TITLE: Lift of homeomorphism is homeomorphism
QUESTION [5 upvotes]: Suppose $p:Y \rightarrow X$ is the universal covering map of $X$.
Given a contiunuous $f: X \rightarrow X$ then a well known theorem for existence of lifts states that there exist a continuous lift
 $$\tilde f : Y \rightarrow Y \text{ with } p \circ \tilde f = f \circ p.$$
If we additionally suppose f is a homeomorphism. Then I think $\tilde f$ is a homeomorphism, too. I tried to prove that but failed. My idea was to take a lift of $f^{-1}$ and compose it with $\tilde f$ in order to obtain a lift of the identity. But that´s not really helping here. Anybody knows how to prove or disprove that?
Thanks in advance

REPLY [2 votes]: If we fix $x_0 \in X$, $y_0\in p^{-1}(x_0)$ and let $y_1 = \tilde f(y_0)$, then there is a unique lifting of $f$ such that $\tilde f(y_0) = y_1$ (for a reference and sketch of proof, see here). By lifting $f^{-1}$ to $Y$ in such a way that $\widetilde {f^{-1}}(y_1) = y_0$, we then obtain the following commuting diagram (of pointed spaces):

The $\mathrm{id}_Y$-arrow follows from the fact that $\mathrm{id}_Y$ is a lifting of $\mathrm{id}_X$ and therefore is the unique such lifting.
It follows that $\widetilde {f^{-1}} \circ \tilde f = \mathrm{id}_Y$ and with a similar diagram we also see that $\tilde f  \circ \widetilde {f^{-1}}= \mathrm{id}_Y$. So $\tilde f$ is indeed a homeomorphism.