TITLE: Intuition of non-concretizable categories?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.