How do I think of the Hom functor?

Question

For an object $X$ in a category $C$, there is a functor $C(-\,,X)$ from $C^{\mathrm{op}}$ to Set that assigns to each object $Z$ the set $C(Z,X)$ and to each morphism $f: Y \to Z$ the pullback $f^*$ to $f$ $$ \require{AMScd} \begin{CD} Y \\ @V{f}VV \\ Z \end{CD} \quad\mapsto \quad \begin{CD} C(Y, X) \\ @AA{f^*}A \\ C(Z, X) \end{CD} $$ (Page 10, A categorical approach to topology)

I am having a lot of difficulty understanding the above definition. After flipping back and forth the book, I think so I get each individual part of the definition (what the hom set is, the pull back etc), but I am unable to think of a relatable realization of it.

Could someone explain the above a bit more simply, perhaps through a metaphor?

I'm looking for something around the lines of how Qiaochu Yuan answered this question.

Answer 1

Alright, this might be a little bit of a stretch, but here’s an attempt at an analogy that might make sense.

Your (locally small) category $\mathcal{C}$ is your city. Think of the objects of $\mathcal{C}$ as different locations around the city. There’s your home, the grocery store, your favorite dive bar, the park you walk your dog Glenn in (this is my analogy so I get to name your dog), the coffee shop you read in on the weekends, etc. For any two locations $a$ and $b$, think of $\text{hom}(a,b)$ as the set of all the different routes you could take to walk from location $a$ to location $b$.

So let’s fix a location. Say, the library. Let $L$ denote the the library. How should we think of the functor $\text{hom}(-,L)$? We can think of this functor as Google maps, in a sense. If you tell this functor some location $a$, it will spit out all the different routes you could take from $a$ to the library $L$. That is, it gives you the set of routes $\text{hom}(a,L)$.

Now what about the functors value on maps? Suppose now we have some arrow $f:a \rightarrow b$ in $\mathcal{C}$. So in the analogy, this is some specific route through your city from $a$ to $b$. By the definition of a contravariant functor, we know that $\text{hom}(f,L)$ should be a set-function from $\text{hom}(b,L)$ to $\text{hom}(a,L)$. So in the analogy, we should get a way to take in a route from $b$ to the library, and spit out a route from $a$ to the library. But we can do this easily by just first taking the pre-chosen route $f:a \rightarrow b$, and then the route from $b$ to $L$.

So in the Google maps analogy, suppose you’re at $H$ := your home, you have to run a quick errand at $S$ := the store, and then head to $L$ := the library. Finally, suppose you know exactly the route you want to take from your home to the store (call it $f$). Then Google maps would tell you all the different ways you could go from your house to the library, starting with your pre-chosen route $f$. Google is asking you to make a choice of a route from $S$ to $L$, and in return you’re getting a route from $H$ to $L$.

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Please note that I don’t think this is a particularly good analogy. In particular, there are many other parts of category theory that are conceptually related to formal paths. This is just the best I could come up with on the fly.

Answer 2

I think Joe's answer is really really nice, but for a diversity of perspectives, here is another way to think about "why care about the hom functor"?

Suppose you are given a category $C$ and an object $c\in C$. You are not told what $C$ is, e.g., we do not know if $C$ is a concrete or abstract category. Nor do we know what $c$ is. All we know is the fact that $C$ is a category and $c\in C$. Suppose someone asks you if you can "pick $c$ out of a crowd", i.e., if you have enough information to identify $c$. It seems like we are stuck here. We know nothing about $c$, nor the category $C$ it lives in.

But, wait! We know something about $C$. Its a category! So, it has to have morphisms. So, by definition, for any other object $x\in C$, we have a set $C(x,c)$ of morphims into $c$. We can in fact do this for every object in $C$. So this defines a map $ob(C)\to Set$ sending $x\mapsto C(x,c)$.

Now, we still don't know much about $C$ or $c$. We don't really know much about the morphisms of $C$, if the are functions or what. But, we do know something about them. They compose! So given any $f\in C(x,c)$ and any $h:y\to x$, we have to have $fh: y \to c$. Hey, this mean we have a map $h^*: C(x,c)\to C(y,c)$! So, we can extend our $ob(C)\to Set$ to a functor $C^{op}\to Set$ by sending any $(f: x\to y)\mapsto f^*:C(y,c)\to C(x,c)$. Thus, we have arrived out our functor $C(-,c)$.

You may be thinking that this does not tell us much about $C$ or $c$. In some sense this is true; we still do not know if $C$ is concrete or abstract or how $c$ was constructed. But, it turns out that we now have all the information we need to "pick $c$ out of a crowd". Hows that? Invoking the yoneda lemma, $C(-,c)$ characterizes $c$ up to isomorphism. That is, if there is some $C(-,x)\cong C(-,c)$, then $x\cong c$. This may not seem like much at first. How important this was did not dawn on me for quite sometime, and I am still appreciating it more each day, but this powerful.

We know nothing about what $C$ is. It could be $Grp$, $Top$, or some weird and abstract category. Nor do we know a lick about $c\in C$. Yet, the structure $C$ has from merely satisfying the definition of a category (which is not a whole lot!) allows us to cook up $C(-,c)$. And this turns out to be enough to essentially define $c$! So maybe we are given another more familiar category $Grp$ and a way to map $F:Grp \to C$. We now have a way (in some sense) to determine if $F(G)\cong c$, for some $G\in Grp$. This in turn gives us (maybe?) enough information to determine if $Grp\simeq C$.

One thing representable functors make me appreciate is how powerful the seemingly simple axioms of category theory are!