# Null-homotopic covering map means the covering space is contractible

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.

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.
• $\tilde X$ connected is in fact sufficient. It is also necessary because contractible spaces are connected. Commented May 2, 2020 at 9:18
• so if $\tilde X$ is connected, the problem is solved. It does not mean this problem is wrong, right? Commented May 2, 2020 at 14:59
• 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. Commented May 2, 2020 at 15:02