How do I think of a representable functor?

Question

For any category C , a (covariant or contravariant) functor $F : C → \text{S e t}$ is said to be representable if there is an object $c$ in $C$ so that for all objects $x$ in $C$ , the elements of $F ( x )$ are really maps $x → c$ (or maps $c → x$ , depending on the "variance" of $F$ ). In this case we say $F$ is represented by the object $c$ .

I'm finding it difficult to understand this definition and it's significance. Could someone share an intuition or picture to understand this?

Note: I've seen this answer but I can't seem to understand the example there... so I am hoping for something lighter.

Answer 1

As far as "understanding the significance" of representable functors goes, I think the main power comes from the ability to apply techniques from some subject to functors related to that subject. Here's a few examples to showcase what I mean.

Oftentimes when working with vector spaces (or modules, more generally), we find ourselves concerned with bilinear maps. For instance, an inner product is a bilinear map $V \times V \to k$. Of course, the theory of vector spaces (resp. modules) isn't set up to handle bilinear maps -- it's set up to handle linear maps. So we find ourselves with a functor $F$ sending a vector space $X$ to the set of bilinear maps $V \times W \to X$, and we want to know if we can study this functor by the techniques of linear algebra (that is, by working inside the category of vector spaces). The answer, of course, is yes since the functor $F$ is represented by the tensor product $V \otimes W$. This lets us study bilinear maps by using the techniques we've built for studying the category of vector spaces and linear maps!

In algebraic topology, we're often interested in the cohomology of a space $X$ with coefficients in an abelian group $G$. Wouldn't it be nice if we could study cohomology by directly using the machinery of algebraic topology? Well, we can! The functor $H^n(-,G)$ is representable, and indeed the Eilenberg-MacLane Space $K(G,n)$ does the representing. Indeed, $H^n(X,G)$ is in natural bijection with (homotopy classes of) maps $X \to K(G,n)$, so by studying $K(G,n)$ we can study all these cohomology groups simultaneously.

In differential geometry, we're often interested in studying vector bundles over a manifold $M$. Again, we're led to consider the functor $B_n$ so that $B_n(M)$ is the set of rank $n$ vector bundles over $M$ (up to isomorphism). Again, this turns out to be representable, and we find that $B_n(M)$ is isomorphic to (homotopy classes of) maps $M \to \text{Gr}(n,\infty)$, the space of $n$ dimensional grassmanians in $\mathbb{R}^\infty$.

As a neat application to show that this really is useful, you might wonder whether every topological rank $n$ vector bundle over $M$ can be upgraded to a smooth rank $n$ vector bundle. A priori, this seems like a hard question to answer, but by using the machinery of classifying spaces, it becomes obvious! A topological rank $n$ vector bundle is represented by a homotopy class of continuous maps $M \to \text{Gr}(n,\infty)$. A smooth vector bundle is represented by a homotopy class of smooth maps $M \to \text{Gr}(n,\infty)$. But since every continuous map between smooth manifolds is homotopic to a smooth map (you might need a compactness assumption here?) we see that these notions coincide (up to isomorphism of topological bundles).

In general, the reasons that people consider high-abstraction objects like schemes (or worse, stacks) is because they allow us to represent solutions that other, more classical, geometric objects cannot represent. Then these objects are useful, via the functors the represent, even if at the end of the day we're only interested in studying classical geometric objects.

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I hope this helps ^_^

Answer 2

There are many different reasons that make representable functors interesting, and your intuition for them will grow as you work with them. Try looking at the Yoneda Lemma, since this tells us quite a lot about what makes representable functors interesting.

But, maybe looking at an example first will help. First, note that another definition of representable is that a covariant functor $F: C \to Set$ is representable if $F\cong C(c,-)$, for some $c\in C$.

Now, consider a category $C$ and two objects $x , y \in C$. Now define a contravarient functor $F : C \to Set$, by $c \mapsto C(c,x)\times C(c,y)$ and $(f:c\to c')\mapsto ((f^*,f^*) : C(c',x)\times C(c',y) \to C(c,x)\times C(c,y))$. Here $C(c,x)$ is a hom set, i.e., the set of all maps from $c$ to $x$. And $(f^*,f^*)$ is defined by $(g,h)\mapsto(gf,hf)$. What would it mean for this functor to be representable? It would mean that there exists an object $X \in C$ such that $F\cong C(-,X)$. So, for any $c\in C$, $Fc = C(c,x)\times C(c,y) \cong C(c,X)$. So, if $F$ is representable, then it must be the case that there is an object $X\in C$ such that maps into $X$ encode pairs of maps into $x$ and into $y$. This means that $X$, the representing object of $F$, has the universal property of the product of $x$ and $y$!

Here are a few other examples. Take the category $C$ to be $Set$ and consider the functor $Id:Set \to Set$. This functor is representable by any one element set. For definiteness take the set $1=\{0\}$. Then $Id\cong Set(1,-)$. So we can think of $Id$ as taking a set and sending it to the set of all its elements. The fact that $Id$ is representable tells us that element of a set $X$ are the same thing as morphisms $1\to X$. Properly, I should have said that elements of $X$ can be identified with morphisms $1\to X$, but we can take these expression to mean the same thing. And I think the emphasis is helpful.

This was mentioned in the comments and it is a good example so I'll repeat it. Take $C=Grp$ and the forgetful functor $U : Grp \to Set$, which sends a group to its underlying set, i.e., its set of elements. This functor is representable since $U \cong Grp(\mathbb{Z},-)$. What this tells us is that an element of a group $G$ is the same thing with a morphism from $\mathbb{Z}\to G$.

These last two examples are nice because it shows that even if a functor is not explicitly sending an object to some set of morphism, if this functor is representable then the we can identify the image of an object under this functor with a set of morphisms.

Later, once you have seen the Yoneda Lemma, there is more to think about here. One interesting thing is that a functor $C(-,x)$ "completely defines $x$", in the sense that if $C(-,x)\cong C(-,y)$ in $Set^{C^{op}}$, then $x\cong y$ in $C$. Additionally, with Yoneda and something called the category of elements in hand, you can prove that every single functor $F:C^{op}\to Set$ is the colimit of representable functors! This is called the density theorem. So, we can sort of think of functors $C(-,x)$ as the building blocks of all other functors $F : C^{op}\to Set$.

Hope this helps. Let me know if I missed something.

Answer 3

A Little Philosophy

There’s a sense in which “real category theory” is what is preserved by equivalences of categories. If you aren’t familiar, an “equivalence” of categories is a weaker notion of “sameness” than equality of categories or isomorphism of categories. A relatively simple definition (among many definitions) of equivalence is as follows.

Definition: Let $C$ and $D$ be categories. We say that $C$ is equivalent to $D$, written $C \simeq D$ If and only if there is a functor $F:C \rightarrow D$ which is full, faithful, and essentially surjective.

I’m assuming you know what full and faithful functors are, but “essential surjectivity” may be new. To say that $F:C \rightarrow D$ is essentially surjective means that for every object $d \in D$, there is an object $c \in C$ such that $F(c) \cong d$. So for an object $d \in D$, it might not the case that $d$ is mapped to by $F$, but at least an object isomorphic to $d$ is mapped to.

It’s worth noting that this is strictly weaker than an isomorphism of categories. An example might be illustrative. Let Ord denote collection of all ordinals. By the axiom of choice, every set can be put in bijection with some ordinal. Let’s turn Ord into a category where morphisms are all set-functions between ordinals. That is, we’re taking a “full subcategory” of Set. We’re picking out a sub collection of objects, and keeping all the morphisms between them. With a little bit of work, one may show:

Theorem: There is an equivalence from Set to Ord.

I’ll leave it to you to prove this equivalence exists, but my remark about every every set being in bijective correspondence with some unique ordinal will be the crux of the proof. It’s even easier to show:

Theorem: Ord is not isomorphic to Set.

I again leave the proof to you.

As mentioned before, “real category theory” is the study of the properties of categories that are invariant under equivalence. One way of looking at this is the “principle of equivalence” (https://ncatlab.org/nlab/show/principle+of+equivalence). As the n-lab page points out, any categorical proof or construction that isn’t equivalence invariant is referred to as “evil”.

One corollary of this perspective is that any discussion of equality of objects is “evil”. When you’re thinking as a category theorist, you should avoid directly referring to equality of objects under as many circumstances as possible. To do otherwise is to be evil! Indeed, the equivalence functor from Set to Ord will take non-equal objects to equal ones.

Isomorphisms of objects, however, are trivial preserved by all functors (and hence by equivalences as well). So talking about isomorphic objects isn’t evil.

That was a long digression, but I promise there was a point. A representable functor $F:C^{\text{op}} \rightarrow \text{Set}$ is “representable” if and only if it is isomorphic to a contravariant hom functor $\text{hom}(-,c)$ for some object $c \in C$. This isomorphism is happening in the category $\text{Set}^{C^{\text{op}}} $ of functors from $C^{\text{op}}$ to Set. From the perspective of the principle of equivalence, being representable is the closest thing we have to being a contravariant hom functor. To get any more fine-grained is to be evil. From this perspective, being representable is to “might as well” be a hom functor, or to be able to be replaced by a hom functor in any evil-free proof or construction.

This is a long way of saying the importance and the definition of a representable functor is a bit subtle. The takeaway is as follows: It’s safe to think of a representable functor as being a hom functor for the purpose of almost any proof or construction. The interesting thing about representable functors is finding out that an otherwise-interesting functor is actually representable. It’s like finding out that something was “secretly” a hom functor all along.

Indeed, this philosophy is in line with what the other answers say. For example, they bring up the fact that every functor $F:C \rightarrow \text{Set}$ can be realized as a colimit of representable functors. Well, as a matter of fact colimits are only defined up to isomorphisms.

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Alright, Now ignoring everything above, I’ll give a shot at an analogy. Keep in mind the “Google maps” intuition I gave for the contravariant hom functor I gave on your question over here (How do I think of the Hom functor?). Once again, I think this analogy is a stretch, but maybe it will be helpful again. This time though, we’re going to have to start with unpacking some definitions.

What Does it Mean For Two Set-Valued Functors to be Isomorphic?

Before we dive into an analogy, I want to make sure you really understand what it means for two set-valued contravariant functors to be isomorphic. Let $\mathcal{C}$ be a category. Let $F,G: \mathcal{C}^\text{op} \rightarrow \text{Set}$ be two set-valued contravariant functors on $\mathcal{C}$. What exactly does it mean for us to say that $F$ and $G$ are isomorphic?

Let’s focus on objects first. Let $c$ be an object of $\mathcal{C}$. Since $F$ and $G$ are isomorphic functors, there is an isomorphism $\phi_c:F(c) \rightarrow G(c)$. Recall that $F(c)$ and $G(c)$ are sets. What is an isomorphism of sets? It’s a bijection! So for every object $c$, we get a bijective set-map $\phi_c:F(c) \rightarrow G(c)$.

This doesn’t say very much though. Any two sets of the same cardinality have a bijection between them. The real “power” of the isomorphism comes from the fact that these bijections $\{\phi_c : c \in \mathcal{C} \}$ have to “play nicely” with the functor’s action of arrows.

So let $c$ and $d$ be two different objects in $\mathcal{C}$, and let $f:c \rightarrow d$ be an arrow in $\mathcal{C}$. What would it mean for these $\{\phi_c : c \in \mathcal{C} \}$ to “play nicely” with $f$? Since $F$ and $G$ are contravariant, we know that applying these functors to $f$ give us a pair of maps: $$\begin{align*}F(f) & :F(d) \rightarrow F(c) \\ G(f) & : G(d) \rightarrow G(c) \end{align*}$$ But we already have maps $\phi_d: F(d) \rightarrow G(d)$ and $\phi_c:F(c) \rightarrow G(c)$. A “reasonable condition” for $f$ to play nicely with the maps $\{\phi_c : c \in \mathcal{C} \}$ would be to demand that: $$\phi_c \circ F(f) = G(f) \circ \phi_d$$ These are both maps from $F(d) \rightarrow G(c)$. Since we have equality on the nose, we know that we’re getting literally the same set-map.

One way to think about what this means is to think of the collection $\{\phi_c : c \in \mathcal{C} \}$ as a dictionary that translates between two different languages. We have the “$F$ language” and the “$G$ language”. And if we start with $F(d)$, translate over to $G(d)$, and then apply the process $G(f)$, we get the exact same thing as if we first applied the process $F(f)$ and then translated. We get exactly the same information out, and nothing is lost or missing from translation.

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An Analogy

Suppose that $F: \mathcal{C}^{\text{op}} \rightarrow \text{Set}$ is a representable functor. This means that there is some fixed object $c \in \mathcal{C}$ such that $F \cong \text{hom}_{\mathcal{C}}(-,c)$.

Recall from before, that we can think of $\mathcal{C}$ as your city, and the functor $\text{hom}(-,c)$ as google maps. For some other location $x$, we have that $\text{hom}(x,c)$ is the collection of routes from $x$ to $c$, and for an arrow $f: x \rightarrow y$ (thought of as a specific route from $x$ to $y$), the set-map $\text{hom}(f,c)$ takes in routes $y \rightarrow c$ adn spits out routes $x \rightarrow $ by “first taking $f$” (post-composition).

What does this tell us about the representable functor $F$? It tells us that for every $x$, $F(x)$ is a set that “encodes” all the different routes $x \rightarrow c$. It’s not literally the set of routes like we get from the hom functor, but it’s a “description” of those routes.

Have you ever had a friend or family member who is incapable of giving directions like an ordinary human being? Instead of being like “Oh yeah, go south on Main Street, then turn left on 1st Ave, and it’s two blocks down on your left”, they’re like “go straight on the big road. When you hit the the tree that kind of looks like Richard Nixon, take a left and keep going until the neighborhood starts looking a little gentrified. If you’d can smell bread you’ve gone too far.” These certainly aren’t directions, but they “encode” directions.

Think of a representable functor like talking to that friend instead of using Google maps. They aren’t describing routes and locations per-se, but they are saying something that can be translated into routes and locations. These translations are perfect — they contain all and exactly the information you need. You can translate back and forth without fail, but translation is needed.