textrm{ if } x \le y \textrm{ and } y \le x \textrm{ then } x = y $$ obviously does

nothold.So, this kind of preorder, very far from being a poset, is good to keep in mind.

Let's use this kind to get a monotone map that's strong monoidal but not strict monoidal.

Answer to Puzzle 82.Let \(X\) and \(Y\) be monoids and let \(f : X \to Y\) be a function that'snota homomorphism: for example,$$ f(x \otimes x') \ne f(x) \otimes f(x') $$ for some \(x,x' \in X\). Examples of this are a dime a dozen: just take your favorite two monoids with lots of elements and take some random idiotic function between them: it probably won't make \(f(x \otimes x') = f(x) \otimes f(x')\) for all \(x,x' \in X\).

Now, make \(X\) into a preorder in the silly way I just described: decree that

everythingis less than or equal toeverything. Do the same for \(Y\).Then it's easy to see that \(X\) and \(Y\) are monoidal preorders. For example \(X\) obeys

$$ x_1 \le x_1' \textrm{ and } x_2 \le x_2' \textrm{ imply } x_1 \otimes x_2 \le x_1' \otimes x_2' $$ because

everything in \(X\) is less than or equal to everything else!Similarly, it's easy to see that \(f: X \to Y\) is lax monoidal. We have

$$ f(x) \otimes f(x) \le f(x \otimes x') \textrm{ and } I \le f(I), $$ because

everything in \(Y\) is less than or equal to everything else!We also know that \(f\) is oplax monoidal:

$$ f(x) \otimes f(x) \ge f(x \otimes x') \textrm{ and } I \ge f(I) $$ because

everything in \(Y\) is greater than or equal to everything else!So \(f\) is strong monoidal for very silly reasons. But it's not

strictmonoidal, because we've set things up to ensure$$ f(x \otimes x') \ne f(x) \otimes f(x') . $$ Get it?

The moral is that

sometimesin a preorder saying that one thing is less than or equal to another is sayingabsolutely nothing, becauseeverybodyis less than or equal toeverybodyelse. So in a typical preorder, you shouldneverexpect to reason from inequalities to equations. For that, you want a poset.`Michael wrote: > I am having a hard time understanding strong monoidal monotone functions. This seems to be a paradox for me (like this water is both hot and cold). Does anyone have an example of one that can clarify how this can happen? I'm guessing what seemed "paradoxical" was having \[ f(x) \otimes f(x) \le f(x \otimes x') \textrm{ and } I \le f(I) \] and also \[ f(x) \otimes f(x) \ge f(x \otimes x') \textrm{ and } I \ge f(I) \] yet still not having \[ f(x) \otimes f(x) = f(x \otimes x') \textrm{ and } I = f(I). \] Presumably this was because you're used to _posets_, where \\(x \le y \\) and \\(y \le x\\) imply \\(x = y\\). But in a _preorder_ this needn't be true. So, you need to understand preorders that aren't posets. Here is the easiest example: take a set \\(X\\) and decree that _everything_ in this set is less than or equal to _everything else_. Then the laws of a preorder hold: check them in your mind, and if you don't instantly remember what these laws are, go to jail and stay there until you do! But the law that makes a preorder a poset: \[ \textrm{ if } x \le y \textrm{ and } y \le x \textrm{ then } x = y \] obviously does _not_ hold. So, this kind of preorder, very far from being a poset, is good to keep in mind. Let's use this kind to get a monotone map that's strong monoidal but not strict monoidal. **Answer to Puzzle 82.** Let \\(X\\) and \\(Y\\) be monoids and let \\(f : X \to Y\\) be a function that's _not_ a homomorphism: for example, \[ f(x \otimes x') \ne f(x) \otimes f(x') \] for some \\(x,x' \in X\\). Examples of this are a dime a dozen: just take your favorite two monoids with lots of elements and take some random idiotic function between them: it probably won't make \\(f(x \otimes x') = f(x) \otimes f(x')\\) for all \\(x,x' \in X\\). Now, make \\(X\\) into a preorder in the silly way I just described: decree that _everything_ is less than or equal to _everything_. Do the same for \\(Y\\). Then it's easy to see that \\(X\\) and \\(Y\\) are monoidal preorders. For example \\(X\\) obeys \[ x_1 \le x_1' \textrm{ and } x_2 \le x_2' \textrm{ imply } x_1 \otimes x_2 \le x_1' \otimes x_2' \] because _everything in \\(X\\) is less than or equal to everything else!_ Similarly, it's easy to see that \\(f: X \to Y\\) is lax monoidal. We have \[ f(x) \otimes f(x) \le f(x \otimes x') \textrm{ and } I \le f(I), \] because _everything in \\(Y\\) is less than or equal to everything else!_ We also know that \\(f\\) is oplax monoidal: \[ f(x) \otimes f(x) \ge f(x \otimes x') \textrm{ and } I \ge f(I) \] because _everything in \\(Y\\) is greater than or equal to everything else!_ So \\(f\\) is strong monoidal for very silly reasons. But it's not _strict_ monoidal, because we've set things up to ensure \[ f(x \otimes x') \ne f(x) \otimes f(x') . \] Get it? The moral is that _sometimes_ in a preorder saying that one thing is less than or equal to another is saying _absolutely nothing_, because _everybody_ is less than or equal to _everybody_ else. So in a typical preorder, you should _never_ expect to reason from inequalities to equations. For that, you want a poset.`

John

I think I will always remember this example John. Thanks. Your example made me realize why I have trouble sometimes with the most basic things. I keep thinking in terms of specific examples while trying to keep tabs on the abstraction which causes problems if the example in my head isn't abstract enough to capture the greater picture.

This opened up a lot for me. Thanks.

`John >take a set X and decree that everything in this set is less than or equal to everything else. I think I will always remember this example John. Thanks. Your example made me realize why I have trouble sometimes with the most basic things. I keep thinking in terms of specific examples while trying to keep tabs on the abstraction which causes problems if the example in my head isn't abstract enough to capture the greater picture. >So in a typical preorder, you should never expect to reason from inequalities to equations. For that, you want a poset. This opened up a lot for me. Thanks.`

I'm glad that helped:

If you do that, which can be dangerous, you need enough examples to keep from getting led astray by the peculiarities of special cases.

In particular, you need to pay special attention to "silly" examples, where the structures are chosen in a "trivial" way. The preorder where everything is \(\le\) to everything else; the preorder where anything is only \(\le\) itself - all other examples stand somewhere between these two.

Another important thing to do is this: whenever someone defines some sort of gadget, like a "preorder", and then defines a nice special case with extra features, like a "poset", you need to look for examples of posets that

aren'tpreorders. Without this, you can't see what the special extra features are buying you! Mathematicians automatically do this whenever they learn new concepts.It's just like if you were a biologist and you'd just learned about "animals" and "vertebrates": you'd have to learn about some animals that aren't vertebrates.

`I'm glad that helped: > I keep thinking in terms of specific examples... If you do that, which can be dangerous, you need enough examples to keep from getting led astray by the peculiarities of special cases. In particular, you need to pay special attention to "silly" examples, where the structures are chosen in a "trivial" way. The preorder where everything is \\(\le\\) to everything else; the preorder where anything is only \\(\le\\) itself - all other examples stand somewhere between these two. Another important thing to do is this: whenever someone defines some sort of gadget, like a "preorder", and then defines a nice special case with extra features, like a "poset", you need to look for examples of posets that _aren't_ preorders. Without this, you can't see what the special extra features are buying you! Mathematicians automatically do this whenever they learn new concepts. It's just like if you were a biologist and you'd just learned about "animals" and "vertebrates": you'd have to learn about some animals that aren't vertebrates.`

I def need more mathematical creatures in my zoo LOL. I will keep this in mind and practice finding examples of new concepts with special features.

`>It's just like if you were a biologist and you'd just learned about "animals" and "vertebrates": you'd have to learn about some animals that aren't vertebrates. I def need more mathematical creatures in my zoo LOL. I will keep this in mind and practice finding examples of new concepts with special features.`

John, this example is a bit confusing: animals are a more generic structure than vertebrates (vertebrates are animals with a specific structure in them - they are animals with vertebrae), so

preorders \(\rightarrow\) animals

posets \(\rightarrow\) vertebrates

So following the logic, to learn more about vertebrates, one should look for vertebrates which

aren'tanimals, with attached interpretation - to learn more about vertebrates, look for some artificial structures which have spine-like support in them.`>It's just like if you were a biologist and you'd just learned about "animals" and "vertebrates": you'd have to learn about some animals that aren't vertebrates. John, this example is a bit confusing: animals are a more generic structure than vertebrates (vertebrates are animals with a specific structure in them - they are animals with vertebrae), so preorders \\(\rightarrow\\) animals posets \\(\rightarrow\\) vertebrates So following the logic, to learn more about vertebrates, one should look for vertebrates which *aren't* animals, with attached interpretation - to learn more about vertebrates, look for some artificial structures which have spine-like support in them.`

Well, every poset

isa preorder, so I believe John meant "you need to look for examples of preorders that aren't posets". If we correct this side of the analogy, everything works well.`Well, every poset _is_ a preorder, so I believe John meant "you need to look for examples of preorders that aren't posets". If we correct this side of the analogy, everything works well.`

Here are a couple programming examples.

In Haskell, in the base language there is a type class for Monoids defined in

`Data.Monoid`

.Haskell has its own notation for Monoids. In Haskell, \(I\) becomes

`mempty`

and \(\otimes\) becomes`(<>)`

.Two of the monoids defined in

`Data.Monoid`

are`Monoid (First a)`

and`Monoid [a]`

.`First a`

is also defined in`Data.Monoid`

.`First`

is a`newtype`

wrapper around`Maybe a`

. Its identity element is`First Nothing`

. Its monoidal operation`(<>)`

has the following definitionThe other monoid instance is for

list, the free monoid over`a`

. Its identity element is`[]`

and monoidal operation islist append(ie,`++`

).There are two monoid homomorphisms between these types.

The first homomorphism is

`First . listToMaybe :: [a] -> First a`

. Here`listToMaybe`

is defined in`Data.Maybe.listToMaybe`

. This function tries to take the first element of a list. If there is such an element`a`

then it gets wrapped in`First (Just a)`

. If it does not exist (because the list is empty), then it returns`First Nothing`

.The

Monoid homomorphisms lawsin Haskell are defined by: $$(\mathsf{First}\; .\; \mathsf{listToMaybe})\; (a\; \mathsf{<>}\; b) \equiv ((\mathsf{First}\; .\; \mathsf{listToMaybe})\; a)\; \mathsf{<>}\; ((\mathsf{First}\; .\; \mathsf{listToMaybe})\; b)$$ $$(\mathsf{First}\; .\; \mathsf{listToMaybe})\; (\mathsf{mempty}\; ::\; \mathsf{[}a\mathsf{]}) \equiv (\mathsf{mempty}\; :: \; \mathsf{First}\; a)$$ Hence this function is a monoid homomorphism.The second homomorphism is in the opposite direction:

`maybeToList . getFirst :: First a -> [a]`

. This is theright inverseof`First . listToMaybe`

, so: $$(\mathsf{First}\; .\; \mathsf{listToMaybe})\; .\; (\mathsf{maybeToList}\; .\; \mathsf{getFirst}) \equiv \mathsf{id}$$ The function`maybeToList`

is defined in`Data.Maybe.maybeToList`

. It obeys the same monoid homomorphism laws as its left inverse.`> **Puzzle 80.** Tell me a few more homomorphisms between monoids that you routinely use, or at least know. Here are a couple programming examples. In Haskell, in the base language there is a type class for Monoids defined in [`Data.Monoid`](https://hackage.haskell.org/package/base-4.11.1.0/docs/Data-Monoid.html). Haskell has its own notation for Monoids. In Haskell, \\(I\\) becomes `mempty` and \\(\otimes\\) becomes `(<>)`. Two of the monoids defined in `Data.Monoid` are `Monoid (First a)` and `Monoid [a]`. `First a` is also defined in `Data.Monoid`. `First` is a [`newtype`](https://wiki.haskell.org/Newtype) wrapper around `Maybe a`. Its identity element is `First Nothing`. Its monoidal operation `(<>)` has the following definition <pre> (First Nothing) <> x = x (First (Just a)) <> _ = First (Just a) </pre> The other monoid instance is for *list*, the [free monoid](https://en.wikipedia.org/wiki/Free_monoid) over `a`. Its identity element is `[]` and monoidal operation is *list append* (ie, [`++`](https://hackage.haskell.org/package/base-4.11.1.0/docs/Prelude.html#v:-43--43-)). There are two monoid homomorphisms between these types. 1. The first homomorphism is `First . listToMaybe :: [a] -> First a`. Here `listToMaybe` is defined in [`Data.Maybe.listToMaybe`](https://hackage.haskell.org/package/base-4.11.1.0/docs/Data-Maybe.html#v:listToMaybe). This function tries to take the first element of a list. If there is such an element `a` then it gets wrapped in `First (Just a)`. If it does not exist (because the list is empty), then it returns `First Nothing`. The *Monoid homomorphisms laws* in Haskell are defined by: $$(\mathsf{First}\; .\; \mathsf{listToMaybe})\; (a\; \mathsf{<>}\; b) \equiv ((\mathsf{First}\; .\; \mathsf{listToMaybe})\; a)\; \mathsf{<>}\; ((\mathsf{First}\; .\; \mathsf{listToMaybe})\; b)$$ $$(\mathsf{First}\; .\; \mathsf{listToMaybe})\; (\mathsf{mempty}\; ::\; \mathsf{[}a\mathsf{]}) \equiv (\mathsf{mempty}\; :: \; \mathsf{First}\; a)$$ Hence this function is a monoid homomorphism. 2. The second homomorphism is in the opposite direction: `maybeToList . getFirst :: First a -> [a]`. This is the *right inverse* of `First . listToMaybe`, so: $$(\mathsf{First}\; .\; \mathsf{listToMaybe})\; .\; (\mathsf{maybeToList}\; .\; \mathsf{getFirst}) \equiv \mathsf{id}$$ The function `maybeToList` is defined in [`Data.Maybe.maybeToList`](https://hackage.haskell.org/package/base-4.11.1.0/docs/Data-Maybe.html#v:maybeToList). It obeys the same monoid homomorphism laws as its left inverse.`

@Matthew – what's the motivation for "wrapping"

`Maybe`

with`First`

? is`Maybe`

not a monoid "already", so to speak?`@Matthew – what's the motivation for "wrapping" `Maybe` with `First`? is `Maybe` not a monoid "already", so to speak?`

`Maybe a`

all on its own doesn’t have a monoid instance. There are a few possible semantics it can have.If

`a`

is a monoid, then there is an instance for`Maybe a`

. It’s worth trying to guess how this is defined in base if you haven the seen it.Apart from

`First`

, there is another new type called`Last`

that gives different monoid semantics for`Maybe`

. You can always look it up with hoogle, but the name is a good hint at how it works if you want to guess...``Maybe a` all on its own doesn’t have a monoid instance. There are a few possible semantics it can have. If `a` is a monoid, then there is an instance for `Maybe a`. It’s worth trying to guess how this is defined in base if you haven the seen it. Apart from `First`, there is another new type called `Last` that gives different monoid semantics for `Maybe`. You can always look it up with hoogle, but the name is a good hint at how it works if you want to guess...`

ah I get it – there's no canonical way of defining a product of Maybes. you have choose one (First or Last).

`ah I get it – there's no canonical way of defining a product of Maybes. you have choose one (First or Last).`

It's a little trickier than that, sadly.

For instance, while

`Maybe a`

is not necessarily a monoid,`Maybe [a]`

is.Feel free to check in the interpreter, but we have:

$$ (\textsf{Just}\; a) <> (\textsf{Just}\; b) \cong \textsf{Just}\; (a ++ b) $$ So we have that

`Just :: [a] -> Maybe [a]`

is a semigroup homorphism...`> ah I get it – there's no canonical way of defining a product of Maybes. you have choose one (First or Last). It's a little trickier than that, sadly. For instance, while `Maybe a` is not necessarily a monoid, `Maybe [a]` *is*. Feel free to check in the interpreter, but we have: $$ (\textsf{Just}\; a) <> (\textsf{Just}\; b) \cong \textsf{Just}\; (a ++ b) $$ So we have that `Just :: [a] -> Maybe [a]` is a semigroup homorphism...`

Also it's important to note that list being the free* (and therefore initial) there is a mapping,

`foldr mempty (<>)`

is a mapping`[a] -> a`

that is automatically a monoid homomorphism, irregardless what the monoid on`a`

is.`Also it's important to note that list being the free* (and therefore initial) there is a mapping, `foldr mempty (<>)` is a mapping `[a] -> a` that is automatically a monoid homomorphism, irregardless what the monoid on `a` is. * technically list append is also infinitely associative. I have read anyway. (When i try to use what I thinkthis means, the left associative form diverges, and the right doesn't. Which makes them non equal. So.. either my memory or my understanding is wrong. ) * the difference between foldr and foldl' only matters if the biop is non associative or you care about /how/ the value is computed.`

Hey I have a terminology question. There are two ways to turn a preorder into a equivalence relation. 1) the symmetric closure:

\[a \sim b := a \le b \text{ or } b \le a \] and 2) the symmetric ??:

\[a \sim b := a \le b \text{ and } b \le a \]

I think I've read "core" used for this in the context of turning groups commutative. Is that a good choice for the general idea? The closure is often the free construction, and I think the "core" would be the other adjoint to the forgetful functor some times.

`Hey I have a terminology question. There are two ways to turn a preorder into a equivalence relation. 1) the symmetric closure: \\[a \sim b := a \le b \text{ or } b \le a \\] and 2) the symmetric ??: \\[a \sim b := a \le b \text{ and } b \le a \\] I think I've read "core" used for this in the context of turning groups commutative. Is that a good choice for the general idea? The closure is often the free construction, and I think the "core" would be the other adjoint to the forgetful functor some times.`

I wrote:

As Jonathan pointed out, that was wrong: there aren't any posets that aren't preorders. I meant to say this:

`I wrote: > Another important thing to do is this: whenever someone defines some sort of gadget, like a "preorder", and then defines a nice special case with extra features, like a "poset", you need to look for examples of posets that aren't preorders. As Jonathan pointed out, that was wrong: there aren't any posets that aren't preorders. I meant to say this: > Another important thing to do is this: whenever someone defines some sort of gadget, like a "preorder", and then defines a nice special case with extra features, like a "poset", you need to look for examples of preorders that aren't posets.`

Christopher wrote:

Alas, that doesn't always work.

Puzzle.Find a preorder such that \(\sim\), defined as above, is not an equivalence relation.This does give an equivalence relation, and it's very important. We usually write \(a \cong b\) in this case, and say \(a\) and \(b\) are

isomorphic.I haven't heard of this, and I don't see how exactly this is related to maing groups commutative (aka "abelian"). There's a forgetful functor from abelian groups to groups, and it has a left adjoint called "abelianization". There's also another way to take a group and get an abelian group, called taking the "center". However, this is not a functor!

Anyway, I guess you're talking about the forgetful functor from the category of "sets with equivalence relation" to the category of preorders, and possible left or right adjoints to this. Method 2) of turning preorders into equivalence relations feels like a right adjoint, though I haven't checked.

Here's another way to turn a preorder into an equivalence relation. Start with a preorder. Take its symmetric closure as you did, defining \(a \sim b\) if \(a \le b\) or \(b \le a\). Then take the transitive closure of that relation\(\sim\), getting a relation that's both symmetric and transitive.

I believe this method gives the left adjoint to the forgetful functor from the category of "sets with equivalence relation" to the category of preorders... though again, I haven't carefully checked this so don't sue me if I'm wrong!

`Christopher wrote: > There are two ways to turn a preorder into a equivalence relation. 1) the symmetric closure: > \\[a \sim b := a \le b \text{ or } b \le a \\] Alas, that doesn't always work. **Puzzle.** Find a preorder such that \\(\sim\\), defined as above, is not an equivalence relation. > and 2) the symmetric ??: > \\[a \sim b := a \le b \text{ and } b \le a \\] This does give an equivalence relation, and it's very important. We usually write \\(a \cong b\\) in this case, and say \\(a\\) and \\(b\\) are **isomorphic**. > I think I've read "core" used for this in the context of turning groups commutative. I haven't heard of this, and I don't see how exactly this is related to maing groups commutative (aka "abelian"). There's a forgetful functor from abelian groups to groups, and it has a left adjoint called ["abelianization"](http://mathworld.wolfram.com/Abelianization.html). There's also another way to take a group and get an abelian group, called taking the ["center"](https://en.wikipedia.org/wiki/Center_(group_theory)). However, this is not a functor! Anyway, I guess you're talking about the forgetful functor from the category of "sets with equivalence relation" to the category of preorders, and possible left or right adjoints to this. Method 2) of turning preorders into equivalence relations feels like a right adjoint, though I haven't checked. Here's another way to turn a preorder into an equivalence relation. Start with a preorder. Take its symmetric closure as you did, defining \\(a \sim b\\) if \\(a \le b\\) or \\(b \le a\\). Then take the [transitive closure](https://en.wikipedia.org/wiki/Transitive_closure) of that relation\\(\sim\\), getting a relation that's both symmetric and transitive. I believe this method gives the left adjoint to the forgetful functor from the category of "sets with equivalence relation" to the category of preorders... though again, I haven't carefully checked this so don't sue me if I'm wrong!`

Christopher: what you actually described were the two main ways to take an arbitrary binary relation and turn it into a symmetric binary relation. These are the left and right adjoints to the forgetful functor from "sets with symmetric relation" to "sets with arbitrary binary relation". As usual the left adjoint is the "permissive, generous" method, which uses "or", while the right adjoint is the "cautious, ungenerous" method which uses "and".

You see, "or" is itself a left adjoint, while "and" is a right adjoint.

`Christopher: what you actually described were the two main ways to take an arbitrary binary relation and turn it into a symmetric binary relation. These are the left and right adjoints to the forgetful functor from "sets with symmetric relation" to "sets with arbitrary binary relation". As usual the left adjoint is the "permissive, generous" method, which uses "or", while the right adjoint is the "cautious, ungenerous" method which uses "and". You see, ["or" is itself a left adjoint, while "and" is a right adjoint](https://forum.azimuthproject.org/discussion/2037/lecture-17-chapter-1-the-grand-synthesis/p1).`

@Matthew -- re:

This seems to be a special case of a more general result, namely that if

`Moo`

is a monoid, then we can define a canonical monoid structure on`Maybe Moo`

.The unit is

`Nothing`

, and we define`(Just x) <> (Just y)`

to be`Just (x <> y)`

.`Just`

here is a good example of a semigroup homomorphism between monoids that isn't a monoid homomorphism (because it doesn't preserve the unit).`@Matthew -- re: > while `Maybe a` is not necessarily a monoid, `Maybe [a]` *is* This seems to be a special case of a more general result, namely that if `Moo` is a monoid, then we can define a canonical monoid structure on `Maybe Moo`. The unit is `Nothing`, and we define `(Just x) <> (Just y)` to be `Just (x <> y)`. `Just`here is a good example of a semigroup homomorphism between monoids that isn't a monoid homomorphism (because it doesn't preserve the unit).`

You are absolutely right!

Haskell's defines the instance

`Semigroup a => Semigroup (Maybe a)`

in the Prelude, following your specification:And the

unitfor this monoid is defined in Data.Monoid, again exactly as you suggest:`> @Matthew -- re: > > while `Maybe a` is not necessarily a monoid, `Maybe [a]` *is* > > This seems to be a special case of a more general result, namely that if `Moo` is a monoid, then we can define a canonical monoid structure on `Maybe Moo`. > > The unit is `Nothing`, and we define `(Just x) <> (Just y)` to be `Just (x <> y)`. You are absolutely right! Haskell's defines the instance `Semigroup a => Semigroup (Maybe a)` in the [Prelude](https://hackage.haskell.org/package/base-4.11.1.0/docs/src/GHC.Base.html#line-406), following your specification: <pre> instance Semigroup a => Semigroup (Maybe a) where Nothing <> b = b a <> Nothing = a Just a <> Just b = Just (a <> b) </pre> And the *unit* for this monoid is defined in [Data.Monoid](https://hackage.haskell.org/package/base-4.11.1.0/docs/src/GHC.Base.html#line-422), again exactly as you suggest: <pre> instance Semigroup a => Monoid (Maybe a) where mempty = Nothing </pre>`