Concrete Categories

In many introductions to category theory, you first learn the notion of a concrete category: A concrete category is a collection of sets, called the objects of the category and, for each pair (X,Y) of objects, a subset of the maps $X \to Y$ . (There are, of course, axioms that these things must obey.) In a concrete category, the objects are sets, and the morphisms are maps that obey certain conditions. So the category of groups is concrete: a map of groups is just a map of the underlying sets such that multiplication is preserved. So are the category of vector spaces, topologicial spaces, smooth manifolds and most of the other most intuitive examples of categories.

Using terminology from a discussion at MO, I’ll call a category concretizable if it is isomorphic to a concrete category. For example, $\mathrm{Set}^{op}$ can be concretized by the functor which sends a set to the set 2^X of subsets of , and sends a map of sets $f:X \to Y$ to the preimage map $f^{-1}: 2^Y \to 2^X$ .

At one point, I learned of a result of Freyd: The category of topological spaces, with maps up to homotopy, is not concretizable. I thought this was an amazing reflection of how subtle homotopy is. But now I think this result is sort of a cheat. As I’ll explain in this post, if you are the sort of person who ignores details of set theory, then you might as well treat all categories as concrete. My view now is that specific concretizations are very interesting; but the question of whether a category has a concretization is not. I’ll also say a few words about small concretizations, and Freyd’s proof.

Let me start by saying exactly what you need to check to see whether a functor is a concretization. Let be a category and a functor from to $\mathrm{Set}$ . Then is a concretization if, for any objects and , and any morphisms and from to , we have F(f)=F(g) only if f=g .

Now, Yoneda’s lemma almost gives us such a functor in every case. Define

$F(X) := \bigsqcup_{S \in \mathrm{Ob}(C)} \mathrm{Hom}_C(S,X).$

Yoneda’s lemma tells us that, if and induce the same map from $\mathrm{Hom}_C(S,X)$ to $\mathrm{Hom}_C(S,Y)$ for every , then f=g . The proof is simply to take S=X .

So, why doesn’t Yoneda’s lemma tell us that all categories are concretizable? Because the collection of objects of our category may not be a set. I assume that you have at some point been introduced to Russell’s paradox, which is resolved by declaring that the collections of all sets is “too big” to be a set. Similarly, you run into trouble if you try to talk about “the set of all vector spaces” or “the set of all groups.” And since people do consider the category of all vector spaces; category theorists have learned to be careful with expressions like $\bigsqcup_{S \in \mathrm{Ob}(C)}$ , which act as if the objects of are a set.

If, like me, you don’t care about this sort of set theoretic issue, then you might as well think that all categories are concretizable. But you should still object to the Yoneda method of concretization. When will you ever be able to check something for all the objects of a category? What you want is some reasonable collection $T \subset \mathrm{Ob}(C)$ of test objects, so that it is enough to see whether f_*=g_* on X(S) for $S \in T$ .

There are several great theorems of this form: In the category of varieties of finite type over $\mathbb{C}$ , the Nullstellansatz tells us that a map is determined by its values on $X(\mathbb{C})$ — so we can think of a variety as made up of its points. ~~In the category of finite CW complexes, with maps up to homotopy, Whitehead’s theorem tells us that it is enough to study $\mathrm{Hom}(S^n, X)$ , for all spheres .~~ The preceeding statement is completely false; thanks to Eric Wofsey for a counter-example. I’ll try to find a better, and more true, example.

My mathematical aesthetic would be to adopt a subjective standard here: the goal of concretization is to find a “nice” set of test objects, and the term “nice” is defined by the judgement of the mathematical community. The choice of a single point, in the Nullstellansatz example, is very nice. The choice of, for example, all Artinian rings, would still be nice, but less so. (PARAGRAPH REWRITTEN due to error in the preceding paragraph.)

For those who seek an objective criterion, Tom Leinster proposes saying that has a small concretization if there is a set of objects of such that $X \mapsto \bigsqcup_{S \in T} \mathrm{Hom}_C(S,X)$ is a concretization. Cautionary exercise: the category whose objects are sets, and whose morphisms are surjections, is a concrete category but has no small concretization.

I don’t want to close the post without saying something about how it is proved that the category of topological spaces, with maps modulo homotopy, is not concretizable. Even though I don’t find concretization interesting, the idea that it can be proved impossible is interesting to me. This is a result of Peter Freyd, whose explanation of the technical points is about as good as it could be, so I’ll leave the details to him.

Suppose that were a concretization. For simplicity, I’ll assume that $F(\{ \mathrm{point} \})$ is the set with a single element. For any connected space , let z(X) be the element of F(X) corresponding to the unique homotopy class of map $\{ \mathrm{point} \} \to X$ .

Freyd constructs a totally ordered set , with cardinality greater than $2^{F(S^2)}$ , two sequences of connected spaces A_p and B_p indexed by , and maps $A_p \to S^2 \to B_p$ . These have the property that the composite $A_p \to B_p$ does not factor through $\{ \mathrm{point} \}$ but $A_p \to B_q$ , with q<p , does. Since is so big, there must be some q<p such that F(A_p) and F(A_q) map to the same subset of F(S^2) . Call this subset .

Then $A_p \to S^2 \to B_q$ factors through $\{\mathrm{point}\}$ , so the map $S^2 \to B_q$ sends to z(B_q) . But then $A_q \to S^2 \to B_q$ also sends F(A_q) to z(B_q) , contradicting that $A_q \to B_q$ does not factor through a point.

To give a little more detail, one first constructs a sequence of groups G_p , with nonzero maps $\mathbb{Z} \to G_p$ , such that, for any map $G_p \to G_q$ with q<p , the composite $\mathbb{Z} \to G_q$ would be zero. Let C_p be a K(G_p,1) . Note that S^1 is a $K(\mathbb{Z}, 1)$ , so we have functorial maps $S^1 \to C_p$ . Then B_p is the suspension $\Sigma(C_p)$ , the map $S^2 \to B_p$ is the suspension of $S^1 \to C_p$ , and A_p is the mapping cone of $S^1 \to C_p$ . If you want more detail than this, you should read Freyd’s paper.