Let $h:M\to M$ be a homeomorphism of a compact manifold. Let $p:\tilde M\to M$ be a covering. 1) Is it always possible to lift $h$ to $H:\tilde M\to \tilde M$ so that everything fits into the commutative diagram?
2) Given such a diagram assume additionally that p is a self-covering. Is it true that $H$ is necessarily homotopic to $h$?


Thanks, Z.


1/I think I can see that the answer to the second question is "no". Any additional assumptions that would make it into a "yes"?


2/A reference to the proof of the statement in Ben's second paragraph is needed.



Lifting a homeomorphism, always possible?
No. Take any homeomorphism that doesn't preserve the subgroup of $\pi\_1$ that lift to closed paths in the covering. For example, take the 2:1 covering $S^1\to S^1$ take the product with the identity map on $S^1$. Let $h$ be the homeomorphism switching the factors.


In general, I believe a homeomorphism will lift if and only if the associated automorphism of $\pi\_1$ send the subgroup of the covering to a conjugate.


Another way of saying this is that the category of coverings is equivalent to the category of $\pi\_1$-sets, and a homeomorphism will lift if the corresponding twist of the $\pi\_1$-set preserves its isomorphism class.


