TITLE: Fiber-preserving homotopy equivalences form a group up to homotopy.
QUESTION [5 upvotes]: Let $p\colon \tilde X \to X$ be a covering space. A homotopy equivalence $\tilde f\colon \tilde X \to \tilde X$ is fiber-preserving if $p(\tilde x) = p(\tilde y)$ implies $p\tilde f(\tilde x) = p\tilde f(\tilde y)$. (Notice: this is different from the notion of a "fiber homotopy equivalence"; in particular we do not have $p\tilde f = p$, just that $\tilde f$ induces a map $f\colon X \to X$.)
Let $\operatorname{fhe}(\tilde X)$ denote the space  of fiber-preserving homotopy equivalences of $\tilde X$ in the compact-open topology. I'd like to see that $\pi_0(\operatorname{fhe}(\tilde X))$ is a group. Since $\operatorname{fhe}(\tilde X)$ is an $H$-space under composition of functions, it's clear that we get a monoid, so the only trouble is showing that elements have inverses.
Concretely: given a fiber-preserving homotopy equivalence $\tilde f$, does there exist a fiber-preserving homotopy equivalence $\tilde g$ such that $\tilde g\tilde f$ and $\tilde f\tilde g$ are homotopic to the identity through fiber-preserving homotopy equivalences?

REPLY [4 votes]: I don't think so. Consider a case where $\tilde{X}$ is contractible and $X$ is not, e.g. the universal covering of $S^1$ by $\mathbb{R}$. A constant map $c\colon\tilde{X}\rightarrow\tilde{X}$ is a homotopy-equivalence, since $\tilde{X}$ is contractible, and trivially fiber-preserving. For any fiber-preserving homotopy equivalence $\tilde{g}$, $\tilde{g}c$ is a constant map all the same. So it suffices to realize a constant map is not homotopic to the identity through fiber-preserving homotopy equivalences. Indeed, a constant map is not even homotopic to the identity through fiber-preserving maps, since such a homotopy would induced a homotopy of a constant map on $X$ to the identity on $X$, contradicting that $X$ is not contractible.