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Let $X$ be a manifold and $Y$ be its universal covering. Is it true that any $\phi \in \mathrm{Aut}(X)$ can be lifted to $\overline{\phi}\in \mathrm{Aut}(Y)$?

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2 Answers 2

Generally, if p is the cover map, the a function $f:Z\rightarrow X$ can be lifted iff $f_*:\pi_1(Z)\rightarrow\pi_1(X): f_*(\varphi)=f\circ\varphi$ satisfies $f_*(\pi_1(Z))\le p_*(\pi_1(Y))$. The universal covering has a trivial fundamental group, so unless $f_*(\pi_1(Z))$ is trivial, it doesn't seem like we can lift the map.

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Isn't $p_{*}(\pi_{1}(Y))$ trivial in your answer? –  M. K. Feb 9 '13 at 23:11
    
@M.K. yes, but so is $f_\ast(\pi_1(Z))$. –  Martin Feb 9 '13 at 23:47
    
I may misunderstand something. We are interested in the case $Z=X$ and $f \in \mathrm{Aut}(X)$. To me the condition looks like $\pi_1(X) \le p_*(\pi_1(Y))=\{0\}$. –  M. K. Feb 10 '13 at 4:21
    
It is, my mistake... –  Idan Feb 10 '13 at 15:49
    
Thanks for clarifying the point. –  M. K. Feb 13 '13 at 19:20

In general, if you have a map $f \colon X \to X'$, then it lifts to the universal covers $\tilde f \colon Y \to Y'$, since the composition $f \circ p \colon Y \to X'$ satisfies the condition in Idan's answer. To show that in your case, the map is in $\mathrm{Aut}(Y)$, check that this lifting behaves well with composition and identity.

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Why the downvote? –  ronno Aug 29 at 11:15

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