4
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\mathbb{R}^2\to T^2$. $\endgroup$ – Kyle Jul 25 '14 at 12:09
  • $\begingroup$ @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. $\endgroup$ – Quang Hoang Jul 26 '14 at 4:06
  • $\begingroup$ 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)? $\endgroup$ – Kyle Jul 26 '14 at 5:03
  • $\begingroup$ 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). $\endgroup$ – Quang Hoang Jul 26 '14 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.