Building Objects with Category Theory

20 Feb 2023

Typically category theory is useful for showing the uniqueness of certain objects by checking that they satisfy some universal property. This makes them unique up to unique isomorphism. However, category theory can also be used in order to show that objects exist at all, usually with the help of the Adjoint Functor Theorems.

The plan is to build a functor which does something useful. Then we’ll show that the functor is representable. In particular, there’s an object $X$ representing the functor, and we’ll have shown the existence of $X$ purely abstractly!

This is a very common mode of argument nowadays in algebraic geometry. You want to build a moduli space for some family of objects, but it can be tricky to build it explicitly. Since it’s a moduli space, though, we know what its “functor of points” should be. So if we can show its functor of points is representable, we’ll be in business! This is one reason to pass to the category of stacks, since oftentimes a moduli functor is representable as a stack but is not representable by a scheme. For more information, see Alper’s excellent notes on Stacks and Moduli.

The algebro-geometric examples are quite intricate, but we can already find a small example in tensor products. After all, we know the tensor product $M \otimes N$ is supposed to represent bilinear maps from $M$ and $N$!

Let’s see what this means, sketch how to build the tensor product purely abstractly, and then talk about the benefits and shortcomings of this approach.

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First, what do we mean when we say the tensor product “represents” bilinear maps? The answer is summed up in the universal property of the tensor product:

This universal property, like all universal properties, pins down $M \otimes N$ up to unique isomorphism if it exists. But it (crucially) doesn’t tell us whether $M \otimes N$ really does exist! Maybe there’s no such object.

In the language of representable functors, we consider the functor $\mathsf{BL}_{M,N} : \mathsf{Vect} \to \mathsf{Set}$ so that

\[\mathsf{BL}_{M,N}(X) = \{ \text{bilinear maps } f : M \times N \to X \}\]

Then the universal property says that there’s a bijection (natural in $X$)

\[\mathsf{Vect}(M \otimes N, X) \cong \mathsf{BL}_{M,N}(X)\]

so that $M \otimes N$ represents $\mathsf{BL}_{M,N}$.

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So the question, generally, is this: If I give you some functor to $\mathsf{Set}$, how can we tell if it’s representable or not? We know from experience that $\mathsf{BL}_{M,N}$ is representable, but if you’d never heard of a tensor product before it would probably be hard to make a guess!

We can answer this general question using the Adjoint Functor Theorems. While this is presented in a lot of category theory classes, I think it’s often left mysterious how and why to use it. Well here’s a prime example of the adjoint functor theorem in action:

The key to proving this (and to figuring out what “nice” means) lies in the adjoint functor theorems. There are a whole bunch of adjoint functor theorems, each of which is impossible to remember in its own special way. For instance:

Condition (2) above is usually called the solution set condition.

Thankfully, in many nice settings we don’t have to remember this condition! Here are two easy to remmeber settings where adjoint functors always exist:

This is great and all, but what do adjoint functors have to do with representables? Here’s the key, which for some reason wasn’t taught to me in any classes I took:

$\ulcorner$ We claim \(L \{ \star \}\) works. Indeed, \(\mathcal{C}(L \{ \star \}, X) \cong \mathsf{Set}(\{ \star \}, FX) \cong FX\), naturally in $X$.

$\lrcorner$

In fact, this is typically the only way for $F$ to be representable:

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Ok, so now we have a strategy! To prove (purely abstractly) that tensor products exist, we’ll show that \(\mathsf{BL}_{M,N} : \mathsf{Vect} \to \mathsf{Set}\) has a left adjoint. Since $\mathsf{Vect}$ and $\mathsf{Set}$ are both essentially algebraic (indeed they’re both algebraic), we just need to show that it preserves limits and filtered colimits.

Thankfully, neither of these is too hard to check! Here’s a sketch, which can be turned into a real proof by someone who doesn’t want to go to sleep soon² :P. They key insight is that for an arrow $f : X \to Y$, $\mathsf{BL}_{M,N}(f)$ is extremely simple: just compose your bilinear map $M \times N \to X$ with $f$! This, paired with the fact that everything here plays nicely with the forgetful functor to $\mathsf{Set}$ makes the computations fairly painless.

$\ulcorner$ We start by showing that $\mathsf{BL}_{M,N}$ preserves limits, and we can do this by checking that it preserves products and equalizers.

First, it’s easy to see that a bilinear map into $\prod X_\alpha$ is the same thing as a family of bilinear maps into each $X_\alpha$. Next, we need to show that \(\mathsf{BL}_{M,N}\) preserves equalizers. So if $f_\alpha : X \to Y$ is a family of maps, we’ll consider the equalizer

\[E = \{ x \in X \mid \forall \alpha, \beta . f_\alpha x = f_\beta x \}\]

Now \(\mathsf{BL}_{M,N}(E)\) is the set of bilinear maps $M \times N \to E$. We would like this to agree with the equalizer in $\mathsf{Set}$ of the \(\mathsf{BL}_{M,N}(f_\alpha) : \mathsf{BL}_{M,N}(X) \to \mathsf{BL}_{M,N}(Y)\). That is, with the set of bilinear maps $g : M \times N \to X$ so that all of the \(f_\alpha \circ g\) are equal. But if all the \(f_\alpha \circ g\) are equal, then $g$ lands in the equalizer of the \(f_\alpha\) (computed in $\mathsf{Set}$) which is the underlying set of $E$!

So $\mathsf{BL}_{M,N}$ preserves limits. Let’s move onto filtered colimits.

Say $\mathcal{D}$ is a filtered category and $X_{(-)}$ is a functor sending each object $i$ of $\mathcal{D}$ to a vector space $X_i$. Then we want to show that bilinear maps $M \times N$ into the colimit of the $X_i$ agrees with the colimit of the sets of bilinear maps.

But what is $\varinjlim X_i$? It’s the disjoint union of the $X_i$, where we glue together the points $x_i$ and $f x_i$ for each arrow $f$ in the diagram. What makes this diagram filtered is that any two arrows are eventually merged. So if we have a bilinear map $M \times N \to \varinjlim X_i$, we’re choosing some equivalence class $[x_i]$ for each $(m,n)$ pair.

But if we have a family of sets of bilinear maps $M \times N \to X_i$, their colimit is the disjoint union of these sets, where we glue together any maps that are eventually merged. So we are left with equivalence classes $[f_i]$ of maps landing in $X_i$.

Now it’s not hard to see that the function sending $[f_i]$ to the $\varinjlim X_i$-valued bilinear map $(m,n) \mapsto [f_i(m,n)]$ is well defined, and really does output a valid equivalence class. It’s also not hard to see that, given some bilinear map $f : M \times N \to \varinjlim X_i$, we can get a family of maps $f_i$ by letting $f_i(m,n)$ be the element of $X_i$ in the equivalence class of $[x_i] = f(m,n)$. Checking that these maps are well defined is where we crucially use filteredness of $\mathcal{D}$.

Then we check these maps are inverse to each other, so that $\mathsf{BL}_{M,N}$ preserves filtered colimits. $\lrcorner$

So \(\mathsf{BL}_{M,N}\) is representable! Call the representing object $M \otimes N$, and we’re done ^_^.

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Now I would argue that the above construction of the tensor product is worse than the explicit definition (by generators and relations) we typically see in an algebra class. It’s very high powered, and gives very little insight into how to make computations with $M \otimes N$.

This, more generally, is the downside to showing the existence of objects in this way. We have very little control over what the resulting object looks like. We only know how to probe it by maps to other objects³. The yoneda lemma tells us that this is enough to fully understand the object we’ve built, and it’s often good enough in the abstract, but sometimes you need to get your hands dirty and compute.

The big benefit to working this way is that it really doesn’t get much harder than this. Yes this was a superpowered proof of the existence of tensor products. But this proof doesn’t get any more complicated no matter what we want to exist! If we want to build something particularly complicated, then frequently this method (or one like it, tailored to the category of interest) will be easier than having to construct the relevant object by hand.

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Anyways, that’s going to do it for today’s post! I have like 4 or 5 posts all most of the way done, and I want to start wrapping them up soon, so hopefully there will be more Content™ for all of you in the near future ^_^.

As a quick story for how addled my brain is, though, I went to mark this post off my todo list of blog post ideas⁴, and I couldn’t find it on the todo list… I knew I’d thought about making this post before, so it was weird it wasn’t there.

It turns out that I never bothered putting it in the todo list, because I had started writing a version of this post in September of last year! I actually remember doing that, because I had this realization about the link between the adjoint functor theorems and representability⁵ while (re)reading Emily Riehl’s Category Theory in Context on a trip to New York. That became the impetus for this entire post, since it makes for a good excuse to clarify the adjoint functor theorems.

In fact, that old draft is still in my git history, and I looked through it after finishing the post, but before writing this epilogue. It’s structurally pretty similar to this one, but longer⁶, and with more examples of moduli spaces in geometry. That’s not so surprising, since only a few months prior I had been doing a reading course on stacks and moduli spaces with Patricio Gallardo Candela. There’s part of me that wants to move some of those examples here (for instance, projective space as a moduli space of lines through the origin), but a bigger part of me that wants to go to sleep, haha. I’m sure I’ll have plenty of time to talk about moduli spaces in some other post!

Thanks as always for reading. Stay warm, and we’ll chat soon⁷ ^_^.

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