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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.

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Proposition 1.33, p.61, of Hatcher's "Algebraic Topology". (Mariano gives the statement below.) – Tom Church Jan 25 '10 at 5:40
Peter May has a really excellent book available for free on his website called "A Concise Introduction to Algebraic Topology". If you're familiar with commutative diagrams and categorical language, you're probably better off with that book than hatcher. (This is with respect to your book request below.) – Harry Gindi Jan 25 '10 at 8:56
@fcqc May's book is terrific,but boy,it would be a struggle to learn from it by May's own admission. I wish he'd write a second edition that had some pictures (not many,but some) and a lot more exercises.That would make the book a lot more usable as a textbook by well-prepared graduate students and still preserve it's sophisticated nature. I was told at this message board that he's actually working on a sequel that should appear soon.Can't wait for that. – The Mathemagician Apr 6 '10 at 23:30
Continued-Indeed-if May combined the 2 texts into a single book with some well chosen pictures and lots of exercises,I believe it would become THE standard text on the subject for graduate students for a generation. – The Mathemagician Apr 6 '10 at 23:31
up vote 8 down vote accepted

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.

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Thank you, I understand the obstruction now. In your second paragraph auto should be changed to homeo. I would appreciate if someone can provide a textbook reference for this statement. – Zarathustra Jan 25 '10 at 3:08
See Prop 1.33 of Hatcher's book. – j.c. Jan 25 '10 at 17:25

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)

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Mariano- you've given the criterion for the lifting to also take a chosen basepoint in the covering space to that basepoint. If you haven't made choices of basepoints, you should never have a criterion involving actual subgroups of $\pi_1$, since those aren't well-defined. – Ben Webster Jan 25 '10 at 8:58

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