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

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    $\begingroup$ I think there is a uniqueness statement in the lifting theorem. Then it should follow. $\endgroup$ – mland Aug 24 '12 at 20:55
  • $\begingroup$ Lifts are in general not unique. For example the group of covering transformations, which is isomorphic to $\pi _1 (X)$, is the set of all lifts of $id_{X}$. $\endgroup$ – drunken_monkey Aug 24 '12 at 21:07
  • $\begingroup$ @drunken_monkey: Yes, but this is the only obstruction to uniqueness. There exists a unique lifting of $f$ such that $f(y_0) = y_1$ for given $y_0, y_1 \in p^{-1}(x_0)$. So we have uniqueness modulo to the group of deck transformations. $\endgroup$ – Sam Aug 24 '12 at 21:21
  • $\begingroup$ That´s true... i just looked it up on Massey and in fact the construction in the proof is unique if we fix points as you told! Thanks a lot $\endgroup$ – drunken_monkey Aug 24 '12 at 21:36

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):

enter image description here

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.

  • $\begingroup$ Typo: The middle $(X,x_0)$ in the diagram should actually read $(X,f(x_0))$, instead. $\endgroup$ – Sam Aug 24 '12 at 21:52

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.

  • $\begingroup$ I dont understand why $q$ and $q'$ do exist. What guarantees that a lift of identity is injective/surjective? $\endgroup$ – drunken_monkey Aug 24 '12 at 21:27
  • $\begingroup$ $q$ and $q'$ are deck transformations. I should clarify that we can choose $g$ and $h$ such that they are bijective (thus so is $g\circ h$ and $h\circ g$), and if we make a choice that is not then we clearly do not get a homeomorphism. $\endgroup$ – Alex Becker Aug 24 '12 at 21:40

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