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?

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.
