Equivariance of lifts to universal cover Let's say you have a homotopy equivalence $f: X \rightarrow Y$. Consider the lift $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ to their universal covers. Let $\gamma$ be an element of the fundamental group of $X$, which can be identified as an element of the deck group of its cover. Why is it that $\widetilde{f} \circ \gamma = f_{*}(\gamma) \circ \widetilde{f}$? Also, is there some analogous equivariance if the cover isn't necessarily normal?
 A: Consider a lift $\widetilde f : \widetilde X \to \widetilde Y$ of $f$.
By definition of the deck group, each of the maps $\widetilde f \circ \gamma$ and $f_*(\gamma) \circ \widetilde f$ is a lift of the map
$$F : \widetilde X \mapsto X \xrightarrow{f} Y
$$
The lifting lemma has a uniqueness provision built into it: given two lifts of $F$, if they have the equal values at one point of $\widetilde X$, then those two lifts are equal. So, all one needs to do is pick a point $p \in \widetilde X$ and prove that $\widetilde f \circ \gamma(x) = f_*(\gamma) \circ \widetilde f(x)$; probably that proof is easy with the appropriate choice of $p$.
At this stage, since your question does not specify base points, and since base points are needed downstairs in $X$ and in $Y$ in order to define the fundamental group, and since base points are needed upstairs in $\widetilde X$ and $\widetilde Y$ in order to define the deck groups, I think I'll end with a hint: With good choices of all these base points, the appropriate point $p$ which makes the proof easy is probably closely associated to the choices of the base points.
