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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.

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    $\begingroup$ Here's an idea for making a counterexample. $X$ is a principal $G$-bundle over $X/G$. If there is such a finite cover, then I believe that the $G$-bundle $X$ should extend to a $G$-bundle over the one-point compactification $(X/G)^*$ of $X/G$. Therefore, to find a counterexample, it suffices to find a locally compact, Hausdorff space $Y$ and a $G$-bundle $P \to Y$ such that $P$ does not extend over the one-point compactification $Y^*$ of $Y$. $\endgroup$ – Rohil Prasad Aug 13 at 20:35
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    $\begingroup$ Thanks for your comment. I realize that I misstated my question: I actually want to know if an open cover with finite order exists. I edited the question. $\endgroup$ – Eusebio Gardella Aug 14 at 13:36
  • $\begingroup$ What does it mean for a cover to have finite order? It's locally finite, or uniformly locally finite, or something else? $\endgroup$ – LSpice Aug 14 at 14:12
  • $\begingroup$ I think a finite cover as desired exists iff a finite-order cover exists. So I went back to the original formulation, this time adding a connection to the Schwarz genus. $\endgroup$ – Eusebio Gardella Aug 17 at 8:27
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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.

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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.

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