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Let $p: \bar{X} \rightarrow X$ be a covering map. Let $Z$ be a connected and locally path connected space. $f: Z \rightarrow X$ and $g: Z \rightarrow X$ are homotopic and $f$ has a lift. Show that $g$ also has a lift.

I thought this is an easy problem and just wanted to make sure if I'm trying to do this properly, because it's a bit suspicious for me, that it was going so smoothly. :)

Here is my solution:

Fix any $z \in Z$. If $f$ and $g$ are homotopic, then in $X$ we have a path from $H(z,0) = f(z)$ to $H(z,1) = g(z)$. Let me call it $\gamma(t) = H(z,t)$. We can fix a set $U \subset X$, which contains $\gamma$. Since $p$ is a covering map, its restriction to an unique set $\bar{U} \subset \bar{X}$ is a homeomorphizm onto $U$. So we have $p^{-1}(\gamma) \subset \bar{U}$. Now, we can define the lift of $g(z)$ for every $z \in Z$ as $\bar{g}(z) = p^{-1}(g(z))$. It's well defined, since $p^{-1}$ is an homeomorpism.

Am I doing it right? In fact I'm not sure how exactly the conectness and local path conectness of Z are used.

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You want to prove that covering maps have the homotopy lifting property. This means that if $p: \bar{X} \rightarrow X$ is a covering map, $H : Z \times I \to X$ is a homotopy (no assumptions on Z!) and $f : Z \to \bar{X}$ is a lift of $H_0 = H(-,0) : Z \to X$, then $H$ has a lift $\bar H : Z \times I \to \bar X$ such that $\bar H_0 = f$.

This is a basic property of covering maps and proved in any textbook dealing with this subject. Therefore I don't think it necessary to give a proof here.

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Here’s a hint in solving the problem.

Let $H:Z\times I\to X$ be a homotopy such that $H(z,0) = f(z)$ and $H(z,1) = g(z)$ for all $z$. Prove the following fact

If $f$ has a lift $\tilde{f}:Z\to\tilde{X}$ then $H$ has a lift $\tilde{H}:Z\times I\to\tilde{X}$ such that $\tilde{H}(x,0) = \tilde{f}(x)$.

and show that the function $\tilde{H}(z,1)$ is a lift of $g$.

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We can fix a set $U \subset X$, which contains $\gamma$. Since $p$ is a covering map, its restriction to an unique set $\bar{U} \subset \bar{X}$ is a homeomorphizm onto $U$.

How do you know that $U$ lifts homeomorphically to $\bar{X}$? Covering maps work locally so when you have a point $z \in Z$ you then get a $U_z$ that lifts homeomorphically. Your path $\gamma(t)$ may not lie entirely inside any such neighborhood and you may not be able to lift the entire thing at once so you have to work point by point.

For example consider $\mathbb{R}$ as a helix covering $S^1$ by projection and the path $f:[0,1] \rightarrow S^1$ tracing out the circle once. Then let $g=f$ and your homotopy be the constant one. $\gamma(t) = f$ in this case but the only $U$ that contains $\gamma$ is $S^1$ which does not lift homeomorphically to $\mathbb{R}$.

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