I have a closed connected manifold $X$, consider the universal cover $p: \tilde X \rightarrow X$. If I recall correctly, any homotopy $F: X \times I \rightarrow X$ with $F(\cdot, 0) = id_X$ lifts to a homotopy $\tilde F : \tilde X \times I \rightarrow \tilde X$ (again with $\tilde F(\cdot,0) = id_{\tilde X}$).
Given any deck transformation $\psi: \tilde X \rightarrow \tilde X$ (i.e. $p \circ \psi = p$), is there a homotopy $F: X \times I \rightarrow X$ with $F(\cdot,0) = id_X$ that lifts to $\tilde F$ with $\tilde F(\cdot, 1) = \psi$?
My motivating example is $X = \mathbb T^2 = \mathbb R^2 / \mathbb Z^2$, where this is true: $\tilde X = \mathbb R^2$, and the group of deck transformations is $\mathbb Z^2 \simeq \pi_1 (X)$. Any deck transformation $\tilde X \rightarrow \tilde X : x \mapsto x + a$ (for $a \in \mathbb Z^2$) can be obtained by the homotopy $F_a: X \times I \rightarrow X : (x,t) \mapsto x + ta$ which lifts to $\tilde F_a : \tilde X \times I \rightarrow \tilde X : x \mapsto x + ta$. Can this be done for any deck transformation of higher-genus surfaces? Orientable manifolds? Arbitrary manifolds? Is there a "nice" sufficient condition for a manifold that makes it have this property?
Edit:
Recall that deck transformations of $p: \tilde X \rightarrow X$ can be identified with $\pi_1(X)$; denote the isomorphism by $\alpha: Deck(p) \stackrel \sim \rightarrow \pi_1(X)$. I show the following claim:
If $\psi \in Deck(p)$ such that there is a homotopy $F: X \times I \rightarrow X$ with $F(\cdot,0) = id_X$ that lifts to $\tilde F$ with $\tilde F(\cdot,1) = \psi$, then $\alpha (\psi)$ is in the center of $\pi_1(X)$. Specifically, if we want this property to hold for all deck transformations, $\pi_1(X)$ is abelian.
Proof: denote the basepoint of $X$ by $x_0$. Let $g: I \rightarrow X$ be $g(t) = F(x_0,t)$, a representative of the class $\alpha(\psi) \in \pi_1(X,x_0)$, with $g(0)=g(1)=x_0$, and let $f: I \rightarrow X$ be a representative of any class $a \in \pi_1(X,x_0)$. Consider the following:
\begin{equation} G: I^2 \rightarrow X \\ G(s,t) = F(f(s),t) \end{equation}
$F$ lifts to $\tilde F$ which has $\tilde F(\cdot,1) = \psi \in Deck(p)$, so $F(\cdot, 1) = id_X$, and $F(\cdot, 0) = id_X$. Thus, $G(\cdot, 0) = G(\cdot, 1) = f$. $G(0, \cdot) = G(1, \cdot) = g$, since $f(0)=f(1)=x_0$, and $F(x_0, \cdot) = g$.
Going around the square Im(G), we see that $a \alpha(\psi) a^{-1} \alpha(\psi)^{-1} = 1$ in $\pi_1(X)$. This was for arbitrary $a \in \pi_1(X)$, therefore $\alpha(\psi)$ is in the center of $\pi_1(X)$.
The answer is no, with counterexamples being given by the projective planes : $S^n\to \mathbb{R}P^n$ is the universal cover of $\mathbb{R}P^n$ for $n\geq 2$, and $-id : S^n\to S^n$ is the only nontrivial deck transformation of this cover.
However, when $n$ is even (e.g. $n=2$), this has degree $-1$ and is thus not homotopic to $id$.