# Null-homotopic covering space map

I'm stuck with the following question, which looks quite innocent.

I'd like to show that if a covering space map $f:\tilde{X}\to X$ between cell complexes is null-homotopic, then the covering space $\tilde{X}$ must be contractible.

Since $f$ is null-homotopic there exists a homotopy $H_t:\tilde{X}\to X$ from $H_0=x_0$ to $H_1=f$ and I would like to use it to construct another homotopy $G:\tilde{X}\to \tilde{X}$ from $G_0=\tilde{x}_0$ to $G_1=Id_{\tilde{X}}$.

By the homotopy lifting property, $H_t$ lifts to a homotopy $\tilde{H}_t:\tilde{X}\rightarrow \tilde{X}$ such that $H_t(x)=f(\tilde{H}_t(x))$ and $\tilde{H}_0(x)=\tilde{x}_0$

So we have a homotopy $\tilde{H}_t:\tilde{X}\rightarrow \tilde{X}$ from $\tilde{H}_0(x)= \tilde{x}_0$ to $\tilde{H}_1(x)$ and besides $f(x)=H_1(x)=f(\tilde{H}_1(x))$.

If $f$ was injective we would be done, but in principle $\tilde{H}_1(x)$ could be any point in $f^{-1}(x_0)$ right?

As you said, we're given a covering $f: \tilde X \to X$ and a homotopy $H_t : \tilde X \to X$ such that $H_0(\tilde x)=f(\tilde x)$ and $H_1(\tilde x)=x_0$ for some fixed $x_0 \in X$. Since $\operatorname{id}_{\tilde X}$ is a lift of $H_0=f$, there is a unique lift $\tilde H_t : \tilde X \to \tilde X$ of the homotopy $H_t$ such that $\tilde H_0= \operatorname{id}_{\tilde X}$. Since $f\circ \tilde H_1=H_1$ is a constant function and $f$ is a local homeomorphism, $\tilde H_1$ is locally constant. Since $\tilde X$ is connected, $\tilde H_1$ is constant, i.e. $\tilde X$ is contractible.
Note: We must assume that $\tilde X$ is connected. To see this, consider any covering $\tilde X= X \sqcup X \to X$, where $X$ is contractible.
Since $f$ is nullhomotopic, $f_*:\pi_n(\tilde X)\to \pi_n(X)$ are trivial for all $n$. Consequently $\pi_n(\tilde X)$ are all trivial. Whitehead theorem implies $\tilde X$ is contractible.
• The induced map $f_*$ is definitely an isomorphism of $\pi_n$ for$n\geq 2$. But for $n=1$, this needn't be the case, right? E.g. the (nullhomotopic) covering $\mathbb{R}^2\to T^2$. – Kyle Jul 25 '14 at 12:09
• @squirrel: Yes, it is only true that $f_*$ is injective on $\pi_1$ and that $\pi_1(X)/\pi_1(\tilde X)$ is equal to $f^{-1}(x_0)$ as sets. – Quang Hoang Jul 26 '14 at 4:06
• Ah, good point. So we still get $\pi_n(\tilde X)=0$ for $n\geq 1$. One other question, since I'm trying to understand Whitehead's theorem: Any constant map will induce the necessary isomorphisms on $\pi_n$, setting up Whitehead's theorem. But do we need to further assume $\tilde X$ is a CW complex (or has the homotopy type of a CW complex)? – Kyle Jul 26 '14 at 5:03
• In this case, we may not need $\tilde X$ being a CW complex (as shown in your answer). However, in general, it is necessary (for example, double comb space). – Quang Hoang Jul 26 '14 at 14:46