Lecture 52 - The Hom-Functor

Now let's take a deeper look at adjoint functors - in our first look at these in Lecture 47, I deliberately downplayed some technicalities. But we need to understand these to truly understand adjoints, and they're actually beautiful in their own right. One of these is the so-called 'hom-functor'

$$ \mathrm{hom}: \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathbf{Set} $$ that every category $\mathcal{C}$ comes born with. Some of you have already figured out how this works. But let me explain it anyway.

Here are two important principles of category theory:

Every collection of things is eager to become the objects of a category.

and

Every map sending things of one kind to things of another kind is eager to become a functor.

But a category is more than a mere collection of objects, and a functor is more than a mere functor. A category, after all, needs morphisms! And a functor must know what to do to morphisms!

True, there's a cheap way to make any set into a category. Just throw in the morphisms that are absolutely required: the identity morphisms! There's also a cheap way to make any function into a functor. But usually we want to do something more interesting.

Now, whenever you've got a category $\mathcal{C}$, there's a set of morphisms from any object $x$ to any object $y$. This is called a hom-set, and people often write it as $\mathrm{hom}(x,y)$. Fong and Spivak call it $\mathcal{C}(x,y)$, and that's good because it reminds you which category $x$ and $y$ are objects of.

Anyway, the hom-functor is what we get when we think about hom-sets using the principles I stated above.

A category $\mathcal{C}$ has a collection of objects $\mathrm{Ob}(\mathcal{C})$, and given any two of these we get a set of morphisms from the first to the second. So, there's a function

$$ \mathrm{hom} : \mathrm{Ob}(\mathcal{C}) \times \mathrm{Ob}(\mathcal{C}) \to \mathrm{Ob}(\mathbf{Set}) $$ sending any pair of objects $ (c,c') \in \mathrm{Ob}(\mathcal{C}) \times \mathrm{Ob}(\mathcal{C}) $ to the set $ \mathcal{C}(c,c') $ of morphisms from $c$ to $c'$. Note that a guy in $\mathrm{Ob}(\mathbf{Set})$ is just a set!

But this function looks like it wants to become a functor

$$ \mathrm{hom} : \mathcal{C} \times \mathcal{C} \to \mathbf{Set}. $$ That doesn't quite work - but if we try, we'll see what goes wrong, and how to fix it.

First of all, what's $\mathcal{C} \times \mathcal{C}$? That actually works fine:

Theorem. For any categories $\mathcal{X}$ and $\mathcal{Y}$, there is a category $\mathcal{X} \times \mathcal{Y}$ for which:

An object is a pair $ (x,y) \in \mathrm{Ob}(\mathcal{X}) \times \mathrm{Ob}(\mathcal{Y}) $.
A morphism from $ (x,y) $ to $ (x',y') $ is a pair of morphisms $ f: x \to y'$ and $g: y \to y'$. We write this as $ (f,g) : (x,y) \to (x',y') $.
We compose morphisms as follows:

$$ (f',g') \circ (f,g) = (f' \circ f, g' \circ g) .$$

Identity morphisms are defined as follows:

$$ 1_{(x,y)} = (1_x, 1_y) .$$

Proof. Just check associativity and the right/left unit laws. $ \qquad \blacksquare $

The problem is that we can't define our would-be functor

$$ \mathrm{hom} : \mathcal{C} \times \mathcal{C} \to \mathbf{Set} $$ on morphisms. We know what we want it to do to objects of $\mathcal{C} \times \mathcal{C}$:

$$ \mathrm{hom}(c,c') = \mathcal{C}(c,c') .$$ But suppose we have a morphism in $\mathcal{C} \times \mathcal{C}$, say

$$ (f,g) : (c,c') \to (d,d') $$ This should get sent to a morphism in $\mathbf{Set}$, that is a function, called

$$ \mathrm{hom}(f,g) : \mathrm{hom}(c,c') \to \mathrm{hom}(d,d') $$ This function should take any morphism $h \in \mathrm{hom}(c,c')$ and give a morphism in $\mathrm{hom}(d,d')$. Can we accomplish this with what we have? Draw a diagram of everything:

$$ \begin{matrix} & & h & & \\ & c & \rightarrow & c' &\\ f & \downarrow & & \downarrow & g\\ & d & \rightarrow & d' &\\ & & ? & & \\ \end{matrix} $$ Can we get a morphism from $d$ to $d'$ from this?

No!

We can compose $h$ with $g$ just fine. But we can't compose it with $f$, because $f$ is pointing the wrong way!

So, we need to turn around an arrow, and for that we need the concept of an 'opposite category'.

Theorem. For any category $\mathcal{C}$ there is a category $\mathcal{C}^{\text{op}}$, called the opposite of $\mathcal{C}$, for which:

The objects of $\mathcal{C}^{\text{op}}$ are the objects of $\mathcal{C}$.
A morphism $f : c \to c'$ in $\mathcal{C}^{\text{op}}$ is a morphism $f : c' \to c$ in $\mathcal{C}$.
The composite $g \circ f $ of morphisms $f : c \to c'$, $g: c' \to c''$ in $\mathcal{C}^{\text{op}}$ is the composite $f \circ g$ of the corresponding morphism $g : c'' \to c' $, $f: c' \to c$ in $\mathcal{C}$.
The identity morphism of an object $c$ of $\mathcal{C}^{\text{op}}$ is the same as its identity morphism in $\mathcal{C}$.

Proof. Again, just check associativity and the left/right unit laws. These are facts that already hold in $\mathcal{C}$; we're just turning them around backwards! $ \qquad \blacksquare$

Now we can succeed in getting our hom-functor! It's really a functor

$$ \mathrm{hom} : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathbf{Set} . $$ We already know what we want it to do to objects:

$$ \mathrm{hom}(c,c') = \mathcal{C}(c,c') .$$ Now suppose have a morphism in $\mathcal{C}^{\text{op}} \times \mathcal{C}$, say

$$ (f,g) : (c,c') \to (d,d') . $$ Thanks to the fiendishly clever 'op', this is the same as a morphism

$$ (f,g) : (d,c') \to (c,d') $$ in $\mathcal{C}\times \mathcal{C}$. Our hom-functor should send this to a morphism in $\mathbf{Set}$, namely a function

$$ \mathrm{hom}(f,g) : \mathrm{hom}(c,c') \to \mathrm{hom}(d,d') $$ This function should take any morphism $h \in \mathrm{hom}(c,c')$ and give a morphism in $\mathrm{hom}(d,d')$. Can we get this to work now? Again, draw everything we've got:

$$ \begin{matrix} & & h & & \\ & c & \rightarrow & c' &\\ f & \uparrow & & \downarrow & g\\ & d & \rightarrow & d' &\\ & & ? & & \\ \end{matrix} $$ Can we get a morphism from $d$ to $d'$ from this?

Yes!

Just compose all the arrows and get

$$ g \circ h \circ f : d \to d'. $$ Now we're on the road to success. Of course we have to check that our would-be hom-functor really is a functor! But I'll let you do that:

Puzzle 161. Prove that for any category $\mathcal{C}$ there is a functor, the hom-functor

$$ \mathrm{hom} : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathbf{Set} $$ that sends any object $ (c,c') $ of $\mathcal{C}^{\mathrm{op}} \times \mathcal{C}$ to the set $\mathcal{C}(x,y)$, and sends any morphism

$$ (f,g) : (c,c') \to (d,d') $$ in $\mathcal{C}^{\text{op}} \times \mathcal{C}$ to the function

$$ \mathrm{hom}(f,g) : \mathrm{hom}(c,c') \to \mathrm{hom}(d,d') $$ that maps any $h \in \mathrm{hom}(c,c') $ to $ g \circ h \circ f \in \mathrm{hom}(d,d')$.

You have to prove it preserves composition and identities!

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