Howdy, Stranger!

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

Options

Lecture 48 - Chapter 3: Adjoint Functors

edited February 2020

Last time I threw the definition of 'adjoint functor' at you. Now let me actually explain adjoint functors!

As we learned long ago, the basic idea is that adjoints give the best possible way to approximately recover data that can't really be recovered.

For example, you might have a map between databases that discards some data. You might like to reverse this process. Strictly speaking this is impossible: if you've truly discarded some data, you don't know what it is anymore, so you can't restore it. But you can still do your best!

There are actually two kinds of 'best': left adjoints and right adjoints.

Remember the idea. We have a functor $$F : \mathcal{A} \to \mathcal{B}$$. We're looking for a nice functor $$G: \mathcal{B} \to \mathcal{A}$$ that goes back the other way, some sort of attempt to reverse the effect of $$F$$. We say that $$G$$ is a right adjoint of $$F$$ if there's a natural one-to-one correspondence between

• morphisms from $$F(a)$$ to $$b$$

and

• morphisms from $$a$$ to $$G(b)$$

whenever $$a$$ is an object of $$\mathcal{A}$$ and $$b$$ is an object of $$\mathcal{B}$$. In this situation we also say $$F$$ is a left adjoint of $$G$$.

The tricky part in this definition is the word 'natural'. That's why I had to explain natural transformations. But let's see how far we can get understanding adjoint functors without worrying about this.

Let's do an example. There's a category $$\mathbf{Set}$$ where objects are sets and morphisms are functions. And there's much more boring category $$\mathbf{1}$$, with exactly one object and one morphism. Today let's call that one object $$\star$$, so the one morphism is $$1_\star$$.

In Puzzle 135 we saw there is always exactly one functor from any category to $$\mathbf{1}$$. So, there's exactly one functor

$$F: \mathbf{Set} \to \mathbf{1}$$ This sends every set to the object $$\star$$, and every function between sets to the morphism $$1_\star$$.

This is an incredibly destructive functor! It discards all the information about every set and every function! $$\mathbf{1}$$ is like the ultimate trash can, or black hole. Drop data into it and it's gone.

So, it seems insane to try to 'reverse' the functor $$F$$, but that's what we'll do. First let's look for a right adjoint

$$G: \mathbf{1} \to \mathbf{Set} .$$ For $$G$$ to be a right adjoint, we need a natural one-to-one correspondence between morphisms

$$m: F(S) \to \star$$ and morphisms

$$n: S \to G(\star)$$ where $$S$$ is any set.

Think about what this means! We know $$F(S) = \star$$: there's nothing else it could be, since $$\mathbf{1}$$ has just one object. So, we're asking for a natural one-to-one correspondence between the set of morphisms

$$m : \star \to \star$$ and the set of functions

$$n : S \to G(\star) .$$ This has got to work for every set $$S$$. This should tell us a lot about $$G(\star)$$.

Well, there's just one morphism $$m : \star \to \star$$, so there had better be just one function $$n : S \to G(\star)$$, for any set $$S$$. This forces $$G(\star)$$ to be a set with just one element. And that does the job! We can take $$G(\star)$$ to be any set with just one element, and that gives us our left adjoint $$G$$.

Well, okay: we have to say what $$G$$ does to morphisms, too. But the only morphism in $$\mathbf{1}$$ is $$1_\star$$, and we must have $$G(1_\star) = 1_{G(\star)}$$, thanks to how functors work.

(Furthermore you might wonder about the 'naturality' condition, but this example is so trivial that it's automatically true.)

So: if you throw away a set into the trash bin called $$\mathbf{1}$$, and I say "wait! I want that set back!", and I have to make up something, the right adjoint procedure says "pick any one-element set". Weird but true. When you really understand adjoints, you'll have a good intuitive sense for why it works this way.

What about the left adjoint procedure? Let's use $$L$$ to mean a left adjoint of our functor $$F$$:

Puzzle 149. Again suppose $$F: \mathbf{Set} \to \mathbf{1}$$ is the functor that sends every set to $$\star$$ and every function to $$1_\star$$. A left adjoint $$L : \mathbf{1} \to \mathbf{Set}$$ is a functor for which there's a natural one-to-one correspondence between functions

$$m: L(\star) \to S$$ and morphisms

$$n: \star \to F(S)$$ for every set $$S$$. On the basis of this, try to figure out all the left adjoints of $$F$$.

Let's also try some slightly harder examples. There is a category $$\mathbf{Set}^2$$ where an object is a pair of sets $$(S,T)$$. In this category a morphism is a pair of functions, so a morphism

$$(f,g): (S,T) \to (S',T')$$ is just a function $$f: S \to S'$$ together with function $$g: T \to T'$$. We compose these morphisms componentwise:

$$(f,g) \circ (f',g') = (f\circ f', g \circ g') .$$ You can figure out what the identity morphisms are, and check all the category axioms.

There's a functor

$$F: \mathbf{Set}^2 \to \mathbf{Set}$$ that discards the second component. So, on objects it throws away the second set:

$$F(S,T) = S$$ and on morphisms it throws away the second function:

$$F(f,g) = f .$$ Puzzle 150. Figure out all the right adjoints of $$F$$.

Puzzle 151. Figure out all the left adjoints of $$F$$.

There's also a functor

$$\times: \mathbf{Set}^2 \to \mathbf{Set}$$ that takes the Cartesian product, both for sets:

$$\times (S,T) = S \times T$$ and for functions:

$$\times (f,g) = f \times g$$ where $$(f\times g)(s,t) = (f(s),g(t))$$ for all $$s \in S, t \in T$$.

Puzzle 152. Figure out all the right adjoints of $$\times$$.

Puzzle 153. Figure out all the left adjoints of $$\times$$.

Finally, there's also a functor

$$+ : \mathbf{Set}^2 \to \mathbf{Set}$$ that takes the disjoint union, both for sets:

$$+ (S,T) = S + T$$ and for functions:

$$+(f,g) = f + g .$$ Here $$S + T$$ is how category theorists write the disjoint union of sets $$S$$ and $$T$$. Furthermore, given functions $$f: S \to S'$$ and $$g: T \to T'$$ there's an obvious function $$f+g: S+T \to S'+T'$$ that does $$f$$ to the guys in $$S$$ and $$g$$ to the guys in $$T$$.

Puzzle 152. Figure out all the right adjoints of $$+$$.

Puzzle 153. Figure out all the left adjoints of $$+$$.

I think it's possible to solve all these puzzles even if one has a rather shaky grasp on adjoint functors. At least try them! It's a good way to start confronting this new concept.

To read other lectures go here.

• Options
1.
edited June 2018

Puzzle 149. Again suppose $$F: \mathbf{Set} \to \mathbf{1}$$ is the functor that sends every set to $$\star$$ and every function to $$1_\star$$. A left adjoint $$L : \mathbf{1} \to \mathbf{Set}$$ is a functor for which there's a natural one-to-one correspondence between functions

$$m: L(\star) \to S$$ and morphisms $$n: \star \to F(S)$$ for every set $$S$$. On the basis of this, try to figure out all the left adjoints of $$F$$.

Demanding that $$L(\star)\to S$$, for all $$S$$ is the same thing as asking for a set $$T$$ such that, for all $$S$$, $$T \to S$$ holds.

The only such set is the empty set, so the left adjoint to $$F: \mathbf{Set} \to \mathbf{1}$$ is a functor taking $$\mathbf{1}$$ to the empty set.

$L(\star) = \varnothing, \\ L(1_\star) = 1_\varnothing$

Comment Source:>**Puzzle 149.** Again suppose \$$F: \mathbf{Set} \to \mathbf{1}\$$ is the functor that sends every set to \$$\star\$$ and every function to \$$1_\star\$$. A _left_ adjoint \$$L : \mathbf{1} \to \mathbf{Set} \$$ is a functor for which there's a natural one-to-one correspondence between functions >$m: L(\star) \to S$ >and morphisms >$n: \star \to F(S)$ >for every set \$$S\$$. On the basis of this, try to figure out all the left adjoints of \$$F\$$. Demanding that \$$L(\star)\to S\$$, for all \$$S\$$ is the same thing as asking for a set \$$T\$$ such that, for all \$$S\$$, \$$T \to S\$$ holds. The only such set is the empty set, so the left adjoint to \$$F: \mathbf{Set} \to \mathbf{1}\$$ is a functor taking \$$\mathbf{1}\$$ to the empty set. \$L(\star) = \varnothing, \\\\ L(1\_\star) = 1\_\varnothing \$
• Options
2.

Perhaps an ill posed question, but here it goes anyway. Does the notion of adjunction in the case of monoids reduce to something formerly known? I mean: When the categories are preorders, an adjoint situation is a Galois connection. In the case when one monoid is homomorphic to another, is there a best-effort inverse, "from above and below", given by a construction named X, that complies with the definition of adjunction for single-object categories?

Comment Source:Perhaps an ill posed question, but here it goes anyway. Does the notion of adjunction in the case of monoids reduce to something formerly known? I mean: When the categories are preorders, an adjoint situation is a Galois connection. In the case when one monoid is homomorphic to another, is there a best-effort inverse, "from above and below", given by a construction named X, that complies with the definition of adjunction for single-object categories?
• Options
3.

Perhaps an ill posed question, but here it goes anyway. Does the notion of adjunction in the case of monoids reduce to something formerly known? I mean: When the categories are preorders, an adjoint situation is a Galois connection. In the case when one monoid is homomorphic to another, is there a best-effort inverse, "from above and below", given by a construction named X, that complies with the definition of adjunction for single-object categories?

I don't think this is an ill-posed question.

In the event of a lattice equipped with a monoid with two adjoints for $$\bullet$$, we call such a structure a residuated lattice.

I believe another example of a residuated lattice is a quantale.

Comment Source:> Perhaps an ill posed question, but here it goes anyway. Does the notion of adjunction in the case of monoids reduce to something formerly known? I mean: When the categories are preorders, an adjoint situation is a Galois connection. In the case when one monoid is homomorphic to another, is there a best-effort inverse, "from above and below", given by a construction named X, that complies with the definition of adjunction for single-object categories? I don't think this is an ill-posed question. In the event of a lattice equipped with a monoid with two adjoints for \$$\bullet\$$, we call such a structure a [*residuated lattice*](https://en.wikipedia.org/wiki/Residuated_lattice#Definition). I believe another example of a residuated lattice is a [quantale](https://en.wikipedia.org/wiki/Quantale).
• Options
4.
edited June 2018

Puzzle 149. Again suppose $$F: \mathbf{Set} \to \mathbf{1}$$ is the functor that sends every set to $$\star$$ and every function to $$1_\star$$. A left adjoint $$L : \mathbf{1} \to \mathbf{Set}$$ is a functor for which there's a natural one-to-one correspondence between functions

$$m: L(\star) \to S$$ and morphisms

$$n: \star \to F(S)$$ for every set $$S$$. On the basis of this, try to figure out all the left adjoints of $$F$$.

Keith wrote:

Demanding that $$L(\star)\to S$$, for all $$S$$ is the same thing as asking for a set $$T$$ such that, for all $$S$$, $$T \to S$$ holds.

The only such set is the empty set, so the left adjoint to $$F: \mathbf{Set} \to \mathbf{1}$$ is a functor taking $$\mathbf{1}$$ to the empty set.

Right!

So we see the beginnings of a nice pattern:

• a right adjoint of the unique functor $$F: \textbf{Set} \to \textbf{1}$$ sends the one object of $$\textbf{1}$$ to a set with one element, sometimes called '1' Such a set is a terminal object in $$\mathbf{Set}$$, meaning that there's exactly one function from any set to this set.

• a left adjoint of the unique functor $$F: \textbf{Set} \to \textbf{1}$$ sends the one object of $$\textbf{1}$$ to the set with no elements, sometimes called '0'. Such a set is an initial object in $$\mathbf{Set}$$, meaning that there's exactly one function to any set from this set.

This makes some sense because right adjoints are defined using morphisms to them, as are terminal objects. Left adjoints are defined using morphisms from them, as are initial objects.

Finally, a discussion of your argument showing that $$L(\star)$$ is empty.

Mathematicians don't say "$$T \to S$$ holds". We say "X holds" when X is a proposition that can be true or false, and X happens to be true. There is no proposition "$$T \to S$$ ". In fact "$$T \to S$$" doesn't mean anything by itself. What makes sense are functions $$f: T \to S$$. Given two sets $$T$$ and $$S$$, we can have one, more than one, or no functions $$f: T \to S$$.

Here's how we prove $$L(\star)$$ must be empty:

If $$L$$ is the left adjoint of $$F$$, there must be a one-to-one correspondence between morphisms $$n: \star \to F(S)$$ and functions $$m: L(\star)\to S$$, for all $$S$$. Since $$F(S) = \star$$, there is exactly one morphism $$n: \star \to F(S)$$ . Thus, there must be exactly one function $$m: L(\star) \to S$$ for any set $$S$$.

This forces $$L(\star)$$ to be the empty set: there's exactly one function from the empty set to any set... and the empty set is the only set with this property.

Comment Source:**Puzzle 149.** Again suppose \$$F: \mathbf{Set} \to \mathbf{1}\$$ is the functor that sends every set to \$$\star\$$ and every function to \$$1_\star\$$. A _left_ adjoint \$$L : \mathbf{1} \to \mathbf{Set} \$$ is a functor for which there's a natural one-to-one correspondence between functions $m: L(\star) \to S$ and morphisms $n: \star \to F(S)$ for every set \$$S\$$. On the basis of this, try to figure out all the left adjoints of \$$F\$$. Keith wrote: > Demanding that \$$L(\star)\to S\$$, for all \$$S\$$ is the same thing as asking for a set \$$T\$$ such that, for all \$$S\$$, \$$T \to S\$$ holds. > The only such set is the empty set, so the left adjoint to \$$F: \mathbf{Set} \to \mathbf{1}\$$ is a functor taking \$$\mathbf{1}\$$ to the empty set. Right! <img src = "http://math.ucr.edu/home/baez/thumbsup.gif"> So we see the beginnings of a nice pattern: * a right adjoint of the unique functor \$$F: \textbf{Set} \to \textbf{1} \$$ sends the one object of \$$\textbf{1}\$$ to a set with one element, sometimes called '1' Such a set is a **terminal object** in \$$\mathbf{Set}\$$, meaning that there's exactly one function _from_ any set _to_ this set. * a left adjoint of the unique functor \$$F: \textbf{Set} \to \textbf{1} \$$ sends the one object of \$$\textbf{1}\$$ to the set with no elements, sometimes called '0'. Such a set is an **initial object** in \$$\mathbf{Set}\$$, meaning that there's exactly one function _to_ any set _from_ this set. This makes some sense because right adjoints are defined using morphisms _to_ them, as are terminal objects. Left adjoints are defined using morphisms _from_ them, as are initial objects. Finally, a discussion of your argument showing that \$$L(\star)\$$ is empty. Mathematicians don't say "\$$T \to S\$$ holds". We say "X holds" when X is a proposition that can be true or false, and X happens to be true. There is no proposition "\$$T \to S\$$ ". In fact "\$$T \to S\$$" doesn't mean anything by itself. What makes sense are functions \$$f: T \to S\$$. Given two sets \$$T\$$ and \$$S\$$, we can have one, more than one, or no functions \$$f: T \to S\$$. Here's how we prove \$$L(\star)\$$ must be empty: If \$$L\$$ is the left adjoint of \$$F\$$, there must be a one-to-one correspondence between morphisms \$$n: \star \to F(S)\$$ and functions \$$m: L(\star)\to S\$$, for all \$$S\$$. Since \$$F(S) = \star\$$, there is exactly one morphism \$$n: \star \to F(S)\$$ . Thus, there must be _exactly one_ function \$$m: L(\star) \to S\$$ for any set \$$S\$$. This forces \$$L(\star)\$$ to be the empty set: there's exactly one function from the empty set to any set... and the empty set is the _only_ set with this property.
• Options
5.
edited June 2018

Puzzle 150. Figure out all the right adjoints of $$F$$, where $$F : \mathbf{Set}^2 \to \mathbf{Set}$$ is a functor that on objects it throws away the second set, $$F(S,T) = S$$, and on morphisms it throws away the second function $$F(f,g) = f$$.

A functor $$F$$ is left adjoint to a functor $$R$$ if there is a one-to-one correspondence between the morphisms $$F(A) \to B$$ and morphisms $$A \to R(B)$$:

$$\mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B))$$ Since $$A \in \mathbf{Set}^2$$, then $$A$$ is a pair $$(A_1, A_2)$$:

$$\mathbf{Set}(F(A_1, A_2), B) \cong \mathbf{Set}^2((A_1, A_2), R(B))$$ We apply the defintion of the functor $$F$$ and write $$R(B) \in \mathbf{Set}^2$$ as a pair $$(B_1, B_2)$$:

$$\mathbf{Set}(A_1, B) \cong \mathbf{Set}^2((A_1, A_2), (B_1, B_2))$$ A function $$(A_1, A_2) \to (B_1, B_2)$$ is a pair of functions, one $$A_1 \to B_1$$, the other $$A_2 \to B_2$$:

$$\mathbf{Set}(A_1, B) \cong \mathbf{Set}(A_1, B_1) \times \mathbf{Set}(A_2, B_2)$$ If we pick $$B_1 = B$$ and $$B_2 = \{\bullet\}$$, then we can map a function $$f : A_1 \to B$$ to the pair of functions $$(f : A_1 \to B, ! : A_2 \to \{\bullet\})$$, where $$!$$ denotes the unique function from any set to the singleton set $$\{\bullet\}$$. Hence, \begin{align} R(B) &= (B, {\bullet}) \\ R(f) &= (f, !). \end{align}

Comment Source:> **Puzzle 150.** Figure out all the right adjoints of \$$F\$$, where \$$F : \mathbf{Set}^2 \to \mathbf{Set}\$$ is a functor that on objects it throws away the second set, \$$F(S,T) = S \$$, and on morphisms it throws away the second function \$$F(f,g) = f \$$. A functor \$$F\$$ is left adjoint to a functor \$$R\$$ if there is a one-to-one correspondence between the morphisms \$$F(A) \to B\$$ and morphisms \$$A \to R(B)\$$: $\mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B))$ Since \$$A \in \mathbf{Set}^2\$$, then \$$A\$$ is a pair \$$(A_1, A_2)\$$: $\mathbf{Set}(F(A_1, A_2), B) \cong \mathbf{Set}^2((A_1, A_2), R(B))$ We apply the defintion of the functor \$$F\$$ and write \$$R(B) \in \mathbf{Set}^2\$$ as a pair \$$(B_1, B_2)\$$: $\mathbf{Set}(A_1, B) \cong \mathbf{Set}^2((A_1, A_2), (B_1, B_2))$ A function \$$(A_1, A_2) \to (B_1, B_2)\$$ is a pair of functions, one \$$A_1 \to B_1\$$, the other \$$A_2 \to B_2\$$: $\mathbf{Set}(A_1, B) \cong \mathbf{Set}(A_1, B_1) \times \mathbf{Set}(A_2, B_2)$ If we pick \$$B_1 = B\$$ and \$$B_2 = \\{\bullet\\}\$$, then we can map a function \$$f : A_1 \to B\$$ to the pair of functions \$$(f : A_1 \to B, ! : A_2 \to \\{\bullet\\})\$$, where \$$!\$$ denotes the unique function from any set to the singleton set \$$\\{\bullet\\}\$$. Hence, \begin{align} R(B) &= (B, \{\bullet\}) \\\\ R(f) &= (f, !). \end{align}
• Options
6.
edited June 2018

Dan: great!

So, if $$F : \mathbf{Set}^2 \to \mathbf{Set}$$ is a functor that discards the second set (or function), then a right adjoint $$R : \mathbf{Set} \to \mathbf{Set}^2$$ attempts to restore this second set by choosing a one-element set. Category theorists like to use "1" to mean any one-element set, so I'd say

$$R(S) = (S, 1)$$ for any set $$S$$ and

$$R(f) = (f, 1_1)$$ for any function $$f: S \to T$$, where $$1_1$$ means the identity function from our one-element set to itself.

This should be compared to something I said in the lecture: if $$F: \mathbf{Set} \to \mathbf{1}$$ is the functor that discards a set, a right adjoint of this functor attempts to restore this set by choosing a one-element set.

By the way, this isn't quite true:

A functor $$F$$ is left adjoint to a functor $$R$$ if there is a one-to-one correspondence between the morphisms $$F(A) \to B$$ and morphisms $$A \to R(B)$$:

$$\mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B))$$

What's true is this:

A functor $$F$$ is left adjoint to a functor $$R$$ if and only if there is a natural one-to-one correspondence between the morphisms $$F(A) \to B$$ and morphisms $$A \to R(B)$$:

$$\mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B)) .$$ This is where natural transformations come into the game. But I am deliberately urging people not to worry about this too much when they're just getting started with adjoints.

What you really used, I think, is this true fact, a consequence of the other one I just stated:

If a functor $$F$$ is left adjoint to a functor $$R$$, then there is a one-to-one correspondence between the morphisms $$F(A) \to B$$ and morphisms $$A \to R(B)$$:

$$\mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B))$$ Using this you can figure out some stuff about what a left adjoint must be like if it exists. In many cases, if you get a nice answer this way, the left adjoint will then exist, because your one-to-one correspondence will turn out to be natural.

Comment Source:Dan: great! So, if \$$F : \mathbf{Set}^2 \to \mathbf{Set}\$$ is a functor that discards the second set (or function), then a right adjoint \$$R : \mathbf{Set} \to \mathbf{Set}^2\$$ attempts to restore this second set by choosing a _one-element set_. Category theorists like to use "1" to mean any one-element set, so I'd say $R(S) = (S, 1)$ for any set \$$S\$$ and $R(f) = (f, 1_1)$ for any function \$$f: S \to T\$$, where \$$1_1\$$ means the identity function from our one-element set to itself. (I'm just restating your answer in different language, to give people a second chance to read it and think about it.) This should be compared to something I said in the lecture: if \$$F: \mathbf{Set} \to \mathbf{1}\$$ is the functor that discards a set, a right adjoint of this functor attempts to restore this set by choosing a one-element set. By the way, this isn't quite true: > A functor \$$F\$$ is left adjoint to a functor \$$R\$$ if there is a one-to-one correspondence between the morphisms \$$F(A) \to B\$$ and morphisms \$$A \to R(B)\$$: > $\mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B))$ What's true is this: A functor \$$F\$$ is left adjoint to a functor \$$R\$$ if and only if there is a _natural_ one-to-one correspondence between the morphisms \$$F(A) \to B\$$ and morphisms \$$A \to R(B)\$$: $\mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B)) .$ This is where natural transformations come into the game. But I am deliberately urging people not to worry about this too much when they're just getting started with adjoints. What you really used, I think, is this true fact, a consequence of the other one I just stated: If a functor \$$F\$$ is left adjoint to a functor \$$R\$$, then there is a one-to-one correspondence between the morphisms \$$F(A) \to B\$$ and morphisms \$$A \to R(B)\$$: $\mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B))$ Using this you can figure out some stuff about what a left adjoint must be like if it exists. In many cases, if you get a nice answer this way, the left adjoint will then exist, because your one-to-one correspondence will turn out to be natural. 
• Options
7.
edited June 2018

Jesus wrote:

Perhaps an ill posed question, but here it goes anyway. Does the notion of adjunction in the case of monoids reduce to something formerly known?

Do you mean if we treat monoids as categories with one object, and ask what adjunctions between categories reduce to in this special case? That's a fun question!

If that's your question, I think the best thing is to just figure out the answer! We can do it.

Matthew wrote:

In the event of a lattice equipped with a monoid with two adjoints for $$\bullet$$, we call such a structure a residuated lattice.

You are interpreting Jesus' question in a different way than I did. You are considering a monoidal poset of a nice sort, and treating this poset $$A$$ as a category, and asking about adjoints to the operations of left and/or right multiplication by a fixed element $$a \in A$$:

$x \mapsto a \bullet x$

and

$x \mapsto x \bullet a$

where $$\bullet$$ is the multiplication in $$A$$.

This is a much more elaborate situation, with a lot more moving parts... but also very fun, and yes, it leads us into the world of logic, and quantales.

Comment Source:Jesus wrote: > Perhaps an ill posed question, but here it goes anyway. Does the notion of adjunction in the case of monoids reduce to something formerly known? Do you mean if we treat monoids as categories with one object, and ask what adjunctions between categories reduce to in this special case? That's a fun question! If that's your question, I think the best thing is to just figure out the answer! We can do it. Matthew wrote: > In the event of a lattice equipped with a monoid with two adjoints for \$$\bullet\$$, we call such a structure a [*residuated lattice*](https://en.wikipedia.org/wiki/Residuated_lattice#Definition). You are interpreting Jesus' question in a different way than I did. You are considering a monoidal poset of a nice sort, and treating this _poset_ \$$A\$$ as a category, and asking about adjoints to the operations of left and/or right multiplication by a fixed element \$$a \in A\$$: \$x \mapsto a \bullet x\$ and \$x \mapsto x \bullet a\$ where \$$\bullet\$$ is the multiplication in \$$A\$$. This is a much more elaborate situation, with a lot more moving parts... but also very fun, and yes, it leads us into the world of logic, and quantales. 
• Options
8.
edited June 2018

Thanks John and Matthew, I was thinking more in what John describes, but will give more thought to what you Matthew wrote.

Do you mean if we treat monoids as categories with one object, and ask what adjunctions between categories reduce to in this special case?

Yes

Comment Source:Thanks John and Matthew, I was thinking more in what John describes, but will give more thought to what you Matthew wrote. > Do you mean if we treat monoids as categories with one object, and ask what adjunctions between categories reduce to in this special case? Yes
• Options
9.
edited June 2018

Okay, let's do it, Jesus! Say we have monoids $$M$$ and $$N$$, which we think of as categories, each with one object which I'll call $$\star$$... we're smart enough to keep track of when we're talking about $$\star_M$$ and $$\star_N$$, I hope. Let's figure out what adjoint functors

$F: M \to N, \quad G: N \to M$

amount to!

First of all, can you (or anyone) guess what a functor $$F: M \to N$$ amounts to in this case? It's something familiar from the study of monoids.

Comment Source:Okay, let's do it, Jesus! Say we have monoids \$$M\$$ and \$$N\$$, which we think of as categories, each with one object which I'll call \$$\star\$$... we're smart enough to keep track of when we're talking about \$$\star_M\$$ and \$$\star_N\$$, I hope. Let's figure out what adjoint functors \$F: M \to N, \quad G: N \to M \$ amount to! First of all, can you (or anyone) guess what a functor \$$F: M \to N\$$ amounts to in this case? It's something familiar from the study of monoids. 
• Options
10.

I should warn people that for some of Puzzles 150-153, the answer is "there are no such adjoints". But don't be scared: this answer will arise as you use the definition of adjoint to figure out what the desired adjoint is like!

You can almost always work out the adjoints of a functor, or the lack of adjoints, by patiently using the definition of adjoint functors.

Comment Source:I should warn people that for some of Puzzles 150-153, the answer is "there are no such adjoints". But don't be scared: this answer will arise as you use the definition of adjoint to figure out what the desired adjoint is like! You can almost always work out the adjoints of a functor, or the lack of adjoints, by patiently using the definition of adjoint functors. 
• Options
11.

John Baez wrote:

First of all, can you (or anyone) guess what a functor $$F: M \to N$$ amounts to in this case? It's something familiar from the study of monoids.

I believe that's just a monoid homomorphism.

Comment Source:John Baez wrote: >First of all, can you (or anyone) guess what a functor \$$F: M \to N\$$ amounts to in this case? It's something familiar from the study of monoids. I believe that's just a monoid homomorphism.
• Options
12.

John – thank you for comment 6 which points out my imprecision! I hope by the end of the course I'll learn to be more careful with my statements :-)

Comment Source:John – thank you for [comment 6](https://forum.azimuthproject.org/discussion/comment/19578/#Comment_19578) which points out my imprecision! I hope by the end of the course I'll learn to be more careful with my statements :-)
• Options
13.
edited June 2018

A functor between monoids should be a homomorphism (or, more precisely, it should send the only object of one monoid to the only object of the other and send the arrows/elements so as to respect the homomorphism conditions). P.S. I just saw that Keith beat me to it.

Comment Source:A functor between monoids should be a homomorphism (or, more precisely, it should send the only object of one monoid to the only object of the other and send the arrows/elements so as to respect the homomorphism conditions). P.S. I just saw that Keith beat me to it.
• Options
14.

That's good then, it means my answer is on the right track.

Comment Source:That's good then, it means my answer is on the right track.
• Options
15.

Dan wrote:

I hope by the end of the course I'll learn to be more careful with my statements :-)

You're pretty good! One of my duties as a teacher, I figure, is to nitpick, in a friendly way, and point out little subtleties.

Comment Source:Dan wrote: > I hope by the end of the course I'll learn to be more careful with my statements :-) You're pretty good! One of my duties as a teacher, I figure, is to nitpick, in a friendly way, and point out little subtleties.
• Options
16.
edited June 2018

Keith and Valter - yes, a functor between one-object categories is the same as a monoid homomorphism!

So, when we have two functors, a left adjoint $$F : M \to N$$ and a right adjoint $$G: N \to N$$, between one-object categories, we should think of these as monoid homomorphisms.

But then there's more! We need a natural one-to-one correspondence between morphisms

$$f: F(m) \to n$$ and morphisms

$$g: m \to G(n)$$ for every pair of objects $$m \in \mathbf{Ob}(M), n \in \mathbf{Ob}(N).$$ That's what makes $$F$$ and $$G$$ adjoints.

But this requirement can be simplified, since we're dealing with monoids. How?

(At first let's not worry about the naturality requirement; that's very important, but it can be put off to the end.)

Comment Source:Keith and Valter - yes, a functor between one-object categories is the same as a monoid homomorphism! So, when we have two functors, a left adjoint \$$F : M \to N\$$ and a right adjoint \$$G: N \to N\$$, between one-object categories, we should think of these as monoid homomorphisms. But then there's more! We need a natural one-to-one correspondence between morphisms $f: F(m) \to n$ and morphisms $g: m \to G(n)$ for every pair of objects \$$m \in \mathbf{Ob}(M), n \in \mathbf{Ob}(N).\$$ That's what makes \$$F\$$ and \$$G\$$ adjoints. But this requirement can be simplified, since we're dealing with monoids. How? (At first let's not worry about the _naturality_ requirement; that's very important, but it can be put off to the end.) 
• Options
17.

$$M$$ and $$N$$ are one-object categories, so we must have $$m = {\star}_M$$ and $$n = {\star}_N$$

Hence we need a bijection between morphisms $$f : F({\star}_M) \rightarrow {\star}_N$$ and morphisms $$g : {\star}_M \rightarrow G({\star}_N)$$

ie, a bijection between elements of $$N$$ and elements of $$M$$

(Now we just have to worry about naturality, at which point I get kinda stuck. I think I can prove that if this bijection is a homomorphism, then it equals $$g$$, and $$f = g^{-1}$$. But I don't believe we can rule out the case where the bijection isn't a homomorphism.)

Comment Source:\$$M\$$ and \$$N\$$ are one-object categories, so we must have \$$m = {\star}_M\$$ and \$$n = {\star}_N\$$ Hence we need a bijection between morphisms \$$f : F({\star}_M) \rightarrow {\star}_N\$$ and morphisms \$$g : {\star}_M \rightarrow G({\star}_N)\$$ ie, a bijection between elements of \$$N\$$ and elements of \$$M\$$ (Now we just have to worry about naturality, at which point I get kinda stuck. I think I can prove that if this bijection is a homomorphism, then it equals \$$g\$$, and \$$f = g^{-1}\$$. But I don't believe we can rule out the case where the bijection _isn't_ a homomorphism.)
• Options
18.
edited June 2018

I agree with Anindya that the hom-sets are the monoids here, but can't go further either.

Comment Source:I agree with Anindya that the hom-sets are the monoids here, but can't go further either.
• Options
19.

A morphism $$f:F(m)\rightarrow n$$ is an element $$f \in N$$ such that $$f\circ_N F(m) = n$$. Likewise, a morphism $$g:m\rightarrow G(n)$$ is an element $$g \in M$$ such that $$g\circ_M m = G(n)$$.

Thus we have $$g \circ_M m = G(n) = G(f \circ_N F(m)) = G(f) \circ_M G(F(m))$$. If we consider the case $$m = 1_M$$ and recall that $$G(F(1_M))=G(1_N)=1_M$$, this becomes $$g=G(f)$$ so we have the mapping in one direction.

In the other direction, using $$g=G(f)$$ and $$n = 1_N$$, I get the condition $$g \circ_M G(F(m)) = g \circ_M m = 1_M$$, i.e., for any $$m \in M$$ we must have $$g \circ_M m = 1_M$$ if and only if $$g \circ_M G(F(m))$$. So for all elements $$m \in M$$ that have a left-inverse $$g$$, this must also be the left-inverse of $$G(F(m))$$. This would surely work if $$G$$ was a left-inverse of $$F$$, but it is not clear to me that it is necessary - nor what monoid concept this corresponds to.

Comment Source:A morphism \$$f:F(m)\rightarrow n\$$ is an element \$$f \in N\$$ such that \$$f\circ_N F(m) = n\$$. Likewise, a morphism \$$g:m\rightarrow G(n)\$$ is an element \$$g \in M\$$ such that \$$g\circ_M m = G(n)\$$. Thus we have \$$g \circ_M m = G(n) = G(f \circ_N F(m)) = G(f) \circ_M G(F(m))\$$. If we consider the case \$$m = 1_M\$$ and recall that \$$G(F(1_M))=G(1_N)=1_M\$$, this becomes \$$g=G(f)\$$ so we have the mapping in one direction. In the other direction, using \$$g=G(f)\$$ and \$$n = 1_N\$$, I get the condition \$$g \circ_M G(F(m)) = g \circ_M m = 1_M\$$, i.e., for any \$$m \in M\$$ we must have \$$g \circ_M m = 1_M\$$ if and only if \$$g \circ_M G(F(m)) \$$. So for all elements \$$m \in M\$$ that have a left-inverse \$$g\$$, this must also be the left-inverse of \$$G(F(m))\$$. This would surely work if \$$G\$$ was a left-inverse of \$$F\$$, but it is not clear to me that it is necessary - nor what monoid concept this corresponds to.
• Options
20.

from what I can make out, if $$\phi$$ is our bijection from $$N$$ to $$M$$, the naturality condition amounts to saying that for all $$m\in M$$ and all $$y, n\in N$$ we have $$\phi(n\circ y\circ F(m)) = G(n)\circ\phi(y)\circ m$$ – but I'm not sure what that entails!

Comment Source:from what I can make out, if \$$\phi\$$ is our bijection from \$$N\$$ to \$$M\$$, the naturality condition amounts to saying that for all \$$m\in M\$$ and all \$$y, n\in N\$$ we have \$$\phi(n\circ y\circ F(m)) = G(n)\circ\phi(y)\circ m\$$ – but I'm not sure what that entails!
• Options
21.
edited June 2018

Nobody with a solution to puzzle 150/151:? I was trying to solve it but wasn't yet successfull so far. We need bijections $$[S \times T,X] \rightarrow [(S,T),G(X)]$$ and $$[G(X),(S,T)] \rightarrow [X,S \times T]$$ for a certain functor $$G: \textbf{Set} \rightarrow \textbf{Set}^2$$.

I think we can use the universal property of the product: If I choose a set X together with maps $$h_S: X \rightarrow S$$ and $$h_T: X \rightarrow T$$ then there is exactly one $$h: X \rightarrow S \times T$$ such that $$h_S = pr_S \circ h$$ and $$h_T = pr_T \circ h$$ whereafter the $$pr_i$$ are the respective projection from the product to its components.

If I choose any X and maps to S and T then $$G(X):=(X_1,X_2)$$ with $$X_1 := h_S^{-1}(S)$$, $$X_2 := h_T^{-1}(T)$$. There is exactly one $$h: X \rightarrow S \times T$$ that maps under G to $$h'=(h'_1,h'_2): G(X) \rightarrow (S,T)$$. So there are so much left adjoints of SxT as there are pairs of maps from X to S and T, for every X. ??? A lot of adjoints....

In the other direction, if we start with a pair $$(X_1,X_2)$$ in $$\textbf{Set}^2$$ and a map $$(h'_1,h'_2)$$, we get a map $$F((h'_1,h'_2)): X_1 \times X_2 \rightarrow S \times T$$ as F is contravariant. But we cannot find the $$h_1$$ and $$h_2$$ which belongs to $$F((h'_1,h'_2))$$. So I guess that there is no right adjoints. But I'm not sure...

Comment Source:Nobody with a solution to <b>puzzle 150/151:</b>? I was trying to solve it but wasn't yet successfull so far. We need bijections \$$[S \times T,X] \rightarrow [(S,T),G(X)] \$$ and \$$[G(X),(S,T)] \rightarrow [X,S \times T] \$$ for a certain functor \$$G: \textbf{Set} \rightarrow \textbf{Set}^2 \$$. I think we can use the universal property of the product: If I choose a set X together with maps \$$h_S: X \rightarrow S \$$ and \$$h_T: X \rightarrow T \$$ then there is exactly one \$$h: X \rightarrow S \times T \$$ such that \$$h_S = pr_S \circ h \$$ and \$$h_T = pr_T \circ h \$$ whereafter the \$$pr_i \$$ are the respective projection from the product to its components. If I choose any X and maps to S and T then \$$G(X):=(X_1,X_2) \$$ with \$$X_1 := h_S^{-1}(S) \$$, \$$X_2 := h_T^{-1}(T) \$$. There is exactly one \$$h: X \rightarrow S \times T \$$ that maps under G to \$$h'=(h'_1,h'_2): G(X) \rightarrow (S,T) \$$. So there are so much left adjoints of SxT as there are pairs of maps from X to S and T, for every X. ??? A lot of adjoints.... In the other direction, if we start with a pair \$$(X_1,X_2) \$$ in \$$\textbf{Set}^2 \$$ and a map \$$(h'_1,h'_2) \$$, we get a map \$$F((h'_1,h'_2)): X_1 \times X_2 \rightarrow S \times T \$$ as F is contravariant. But we cannot find the \$$h_1 \$$ and \$$h_2 \$$ which belongs to \$$F((h'_1,h'_2)) \$$. So I guess that there is no right adjoints. But I'm not sure... ![](https://svgshare.com/i/7AE.svg)
• Options
22.

I though I had managed to prove that the right adjoint was trying to go between sets of different cardnalities, but I had misremembered my algebra.

Comment Source:I though I had managed to prove that the right adjoint was trying to go between sets of different cardnalities, but I had misremembered my algebra.
• Options
23.
edited June 2018

Anindya wrote:

from what I can make out, if $$\phi$$ is our bijection from $$N$$ to $$M$$, the naturality condition amounts to saying that for all $$m\in M$$ and all $$y, n\in N$$ we have $$\phi(n\circ y\circ F(m)) = G(n)\circ\phi(y)\circ m$$ – but I'm not sure what that entails!

I think you're on the right track! If we set $$m=1_M$$ and $$y=1_N$$, we get the equation $$\phi(n) = G(n)\circ \phi(1_N),$$ so $$\phi$$ is completely determined by where it sends $$1_N$$. It must be an invertible element of $$M$$ for $$\phi$$ to be a bijection, so let's call that element $$u_0$$. Then we know that $$\phi(n) = G(n)\circ u_0$$, and the equation $$\phi(n\circ y\circ F(m)) = G(n)\circ\phi(y)\circ m$$ becomes $$G(n\circ y\circ F(m)) \circ u_0= G(n)\circ (G(y)\circ u_0) \circ m.$$ This is true for all $$n,y,m$$ if and only if for all $$m\in M$$ we have $$G(F(m))\circ u_0 = u_0 \circ m$$ or $$G(F(m)) = u_0 \circ m \circ u_0^{-1}.$$ So this means that $$F$$ and $$G$$ are bijections (and therefore isomorphisms) and they compose to conjugation by a unit. That does seem like a slightly more general version of being inverses!

Edit: I've just realized that all we can deduce from the fact that $$\phi(n) = G(n)\circ u_0$$ is a bijection is that $$u_0$$ has a left inverse: $$G$$ of whatever $$\phi$$ sends to $$1_M$$. That makes the very last step to $$u_0 \circ m \circ u_0^{-1}$$ not work. (I'm too used to working with groups and commutative monoids!) But we can say instead that if $$l_0$$ is a left inverse to $$u_0$$, then $$l_0 \circ G(F(m)) \circ u_0 = m.$$ So whatever composite $$G\circ F$$ is, it has to sort-of-conjugate to the identity.

Comment Source:Anindya wrote: > from what I can make out, if \$$\phi\$$ is our bijection from \$$N\$$ to \$$M\$$, the naturality condition amounts to saying that for all \$$m\in M\$$ and all \$$y, n\in N\$$ we have \$$\phi(n\circ y\circ F(m)) = G(n)\circ\phi(y)\circ m\$$ – but I'm not sure what that entails! I think you're on the right track! If we set \$$m=1_M\$$ and \$$y=1_N\$$, we get the equation $\phi(n) = G(n)\circ \phi(1_N),$ so \$$\phi\$$ is completely determined by where it sends \$$1_N\$$. *It must be an invertible element of \$$M\$$ for \$$\phi\$$ to be a bijection, so* let's call that element \$$u_0\$$. Then we know that \$$\phi(n) = G(n)\circ u_0\$$, and the equation \$$\phi(n\circ y\circ F(m)) = G(n)\circ\phi(y)\circ m\$$ becomes $G(n\circ y\circ F(m)) \circ u_0= G(n)\circ (G(y)\circ u_0) \circ m.$ This is true for all \$$n,y,m\$$ if and only if for all \$$m\in M\$$ we have $G(F(m))\circ u_0 = u_0 \circ m$ *or* $G(F(m)) = u_0 \circ m \circ u_0^{-1}.$ *So this means that \$$F\$$ and \$$G\$$ are bijections (and therefore isomorphisms) and they compose to conjugation by a unit. That does seem like a slightly more general version of being inverses!* **Edit:** I've just realized that all we can deduce from the fact that \$$\phi(n) = G(n)\circ u_0\$$ is a bijection is that \$$u_0\$$ has a *left* inverse: \$$G\$$ of whatever \$$\phi\$$ sends to \$$1_M\$$. That makes the very last step to \$$u_0 \circ m \circ u_0^{-1}\$$ not work. (I'm too used to working with groups and commutative monoids!) But we *can* say instead that if \$$l_0\$$ is a left inverse to \$$u_0\$$, then $l_0 \circ G(F(m)) \circ u_0 = m.$ So whatever composite \$$G\circ F\$$ is, it has to sort-of-conjugate to the identity.
• Options
24.
edited June 2018

Peter wrote:

Nobody with a solution to Puzzle 150/151? I was trying to solve it but wasn't yet successful so far. We need bijections $$[S \times T,X] \rightarrow [(S,T),G(X)]$$ and $$[G(X),(S,T)] \rightarrow [X,S \times T]$$ for a certain functor $$G: \textbf{Set} \rightarrow \textbf{Set}^2$$.

I'm glad you're trying these. I'm a little worried: it looks here like you're trying to make the same functor $$G$$ into both the left and right adjoint. That won't work. Maybe you're not really wanting this, but I think other readers will be less confused if you said:

For $$G: \textbf{Set} \rightarrow \textbf{Set}^2$$ to be a right adjoint of $$\times: \textbf{Set}^2 \to \textbf{Set}$$, we need a bijection $$[S \times T,X] \rightarrow [(S,T),G(X)]$$. For $$H$$ to be a left adjoint of $$\times: \textbf{Set}^2 \to \textbf{Set}$$, we need a bijection $$[H(X),(S,T)] \rightarrow [X,S \times T]$$...

... where, clearly, you are using square brackets $$[A,B]$$ to denote the set of morphisms from an object $$A$$ to an object $$B$$.

I will now read further into your comment: I just had to get this off my chest.

But I have a hint that may be helpful. In my formulation of the problem, only one of $$G$$ and $$H$$ will actually exist! The functor $$\times: \textbf{Set}^2 \to \textbf{Set}$$ does not have both a left and right adjoint. It has just one of these.

Comment Source:Peter wrote: > Nobody with a solution to <b>Puzzle 150/151</b>? I was trying to solve it but wasn't yet successful so far. We need bijections \$$[S \times T,X] \rightarrow [(S,T),G(X)] \$$ and \$$[G(X),(S,T)] \rightarrow [X,S \times T] \$$ for a certain functor \$$G: \textbf{Set} \rightarrow \textbf{Set}^2 \$$. I'm glad you're trying these. I'm a little worried: it looks here like you're trying to make the same functor \$$G\$$ into both the left and right adjoint. That won't work. Maybe you're not really wanting this, but I think other readers will be less confused if you said: > For \$$G: \textbf{Set} \rightarrow \textbf{Set}^2\$$ to be a right adjoint of \$$\times: \textbf{Set}^2 \to \textbf{Set}\$$, we need a bijection \$$[S \times T,X] \rightarrow [(S,T),G(X)] \$$. For \$$H\$$ to be a left adjoint of \$$\times: \textbf{Set}^2 \to \textbf{Set}\$$, we need a bijection \$$[H(X),(S,T)] \rightarrow [X,S \times T] \$$... ... where, clearly, you are using square brackets \$$[A,B]\$$ to denote the set of morphisms from an object \$$A\$$ to an object \$$B\$$. I will now read further into your comment: I just had to get this off my chest. But I have a hint that may be helpful. In my formulation of the problem, only _one_ of \$$G\$$ and \$$H\$$ will actually _exist!_ The functor \$$\times: \textbf{Set}^2 \to \textbf{Set}\$$ does not have both a left and right adjoint. It has just one of these. 
• Options
25.
edited June 2018

In $$R\vdash L$$ we need a natural bijection between $$[L(a),b] \leftrightarrow [a,R(b)]$$.

Okay so looking at the bijection's two sides for the left adjoint of $$\times$$ $[L(X),(S,T)] \leftrightarrow [X,S \times T]$, which requires $|S|^{|L_0(X)|} \ast |T|^{|L_1(X)|} = (|S|\ast|T|)^{|X|}$

Setting those equal and using that it has to work for all $$X$$ , we get $$L_0(X)=L_1(X)=X$$.

So L must be equivalent to the diagonal/duplication/$$\delta$$ and I am going to take Johns advice and just assume that if I make L be $$\delta$$ then it ends up being natural. (My in my head calculations/gut feel says its plausible at least)

so the bijection has type [(X,X),(S,T)] <-> [X,S x T], and the forward direction is basically just the pair (term) constructor, so yeah it's natural and a bijection. (I know this because of some programming language theory stuff and a little bit of hand waving. This is an engineers approach not a mathematician's)

Now I have a hunch here. Let's see if it's also the right adjoint of + !

[S+T,X] <-> [(S,T),(X,X)] is the type of the bijection we need. And yeah, the back direction is (basically) just definition by cases which better be a natural bijection. (for the same reasons as the above.)

We can use some algebra and prove that the right adjoint must be equivalent to $$\delta$$ the same way as above. So this forms an adjoint triplet $$\times \vdash \delta \vdash +$$, which I think we saw forall, there exists, and ..something forming back when we were working with preorders.

Comment Source:In \$$R\vdash L\$$ we need a natural bijection between \$$$L(a),b$ \leftrightarrow $a,R(b)$\$$. Okay so looking at the bijection's two sides for the left adjoint of \$$\times\$$ \$[L(X),(S,T)] \leftrightarrow [X,S \times T]\$, which requires \$|S|^{|L_0(X)|} \ast |T|^{|L_1(X)|} = (|S|\ast|T|)^{|X|}\$ Setting those equal and using that it has to work for all \$$X\$$ , we get \$$L_0(X)=L_1(X)=X\$$. So L must be equivalent to the diagonal/duplication/\$$\delta\$$ and I am going to take Johns advice and just assume that if I make L _be_ \$$\delta\$$ then it ends up being natural. (My in my head calculations/gut feel says its plausible at least) so the bijection has type [(X,X),(S,T)] <-> [X,S x T], and the forward direction is basically just the pair (term) constructor, so yeah it's natural and a bijection. (I know this because of some programming language theory stuff and a little bit of hand waving. This is an engineers approach not a mathematician's) Now I have a hunch here. Let's see if it's also the *right* adjoint of *+* ! [S+T,X] <-> [(S,T),(X,X)] is the type of the bijection we need. And yeah, the back direction is (basically) just definition by cases which better be a natural bijection. (for the same reasons as the above.) We can use some algebra and prove that the right adjoint must be equivalent to \$$\delta\$$ the same way as above. So this forms an adjoint triplet \$$\times \vdash \delta \vdash +\$$, which I think we saw forall, there exists, and ..something forming back when we were working with preorders. 
• Options
26.

@christopher Oh I see! Of course L(X)=(X,X) as a left adjoint makes sense. Thank you for ýour comments.

Comment Source:@christopher Oh I see! Of course L(X)=(X,X) as a left adjoint makes sense. Thank you for ýour comments.
• Options
27.

@Owen – yeah, that's pretty much as far as I got.

Specifically, we can pick $$p\in M$$ and $$q\in N$$ such that $$\phi(1_N) = p$$ and $$\phi(q) = 1_M$$.

We then get $$\phi = n \mapsto G(n)\circ p$$ and $$\phi^{-1} = m \mapsto q\circ F(m)$$.

This implies that $$F$$ and $$G$$ are both injective, also that $$m \mapsto m\circ p$$ and $$n \mapsto q\circ n$$ are both surjective.

But I'm not sure there's much more we can say.

Comment Source:@Owen – yeah, that's pretty much as far as I got. Specifically, we can pick \$$p\in M\$$ and \$$q\in N\$$ such that \$$\phi(1_N) = p\$$ and \$$\phi(q) = 1_M\$$. We then get \$$\phi = n \mapsto G(n)\circ p\$$ and \$$\phi^{-1} = m \mapsto q\circ F(m)\$$. This implies that \$$F\$$ and \$$G\$$ are both injective, also that \$$m \mapsto m\circ p\$$ and \$$n \mapsto q\circ n\$$ are both surjective. But I'm not sure there's much more we can say. 
• Options
28.

Is an adjunction of monoids just an isomorphism between them?

Comment Source:Is an adjunction of monoids just an isomorphism between them?
• Options
29.

An isomorphism of monoids is certainly an adjunction between them – but I don't think the converse is true.

Comment Source:An isomorphism of monoids is certainly an adjunction between them – but I don't think the converse is true. 
• Options
30.
edited June 2018

Taking the concrete example of

$\mathbb{Z}_2[n \mod 2, b] \leftrightarrow \mathbb{N}[n, \iota(b)],$

we get,

$\begin{matrix} \mathbb{N} & \overset{- \mod 2}\rightarrow & \mathbb{Z}_2\\ s\downarrow & & \downarrow \neg\\ \mathbb{N} & \underset{\iota}\leftarrow & \mathbb{Z}_2 \end{matrix}$

The induced functor $$\iota \circ \neg \circ (- \mod n)$$ then sends even numbers to $$0$$ in $$\mathbb{N}$$ and odd numbers to $$1$$ in $$\mathbb{N}$$. This functor is a monad.

Comment Source:Taking the concrete example of \$\mathbb{Z}_2[n \mod 2, b] \leftrightarrow \mathbb{N}[n, \iota(b)], \$ we get, \$\begin{matrix} \mathbb{N} & \overset{- \mod 2}\rightarrow & \mathbb{Z}\_2\\\\ s\downarrow & & \downarrow \neg\\\\ \mathbb{N} & \underset{\iota}\leftarrow & \mathbb{Z}\_2 \end{matrix} \$ The induced functor \$$\iota \circ \neg \circ (- \mod n) \$$ then sends even numbers to \$$0\$$ in \$$\mathbb{N}\$$ and odd numbers to \$$1\$$ in \$$\mathbb{N}\$$. This functor is a monad. 
• Options
31.

@Christopher In your comment what do you mean with $$|S|$$ and with the star operation $$|S| * |T|$$ ?

Comment Source:@Christopher In your comment what do you mean with \$$|S| \$$ and with the star operation \$$|S| * |T| \$$ ?
• Options
32.
edited June 2018

He may mean cardinal multiplication.

What he says is clearly valid for finite categories and should also be true for infinite categories.

Comment Source:@ [Peter](https://forum.azimuthproject.org/discussion/comment/19654/#Comment_19654) He may mean [cardinal multiplication](https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_multiplication). What he says is clearly valid for finite categories and should also be true for infinite categories.
• Options
33.

Found this paper on arxiv.org – it's from 2005 by Vladimir Molotkov, and proves that there are adjunctions between monoids that aren't isomorphisms. Apparently this was an open question until then! https://arxiv.org/pdf/math/0504060.pdf

Comment Source:Found this paper on arxiv.org – it's from 2005 by Vladimir Molotkov, and proves that there are adjunctions between monoids that aren't isomorphisms. Apparently this was an open question until then! https://arxiv.org/pdf/math/0504060.pdf
• Options
34.

That's an interesting finding for me, Anindya, food for thought.

Comment Source:That's an interesting finding for me, Anindya, food for thought.
• Options
35.
edited July 2018

@John wrote in the lecture:

In this situation we also say $$F$$ is a left adjoint of $$G$$.

I think this should be right adjoint?

Edit: Read this incorrectly. Sorry no typo...

Comment Source:@John wrote in the lecture: >In this situation we also say \$$F\$$ is a **left adjoint** of \$$G\$$. I think this should be right adjoint? Edit: Read this incorrectly. Sorry no typo...
• Options
36.

As Christopher wrote in his answers to the puzzles, $$\times \vdash \delta \vdash +$$ should make sense but I am having trouble with these puzzles still...

I am trying these puzzles and its looks to me that the adjoints for both $$\times$$ and + would be the same and not in opposite directions since they are both pairs becoming single entities in objects. What is the difference between the two that is causing the switch?

Comment Source:As Christopher wrote in his answers to the puzzles, \$$\times \vdash \delta \vdash +\$$ should make sense but I am having trouble with these puzzles still... I am trying these puzzles and its looks to me that the adjoints for both \$$\times\$$ and + would be the same and not in opposite directions since they are both pairs becoming single entities in objects. What is the difference between the two that is causing the switch?
• Options
37.
edited July 2018

The thing is that a product $$\times$$ isn't just a single object, it's a single object and a pair of projection maps from that object to the original pair.

Dually the coproduct $$+$$ is a single object and a pair of inclusion maps from the original pair to that object.

It's the fact that these projection/inclusion maps point in different directions that makes one a right adjoint and the other a left adjoint.

Comment Source:The thing is that a product \$$\times\$$ isn't just a single object, it's a single object _and a pair of projection maps from that object to the original pair_. Dually the coproduct \$$+\$$ is a single object _and a pair of inclusion maps from the original pair to that object_. It's the fact that these projection/inclusion maps point in different directions that makes one a right adjoint and the other a left adjoint. 
• Options
38.
edited July 2018

Think of it this way, the functor $$\Delta$$ is a category copying machine. It takes a category and copies or duplicates it.

$\Delta: \mathcal{C} \to \mathcal{C} \times \mathcal{C},$

now ask yourself, "what is a left- and right-adjoint to copying something?"

Comment Source:Think of it this way, the functor \$$\Delta\$$ is a category copying machine. It takes a category and copies or *duplicates* it. \$\Delta: \mathcal{C} \to \mathcal{C} \times \mathcal{C}, \$ now ask yourself, "what is a left- and right-adjoint to copying something?"
• Options
39.

Thanks Anindya and Keith for you answers and tips.

I think I have managed to figure out why the direction changes; it basically comes down to the fact that the way you count direct sums and products is different. So Christopher pointed out that for products :

$|S|^{|L_0(X)|} \ast |T|^{|L_1(X)|} = (|S|\ast|T|)^{|X|}$ must be true so therefore $|S|^{X|} \ast |T|^{|X|} = (|S|\ast|T|)^{|X|}$.

But for sums we need $$|S|+|T|$$ and not $$|S|\ast|T|$$. So the arrows need to be switched around so that everything counts up :

$|X|^{|S|} \ast |X|^{|T|} = |X|^{||S|+|T||}$

Now that I have that straight I am still trying to figure out Anindya's answer... Why is there a projection map for products and inclusion map for coproducts?

Comment Source:Thanks Anindya and Keith for you answers and tips. I think I have managed to figure out why the direction changes; it basically comes down to the fact that the way you count direct sums and products is different. So Christopher pointed out that for products : \$|S|^{|L_0(X)|} \ast |T|^{|L_1(X)|} = (|S|\ast|T|)^{|X|}\$ must be true so therefore \$|S|^{X|} \ast |T|^{|X|} = (|S|\ast|T|)^{|X|}\$. But for sums we need \$$|S|+|T|\$$ and not \$$|S|\ast|T|\$$. So the arrows need to be switched around so that everything counts up : \$|X|^{|S|} \ast |X|^{|T|} = |X|^{||S|+|T||}\$ Now that I have that straight I am still trying to figure out Anindya's answer... Why is there a projection map for products and inclusion map for coproducts?
• Options
40.
edited July 2018

Keith wrote:

Think of it this way, the functor $$\Delta$$ is a category copying machine. It takes a category and copies or duplicates it.

$\Delta: \mathcal{C} \to \mathcal{C} \times \mathcal{C},$

Level slip? I wouldn't say this $$\Delta$$ is duplicating the category. I'd say it's duplicating each object in the category. Let's call it $$\Delta_{\mathcal{C}}$$. It has

$$\Delta_{\mathcal{C}}(c) = (c,c)$$ for each object $$c$$ in $$\mathcal{C}$$.

Of course, there is also something that's duplicating the category $$\mathcal{C}$$. There's a functor

$$\Delta_{\mathbf{Cat}} : \mathbf{Cat} \to \mathbf{Cat} \times \mathbf{Cat}$$ and we have

$$\Delta_{\mathbf{Cat}}(\mathcal{C}) = (\mathcal{C}, \mathcal{C})$$ There's also a functor

$$\times_{\mathbf{Cat}} : \mathbf{Cat} \times \mathbf{Cat} \to \mathbf{Cat}$$ that takes the product of any two categories:

$$\times_{\mathbf{Cat}} (\mathcal{C}, \mathcal{D}) = \mathcal{C} \times \mathcal{D} .$$ We can compose these two functors and get a functor

$$\times_{\mathbf{Cat}} \circ \Delta_{\mathbf{Cat}} : \mathbf{Cat} \to \mathbf{Cat}$$ which does this:

$$\times_{\mathbf{Cat}} \circ \Delta_{\mathbf{Cat}} (\mathcal{C}) = \mathcal{C} \times \mathcal{C}$$ So, it sends any category $$\mathcal{C}$$ to the category $$\mathcal{C} \times \mathcal{C}$$.

Finally, there's a natural transformation from the identity functor

$$1_\mathbf{Cat} : \mathbf{Cat} \to \mathbf{Cat}$$ to this composite functor

$$\times_{\mathbf{Cat}} \circ \Delta_{\mathbf{Cat}} : \mathbf{Cat} \to \mathbf{Cat}$$ To each category $$\mathcal{C}$$, this natural transformation assigns a functor from $$\mathcal{C}$$ to $$\mathcal{C} \times \mathcal{C}$$. And what is this functor? It's

$$\Delta_{\mathcal{C}}: \mathcal{C} \to \mathcal{C} \times \mathcal{C}.$$ Hey, we've come full circle!

Comment Source:Keith wrote: > Think of it this way, the functor \$$\Delta\$$ is a category copying machine. It takes a category and copies or *duplicates* it. > \$\Delta: \mathcal{C} \to \mathcal{C} \times \mathcal{C}, \$ Level slip? I wouldn't say this \$$\Delta\$$ is duplicating the category. I'd say it's duplicating each object _in_ the category. Let's call it \$$\Delta_{\mathcal{C}}\$$. It has $\Delta_{\mathcal{C}}(c) = (c,c)$ for each object \$$c\$$ in \$$\mathcal{C}\$$. <img width = "100" src = "http://math.ucr.edu/home/baez/mathematical/warning_sign.jpg"> Of course, there is also something that's duplicating the category \$$\mathcal{C}\$$. There's a functor $\Delta_{\mathbf{Cat}} : \mathbf{Cat} \to \mathbf{Cat} \times \mathbf{Cat}$ and we have $\Delta_{\mathbf{Cat}}(\mathcal{C}) = (\mathcal{C}, \mathcal{C})$ There's also a functor $\times_{\mathbf{Cat}} : \mathbf{Cat} \times \mathbf{Cat} \to \mathbf{Cat}$ that takes the product of any two categories: $\times_{\mathbf{Cat}} (\mathcal{C}, \mathcal{D}) = \mathcal{C} \times \mathcal{D} .$ We can compose these two functors and get a functor $\times_{\mathbf{Cat}} \circ \Delta_{\mathbf{Cat}} : \mathbf{Cat} \to \mathbf{Cat}$ which does this: $\times_{\mathbf{Cat}} \circ \Delta_{\mathbf{Cat}} (\mathcal{C}) = \mathcal{C} \times \mathcal{C}$ So, it sends any category \$$\mathcal{C}\$$ to the category \$$\mathcal{C} \times \mathcal{C} \$$. Finally, there's a natural transformation from the identity functor $1_\mathbf{Cat} : \mathbf{Cat} \to \mathbf{Cat}$ to this composite functor $\times_{\mathbf{Cat}} \circ \Delta_{\mathbf{Cat}} : \mathbf{Cat} \to \mathbf{Cat}$ To each category \$$\mathcal{C}\$$, this natural transformation assigns a functor from \$$\mathcal{C}\$$ to \$$\mathcal{C} \times \mathcal{C} \$$. And what is this functor? It's $\Delta_{\mathcal{C}}: \mathcal{C} \to \mathcal{C} \times \mathcal{C}.$ Hey, we've come full circle! 
• Options
41.
edited July 2019

Still trying to figure out Puzzles 150 - 155 (note that the numbering of the last two puzzles is repeated in the original post, I am extending it here). To wrap up the solutions so far:

Puzzle 150:

$$R(S) = (S, 1)$$ for any set $$S$$ and $$R(f) = (f, 1_1)$$ for any function $$f: S \to T$$, where $$1_1$$ means the identity function from our one-element set to itself.

By the way, it seems to me that this identity differs from Dan's solution, where

$$!$$ denotes the unique function from any set to the singleton set $$\{\bullet\}$$.

Puzzle 151: ???

Puzzle 152: Does not exist?

Puzzle 153: The diagonal functor $$\Delta$$.

Puzzle 154: Also the diagonal functor $$\Delta$$.

Puzzle 155: ???

I think that these examples are really great because they are simple and concrete. But I still find it very hard to start reasoning about adjunctions, I can only think: "what would be a generous/selfish way to come back?". And once I have a candidate, it is hard to start a proving that it is an adjoint. Is there any reference where we can find complete proofs of all this?

Comment Source:Still trying to figure out Puzzles 150 - 155 (note that the numbering of the last two puzzles is repeated in the original post, I am extending it here). To wrap up the solutions so far: **Puzzle 150:** > \$$R(S) = (S, 1) \$$ for any set \$$S \$$ and \$$R(f) = (f, 1_1) \$$ for any function \$$f: S \to T\$$, where \$$1_1\$$ means the identity function from our one-element set to itself. By the way, it seems to me that this identity differs from Dan's solution, where > \$$!\$$ denotes the unique function from any set to the singleton set \$$\\{\bullet\\}\$$. **Puzzle 151:** ??? **Puzzle 152:** Does not exist? **Puzzle 153:** The diagonal functor \$$\Delta \$$. **Puzzle 154:** Also the diagonal functor \$$\Delta \$$. **Puzzle 155:** ??? I think that these examples are really great because they are simple and concrete. But I still find it very hard to start reasoning about adjunctions, I can only think: "what would be a generous/selfish way to come back?". And once I have a candidate, it is hard to start a proving that it is an adjoint. Is there any reference where we can find complete proofs of all this?