# Lecture 48 - Adjoint Functors

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.

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