Local cross-sections for free actions of finite groups

Question

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.

Answer 1

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.

Answer 2

There is a general cohomological lower bound for the Schwarz genus of a map $p:E\to B$. Namely, if there are cohomology classes $x_1,\ldots , x_k\in H^*(B)$ such that $0=p^*(x_i)\in H^*(E)$ for all $i=1,\ldots , k$ and $x_1\cup\cdots \cup x_k \neq 0$, then the genus of $p$ is greater than $k$. Here the coefficients are completely arbitrary, in particular can be twisted. (This is a generalisation of the cup-length lower bound for Lusternik-Schnirelmann category, since the LS-category of a space $X$ is equal to the genus of any fibration over $X$ with contractible total space.)

So you can get many counter-examples using this cohomological criterion. In fact, whenever $X$ is a contractible CW-complex then it is a model for $EG$, and $X/G$ is a model for $BG$. The cup-length of $BG$ is always infinite for a finite group $G$ (with appropriately chosen, possibly twisted coefficients). This generalises the example in Hannes Thiel's answer.