Let $G$ be a finite group, let $X$ be a locally compact Hausdorff space, and let $G$ act freely on $X$. It is well-known that the canonical quotient map $\pi\colon X\to X/G$ onto the orbit space $X/G$ admits local cross-sections. More precisely, for every $z\in X/G$ there are an open set $U$ in $X/G$ containing $z$, and a continuous function $s\colon U\to X$ such that $\pi\circ s$ is the identity on $U$. In particular, there is an open cover of $X/G$ consisting of sets where a local cross-section can be defined.


**Question**: is there a *finite* open cover of $X/G$ consisting of sets where a local cross-section can be defined?


(This is the same as asking whether the Schwarz genus of the fiber map $X\to X/G$ is finite.)


The answer is "yes" if $X$ (or at least $X/G$) is finitistic, so in particular whenever $X$ has finite covering dimension, and clearly also whenever $X$ is compact. I wonder if it is true in general.



Local cross-sections for free actions of finite groups
Let $X=[-1,1]^\infty\setminus\{0\}$, which is a metrizable, locally compact space. Consider the two-element group $G$, and the free $G$-action on $X$ given by $(x\_j)\_{j=1}^\infty\mapsto (-x\_j)\_{j=1}^\infty$. We show that the fibration $X\to X/G$ has infinite Schwarz genus.


Consider the $n$-sphere $S^n$ with the antipodal $G$-action. Then $S^n$ can be embedded equivariantly into $X$ for all $n$. (Use an equivariant map $S^n\to [-1,1]^{n+1}\setminus\{0\}$.) By the Lusternik–Schnirelmann theorem (a strengthening of the Borsuk-Ulam theorem), $S^n$ cannot be covered by $n+1$ closed sets that do not contain antipodal points. It follows that the Schwarz genus of $S^n\to S^n/G$ is at least $n+2$. Since the Schwarz genus of $X\to X/G$ is an upper bound for the Schwarz genus of $S^n\to S^n/G$, it follows that $X\to X/G$ has infinite Schwarz genus.


