# Does a group action lifted to the universal cover commute with the fundamental group action?

Question: Let $\varphi \colon G \to \text{Homeo}(X)$ be a group action on a topological space $X$ with basepoint $x_0$ and universal covering $\pi \colon \widetilde{X} \to X$. Then the subgroup of lifts $$L = \{ (\widetilde{f},g) \in \text{Homeo}(\widetilde{X}) \times G : \widetilde{f} \circ \pi = \pi \circ \varphi(g)\}$$ constitutes a group extension $$1 \to \pi_1(X,x_0) \to L \to G \to 1,$$ where we map the homotopy class of a loop to the associated deck transformation in $L$.

Now a right split $\sigma \colon G \to L$ of this extension is equivalent to a lift of the group action $\varphi$ to $\widetilde X$.

If a lifted group action exists and furthermore the extension is trivial (i.e. a product), does the lifted group action commute with the action of the fundamental group by deck transformations?

Motivation/Background: If $\varphi(g)$ is homotopic to the identity for all $g$ and the fundamental group has trivial center, then one can construct a lifted group action using the homotopy lifting lemma and see that the lifted action commutes with the action of $\pi_1(X,x)$. Also in this case the extension is a product. So I wonder if this is just a special case for homotopy lifts or a general thing.

In this case $L= \pi_1(X,x_0)\times \sigma(G)$, which implies that $\sigma(G)$ commutes with $\pi_1(X,x_0)$. One remark: You want $X$ to be path connected and admit the universal cover.