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

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?

• $\gamma$ should be a loop and $\tilde{f}(x)$ should belong to the fiber of the basepoint of the loop $\gamma$, isn't it ?
– paf
Commented Jul 29, 2016 at 9:38
• I agree about the first point, but the fundamental group acts on the whole of the universal covering, if we fix a base point we get a unique isomorphism between the fundamental group and the group of deck transformations. Commented Jul 29, 2016 at 9:40
• Yes, you're right. You're assuming that $X$ and $Y$ are path-connected, I think.
– paf
Commented Jul 29, 2016 at 9:42
• Oh yes, you are right. Good point. I was a bit vague with saying "nice" spaces. Commented Jul 29, 2016 at 11:41

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.