Lifting a homeomorphism, always possible? 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.
 A: 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.
A: A necessary and sufficient condition $h$ to lift is that $h_*(p_*(\pi_1(\tilde M)))\subseteq p_*(\pi_1(\tilde M))$. This follows from the usual conditions for a map (in this case $h\circ p:\tilde M\to M$) to lift along a covering, as given in Hatcher's book, for example.
(Notice that this condition does not say that $h$ sends $p_*(\pi_1(\tilde M))$ to a conjugate, but into itself)
