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.

• 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 $Y$. – Rohil Prasad Aug 13 at 20:35
• 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. – Eusebio Gardella Aug 14 at 13:36
• What does it mean for a cover to have finite order? It's locally finite, or uniformly locally finite, or something else? – LSpice Aug 14 at 14:12
• 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. – Eusebio Gardella Aug 17 at 8:27

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.
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.