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Let $\widetilde{X}$ be a covering space of $X$ with projection map $p:\widetilde{X}\to X$. We know by definition that, for all $x$ in $X$, there exists a neighborhood $x\in U\subset X$ such that $p^{-1}(U)$ is a collection of disjoint open sets in $\widetilde{X}$, and further each of these open sets are mapped homeomorphically to $U$ by $p$. I would like to know if the same can be said for a path in $X$. The reason I think this could be possible is as follows: Let $f:[0,1]\to X$ be a path in $X$. Then for $0\leq t\leq 1$, $f(t)$ contains a neighborhood about it that lifts homeomorphically to $\widetilde{X}$. The path is compact, so we can cover the path with finitely many of these neighborhoods. If we then look at the inverse image of all of these neighborhoods, it seems as if we should get the desired result.

My main problem with the above reasoning is that if you take $X$ to be the circle $S^1$ with covering space $\mathbb{R}$, using this process on a loop which goes around the circle once would cause your inverse image to be all of $\mathbb{R}$, in which case we don't get a homeomorphic lift.

So I guess my main question is, to what extent can we ask for such homeomorphic lifts?

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  • $\begingroup$ Your reasoning is fine apart from your conclusion that the "process" leads to a contradiction. The inverse image of the loop in your example is indeed all of $\Bbb{R}$ and is indeed covered by open intervals (say of the form $(2n\pi, 2(n+1)\pi)$ and $((2n+1)\pi, (2n+3)\pi)$) that are mapped homeomorphically by $p$. The path can be lifted to $\Bbb{R}$, but a lifting doesn't map $[0, 1]$ onto $\Bbb{R}$, instead, for example, it could map $[0, 1]$ onto $[0, 2\pi]$. (I'm assuming you're mapping $\Bbb{R}$ to $S^1 \subseteq \Bbb{C}$ by $x \mapsto e^x$) $\endgroup$ – Rob Arthan Feb 21 '16 at 19:18
  • $\begingroup$ That should read $x \mapsto e^{ix}$. $\endgroup$ – Rob Arthan Feb 21 '16 at 19:24
  • $\begingroup$ I know what it means to lift a subset $A \subset X$ homeomorphically to $\widetilde X$, namely, a subset $\widetilde A \subset \widetilde X$ such that $p : \widetilde X \to X$ restricts to a homeomorphism $\widetilde A \to A$. I also know what it means to lift a path $f : [0,1] \to X$ to $\widetilde X$, namely, a path $\tilde f : [0,1] \to \widetilde X$ such that $p \circ \tilde f = f$. However, what does it mean to lift a path $f$ homeomorphically to $\widetilde X$? $\endgroup$ – Lee Mosher Feb 21 '16 at 20:41

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