Are different categories of manifolds non-equivalent (as abstract categories)?

Question

Consider, for instance, the categories of $C^k$-manifolds, where $k=0,1,2,...,\infty,\omega$. ($C^\omega$ means real analytic.) Are these categories pairwise non-equivalent?

Of course, the obviuos forgetful functor is not an equivalence, because it is not full: for $k<l$, there are $C^k$ functions which are not $C^l$.

I was just thinking about possible ways of proving non-equivalence of categories and came up with this question. For example, one can prove non-equivalence by showing that limits or colimits of certain shape exist in one category but not in the other. I don't know whether one can deal with manifolds in this way.

What I can prove is that the category of complex-analytic manifolds is not equivalent to any of the above.

This follows from the fact that any real manifold other than the point (which is the terminal object) has non-trivial automorphisms. For $C^k$-manifolds with $k=0,1,...,\infty$ this is very easy to show, as there is a $C^\infty$-diffeomorphism of the ball $|x|<1$ which is the identity near its boundary sphere $|x|=1$. (One can construct such a diffeomorphism uning bump functions.) This does not work for the real analytic case. However, by a theorem of Grauert, for any real analytic manifold $M$ there is a closed analytic embedding $i$ of $M$ into some $\mathbb R^n$. Then one can choose a non-zero vector tangent to $i(M)$ at some point, extend it to a constant vector field along $i(M)\subset\mathbb R^n$ and project this vector field orthogonally to get a bounded vector field on $M$ which is not identically zero. As $M$ is complete in the induced metric (because $\mathbb R^n$ is complete and $i$ is closed), the flow of this vector field will be globally defined and thus it will contain a non-trivial analytic self-diffeomorphism of $M$.

On the other hand, there are lots of complex analytic manifolds with no non-trivial automorphisms. For example, a generic compact Riemann surface of genus $\geq 3$ is known to have this property.

Answer 1

Your method of looking for automorphisms of objects works for $C^k$ manifolds as well, $0 < k \le \infty$. This follows from a result of Filipkiewicz (which I found in Kathryn Mann's excellent survey):

If $M, N$ are smooth manifolds without boundary and $\varphi: \text{Diff}^p(M) \to \text{Diff}^q(N)$ is an isomorphism of groups (with $1 \le p, q \le \infty$), then $p = q$ and $\varphi$ is induced by a $C^p$ diffeomorphism $w: M \to N$.

(Recall that every $C^p$ manifold for $p > 0$ carries a compatible smooth structure, so the condition that $M,N$ are smooth manifolds is harmless --- cf. Hirsch, Differential Topology, chapter 2).

If $F: \mathsf{Man}_p \to \mathsf{Man}_q$ is an equivalence of categories, then $F$ induces an isomorphism $\text{Diff}^p(M) \cong \text{Diff}^q(F(M))$, and hence $p = q$ and $M \cong F(M)$. Not only do we have $p = q$, we also see that any autoequivalence of $\mathsf{Man}_p$ is the identity on isomorphism classes. Probably one can say something stronger (maybe they're all naturally isomorphic to the identity).

This handles all but the case where one of $p,q$ is zero or $\omega$. It seems plausible to me that Filipkiewicz' proof goes through when $p = 0$ (though I haven't read carefully enough to say this confidently). I don't know about the case of $\omega$, and Filipkiewicz asks about this as the final question in his paper.