Non-canonicity of skeleta

Question

I was mulling on this previous question of mine, and I think I'd like to play the devil's advocate a bit more. I am now convinced that skeleta do not make category theory any simpler, and this is mostly due to the fact that there is no canonical way to construct a skeleton for a given category.

I am now wondering whether this can be made into a precise theorem. I was thinking of something about this lines (this may not be the right formulation):

Let $Cat$ be the $2$-category of small categories. There is no pseudofunctor $S \colon Cat \to Cat$ such that for every $C \in Cat$, $S(C)$ is a skeletal category equivalent to $C$.

Is this true? I have no clue how to prove this.

And if not, is there some variation which puts the intuition that skeleta involve arbitrary choices on a sound basis?

Answer 1

As David says, such pseudofunctors do exist. Moreover, any such pseudofunctor is pseudonaturally equivalent to the identity. However, I think that's the wrong question to be asking, because working with pseudofunctors and pseudonatural transformations means that equivalent categories will be essentially indistinguishable. Just as any functor can be modified by replacing each of its values by an isomorphic object along a specified isomorphism, any pseudofunctor can be so modified along specified equivalences. If you want to be able to distinguish between a category and its skeleton, then you need to talk about something stricter. For instance, here's one statement which may be more along the lines of what you're looking for.

Claim: There is no pseudofunctor $skel:Cat\to Cat$ such that each category $skel(C)$ is skeletal and there is a strict 2-natural transformation $\alpha:skel\to Id$ whose components are equivalences.

Proof Let $C$ be the terminal category and let $D$ be the walking isomorphism with two objects $x,y$. Then $skel(C)\cong C$ and $skel(D)\cong C$, and so the inclusion $\alpha_D : C \to D$ must either pick out $x$ or $y$. Let $f:C \to D$ pick out the other one; then $skel(f)=id$ and so $\alpha_D \circ skel(f) \neq f \circ \alpha_C$, i.e. $\alpha$ is not strictly natural.

One could also ask about strictifying things in other ways, such as looking for a 2-natural transformation $Id\to skel$, or asking whether $skel$ could be made a strict 2-functor.

Answer 2

I believe that your theorem is false. Suppose that for every small category $C$ I make an arbitrary choice of representing object for each isomorphism class. Then, the fullsubcategory on these chosen objects S(C) is skeletal and the inclusion functor i_C is also essentially surjective, hence an equivalence. Choose for each $C$ also a functor $s_C:C \to S(C)$ which is "inverse" to $i_C$- that is also choose two natural isomorphisms $\alpha_C:i_C \circ s_C \to id_C$ and $\beta_C:id_{S(C)}\to s_C \circ i_C$. After these choices are made, suppose I have a functor $F:C \to D$. Then, $s_D \circ F \circ i_C:S(C) \to S(D)$. Call this composite $S(F)$. Now suppose we also have $G:D \to E$. Then $S(GF)=S_E GF i_C$ and by composing with $\alpha_D$ we get a canonical (after these choices have already been made) isomorphism of functors between this and $S(G)S(F)=s_E G i_D s_D F i_C$. The pentagon that needs to commute to show that this is a pseudofunctor is a tautology. Nonetheless, the answers to your last question, should convince you that working with skeleta gains you nothing.