Skeleton is not a functorial construction How can I prove that there is no functor $sk(-): CAT \rightarrow CAT$, that sends a category to 'its skeleton'?
I mean by 'its skeleton' a skeletal category that is equivalent to $C$.
Edit: As a comment says, I know that skeleton defines a pseudo-functor; my question is about why it can't be restricted to define a functor.
 A: To obtain the skeleton construction, we can fix an object in each isomorphism class and fix isomorphisms $\varphi_b:b\to b_0$ to these fixed objects, and these determine the equivalence functor by sending $\beta:b\to y$ to $\varphi_y\circ\beta\circ\varphi^{-1}_b:b_0\to y_0$.
Then, for a functor $F:A\to B$ and arrow $\alpha:a_0\to x_0$ in ${\rm sk}(A)$ we have
$${\rm sk}(F)(\alpha)=\varphi_{Fx_0}\circ F(\alpha)\circ\varphi^{-1}_{Fa_0}\,.$$
Note that there's no interaction assumed between $F$ and the skeleton functors, i.e. $Fa_0$ need not be in the skeleton of $B$ even though $a_0$ is taken from the skeleton of $A$.
For simplicity of a counterexample, assume $A$ and $C$ are skeletonial categories, so all $\varphi_x$ are the identities for $x$ in $A$ or $C$, and consider functors $F:A\to B$ and $G:B\to C$.
We then have ${\rm sk}(GF)=GF$.
On the other hand, for an arrow $\alpha:a\to x$, consider the commutative square and its image under $G$.
$$\matrix{Fa&\overset{\varphi_{Fa}}\to&(Fa)_0\\ \downarrow&&\downarrow\\ 
Fx &\underset{\varphi_{Fx}}\to&(Fx)_0}$$
The point is that $G(\varphi_{Fa})$ and $G(\varphi_{Fx})$ can be any automorphisms $GFa=G(Fa)_0\to GFa$ and $GFx\to GFx$, so we obtain
$${\rm sk}(G)\,{\rm sk}(F)\,(\alpha)\ =\ 
G(\varphi_{Fx}\circ F(\alpha)\circ \varphi^{-1}_{Fa})$$
which might well differ from $GF(\alpha)$ by automorphisms on both ends.
