Does lifting commute with the right action of the fundamental group?

Question

Let $f \colon X \to Y$ be a map of nice pointed spaces and $\widetilde{f} \colon \widetilde{X} \to \widetilde{Y}$ the lift to the universal covers making $$ \widetilde{X} \xrightarrow{\widetilde{f}} \widetilde{Y} \\ \downarrow \qquad \downarrow \\ X \,\xrightarrow{f} \,Y $$ commute and mapping the base points to each other. Let $x \in \widetilde{X}$ and $\gamma \colon [0,1] \to X$ be a loop in $X$. Do we have

$$ \widetilde{f}(x) \,\cdot\, [f \circ \gamma] = \widetilde{f}(x \cdot [\gamma]), $$ where the dot denotes the right action of the fundamental group on the universal covering?

Answer 1

It suffices to prove that the lift $\tilde{\delta}$ of $\delta := f\circ\gamma$ which begins at $\tilde{f}(x)$ is equal to $\tilde{f}\circ\tilde{\gamma}$, where $\tilde{\gamma} : [0;1]\to \tilde{X}$ is the lift of the loop $\gamma$ which begins at $x$. Once we will have proved this, we'll obtain \begin{eqnarray} \tilde f(x).[f\circ\gamma] &=& \text{endpoint of the lift }\tilde\delta\text{ of }f\circ \gamma\text{ which begins at }\tilde f(x)\\ &= &\text{endpoint of }\tilde{f}\circ\tilde{\gamma}\\ &=& \tilde f(\text{endpoint of }\tilde\gamma)\\ &=& \tilde f(\text{endpoint of the lift of }\gamma\text{ which begins at }x)\\ &=& \tilde f(x.[\gamma]). \end{eqnarray}

But $$\pi_Y\circ (\tilde{f}\circ\tilde{\gamma}) = f\circ \pi_X\circ \tilde{\gamma} = f\circ \gamma$$ hence $\tilde{f}\circ \tilde\gamma$ is also a lift of $f\circ \gamma$ beginning at $f(x)$, thus $\tilde{f}\circ \tilde\gamma = \tilde{\delta}$ because the lift of $f\circ \gamma$ beginning at $f(x)$ is unique.