Lift of homeomorphism is homeomorphism 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
 A: 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.
A: Suppose $f:X\to X$ is a homeomorphism. Let $g$ be a lift of $f$ and $h$ a lift of $f^{-1}$. Then $$p\circ h\circ g=f^{-1}\circ p\circ g=f^{-1}\circ f\circ p=p$$
so $h\circ g$ is a lift of the identity. Thus we have functions $q:Y\to Y$ such that $\mathrm{id}_Y=q\circ h\circ g$ and $p\circ q=p$ (this would be better drawn out as a diagram). Then $q\circ h$ is a left-inverse for $g$. Note that
$$p\circ g\circ h=f\circ p\circ h=f\circ f^{-1}\circ p=p$$
so $g\circ h$ is also a lift of the identity. Thus we have a function $q':Y\to Y$ such that $\mathrm{id}_Y=g\circ h\circ q'$ and $p\circ q'=p$. Then $h\circ q'$ is a right-inverse for $g$, hence $q\circ h = h\circ q'$ and these are the inverse of $g$, so $g$ is a homeomorphism.
