Lifting an automorphism to the universal covering space..

Question

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 $\overline{\phi}\in \mathrm{Aut}(Y)$?

score 4 · Answer 1 · answered Jun 03 '13 at 12:34

4

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.

answered Jun 03 '13 at 12:34

ronno

11,390

Why the downvote? – ronno Aug 29 '14 at 11:15

Idan · Answer 2 · 2013-02-10T15:51:44.317

1

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.

edited Feb 10 '13 at 15:51

answered Feb 09 '13 at 20:40

Idan

782

Isn't $p_{*}(\pi_{1}(Y))$ trivial in your answer? – M. K. Feb 09 '13 at 23:11
@M.K. yes, but so is $f_\ast(\pi_1(Z))$. – Martin Feb 09 '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 04:21
It is, my mistake... – Idan Feb 10 '13 at 15:49
Thanks for clarifying the point. – M. K. Feb 13 '13 at 19:20

Lifting an automorphism to the universal covering space..

2 Answers2