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I am working on problems in the past Qual exams.

"Let $f:\tilde X\to X$ be a covering map between path-connected and locally path-connected spaces. Suppose $p$ is homotopic to a constant map. Prove that $\tilde X$ is contractible."

I remember doing this before, but I forgot the final touch: we have a homotopy $H_t : \tilde X \to X$ s.t. $H_0(\tilde x)=f(\tilde x)$ and $H_1(\tilde x)=x_0$ constant. We observe that $\operatorname{id}_{\tilde X}$ is a lift of $H_0=f$, hence there is a unique lift $\tilde H_t : \tilde X \to \tilde X$ of the homotopy $H_t$ s.t. $\tilde H_0= \operatorname{id}_{\tilde X}$ and $f\circ \tilde H_1=H_1=x_0$.\

I'm stuck here. It suffices to prove $\tilde H_1$ is constant. But I don't have much information about $\tilde H_1$ except that $\tilde H_t$ is a unique lift of the whole $H_t$. How do I conclude?

I don't know how to use the hypothesis "path-connected and locally path-connected" here since in Hatcher's book, these hypotheses usually needed in the Galois corespondence, which I don't think is relevant here. Even the lift above is due to Proposition 1.30 which does not need these requirements.

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To be honest, I am not entirely sure why $\tilde{X}$ path-connected and $X$ locally path connected are assumed here. It seems like $\tilde{X}$ being connected is sufficient.

As you said, it is sufficient to show that $\tilde{H}_1$ is constant. Now note that since $f \circ \tilde{H}_1 = x_0$ is constant, we have that the image of $\tilde{H}_1$ is contained in the fiber of $x_0$ under the map $f$. But since $f$ is a covering map, this fiber is discrete and since $\tilde{X}$ is connected, the map needs to map into one connected component i.e. $\tilde{H}_1$ is constant.

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  • $\begingroup$ $\tilde X$ connected is in fact sufficient. It is also necessary because contractible spaces are connected. $\endgroup$
    – Paul Frost
    May 2, 2020 at 9:18
  • $\begingroup$ so if $\tilde X$ is connected, the problem is solved. It does not mean this problem is wrong, right? $\endgroup$ May 2, 2020 at 14:59
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    $\begingroup$ Yes, the problem is correct. I'm just saying that it is not necessary to assume that $\tilde{X}$ is path connected, but connected suffices. $\endgroup$ May 2, 2020 at 15:02
  • $\begingroup$ oh right, i thought connected implies path-con, i clearly misremember this $\endgroup$ May 2, 2020 at 15:25

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