Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer
    
Moreover (although I haven't checked that everything that needs to commute does), given any OTHER collecton of such choices, you can construct another pseudofunctor S', and this pseudofunctor will be canonically 2-iso to S (if you stare at the lax-naturality square you'd need, the choice for the 2-cell has already been made). –  David Carchedi Jun 10 '10 at 14:39
    
Yes, I'm definitely convinced that using skeleta does not help. I just wanted to know how far they are from being canonical. Can you find any possible variant of the stated theorem which expresses this AND is true? –  Andrea Ferretti Jun 10 '10 at 14:49
    
My gut would be no. The "problem" with working with skeleta isn't mathematical, it is out or practicality (as far as my understanding goes). They are just inconvenient since you have to keep track of all these arbitrary choices you made. Unless however, you don't believe in the axiom of choice, because then, doing what I suggested would be impossible. With choice around, it seems skeleta will always exist as a choice, but perhaps not a good one :-). –  David Carchedi Jun 10 '10 at 14:57
    
P.S. How far are they from canonical? Look at my first comment. ANY collection of completely arbitrary choices as in my answer, will yield such a pseudofunctor, and they will all be 2-iso. This is about as far from canonical as you can get. –  David Carchedi Jun 10 '10 at 15:01
1  
One should be careful to distinguish between "natural/functorial," in the precise sense of category theory, and "canonical" in the intuitive sense. As this example shows, they can be completely different notions. –  Mike Shulman Jun 10 '10 at 18:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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