A concretizable category is a category that admits a faithful functor to the category of sets.

Apparently, there are also non-concretizable categories, but I find it hard to imagine what this would look like, since I think of objects in a category as sets (although, we treat the sets as black boxes in category theory), and morphisms as mappings between them. It seems therefore intuitive to me that any category can be embedded in the category of sets.

Could you give an intuitive example of a category that is not concretizable, where it is clear/obvious/intuitive why it is not?

ps. wikipedia states $\textbf {hTop}$ as an example, but it is not obvious to me why it is not concretizable.

  • $\begingroup$ I think the last requirement is hard to fulfill $\endgroup$ – Maxime Ramzi Jun 24 '18 at 8:23
  • $\begingroup$ What about the opposite category of sets $\endgroup$ – Crostul Jun 24 '18 at 8:37
  • 3
    $\begingroup$ @Crostul: The contravariant powerset functor is faithful. $\endgroup$ – user14972 Jun 24 '18 at 8:45
  • 1
    $\begingroup$ It seems that all known examples are really non-trivial. See arxiv.org/pdf/1704.00303.pdf where you can find more examples and a remark that "almost everything is concrete" (Remark 2.2). $\endgroup$ – Paul Frost Jun 24 '18 at 9:49
  • 2
    $\begingroup$ Any locally small category $\mathcal{C}$ with a set $G$ of generators is concretizable by the functor $U(-) = \coprod_{g \in G} \hom_{\mathcal{C}}(g, -)$. So any locally small non-concretizable category will have to be strange enough so as to not have a set of generators. $\endgroup$ – user14972 Jun 24 '18 at 17:20

I don't think that any such example exists, and for good reason. The failure of concretizability is a subtle phenomenon that has nothing to do with abstract algebraic ideas and is instead entirely about size issues (that is, the distinction between sets and proper classes).

To illustrate this last point, if you ignore size issues, then every category is concretizable. Given a category $C$, you can get a faithful functor to sets by sending each object $a$ to the set $F(a)$ of all morphisms to $a$. Given a morphism $a\to b$, you get a map $F(a)\to F(b)$ by taking any morphism to $a$ and just composing it with the morphism $a\to b$. (Note that this construction is exactly the generalization of the usual proof of Cayley's theorem on groups to general categories.)

Now, this doesn't actually work in general if $C$ is a large category, since the "set" $F(a)$ of all morphisms to $a$ may actually be a proper class. But this does show that any small category is concretizable, and any failure of concretizability has to be about size issues. In other words, it has nothing do with sets and their structure per se (any category would be concretizable if you were allowed to use sufficiently large "sets"). Rather, it must have to do with the structure of the category being in some sense "too large" to embed in the category of small sets.


In a category with finite limits being concretizable is equivalent to the fact that RegSub(A) (regular subobjects) is a set for any object A.

This appears in the paper “Concreteness” by Freyd.

Of course this occours just very often, thus a non concretizable category must have “very huge” objects.

For example, a category with a generator is always concrete, this is the case of all the categories you have in mind.

To get natural counterexample to concreteness you can consider the category of presheaves over a big category. The fact that it is not locally small implies that it is not concretizable (in this case).


Isbell's example of a (locally small) non-concretizable category, although arguable artificial as Fosco says in his answer, is fairly simple and it is quite easy to see concretely (sorry!) what makes it work.

The category has two classes $\left\{A_\alpha\right\}$ and $\left\{B_\alpha\right\}$ of objects indexed by the same proper class, and two more objects $X$ and $Y$.

There are morphisms $A_\alpha\to X$, $A_\alpha\to Y$, $X\to B_\alpha$ and $Y\to B_\alpha$ for each $\alpha$, and the only morphisms are the identity morphisms and compositions of these, with the compositions $A_\alpha\to X\to B_\alpha$ and $A_\alpha\to Y\to B_\alpha$ equal (but not the compositions $A_\alpha\to X\to B_\beta$ and $A_\alpha\to Y\to B_\beta$ for $\alpha\neq\beta$).

This is a locally small category, with at most two morphisms between any pair of objects.

Suppose $F$ were a faithful functor from this category to sets, and for a given $\alpha$ let $f_\alpha:F(X)\amalg F(Y)\to F(B_\alpha)$ be the function induced by $X\to B_\alpha$ and $Y\to B_\alpha$.

There is only a set of equivalence relations on $F(X)\amalg F(Y)$, so for some $\alpha\neq\beta$ the equivalence relation induced by $f_\alpha$ (i.e., $\left\{(s,t)\in F(X)\amalg F(Y)\middle\vert f_\alpha(s)=f_\alpha(t)\right\}$) must be the same as that induced by $f_\beta$.

But then, since $F(A_\alpha)\to F(X)\to F(B_\alpha)$ is equal to $F(A_\alpha)\to F(Y)\to F(B_\alpha)$, we must also have that $F(A_\alpha)\to F(X)\to F(B_\beta)$ is equal to $F(A_\alpha)\to F(Y)\to F(B_\beta)$, contradicting faithfulness.


Isbell constructed an example of a non-concrete category, but it was pretty artificial.

Freyd showed later that the category $\bf hTop$ of spaces and homotopy classes of maps is not concrete.

Our work mentioned by Paul Frost in the comments, shows that there is nothing special about spaces. What destroys concreteness is "having a sufficiently powerful homotopy theory", i.e. (almost equivalently) the fact that a category has a non-trivial model structure.

Hope it helps!

  • $\begingroup$ You did not reference what "our work" meant, since I have already seen you mention it, perhaps you could give a link to it ? $\endgroup$ – Maxime Ramzi Jun 24 '18 at 16:16
  • $\begingroup$ @Max I think Fosco is talking about the paper linked by Paul Frost in the comments, of which he is a coauthor. But it would be good to add the link directly in the answer. $\endgroup$ – Arnaud D. Jun 24 '18 at 16:26
  • $\begingroup$ @ArnaudD. My bad ! (though you are right, it would be good to add the link to the answer) $\endgroup$ – Maxime Ramzi Jun 24 '18 at 16:30
  • $\begingroup$ My bad! I could have added the link to our paper but thought it was redundant since Paul Frost's comment already mentioned it. $\endgroup$ – fosco Jun 24 '18 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.