Let $C$ be a locally CAT(0) cube complex. Let $f\colon \Sigma_g \rightarrow C$ be a continuous map from a closed oriented genus $g \geq 1$ surface to $C$. Is it always possible to find a cube complex structure on $\Sigma_g$ such that $f$ is homotopic to a map taking cubes to cubes?
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1$\begingroup$ Do you mean locally CAT(0)? $\endgroup$– Moishe KohanCommented Feb 15, 2023 at 18:02
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1$\begingroup$ @MoisheKohan: Whoops, yes, that's definitely what I meant. I'll change the question to reflect this. $\endgroup$– UrsulaCommented Feb 15, 2023 at 18:19
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1$\begingroup$ @SamNead: Yes, I'm totally fine with collapsing a cube to a lower-dimensional cube, so a constant map is fine (just choose any cube complex structure on $\Sigma_g$, and then homotope $f$ to a constant map taking $\Sigma_g$ to a vertex). $\endgroup$– UrsulaCommented Feb 15, 2023 at 22:46
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1$\begingroup$ I think that there are examples where there exists a pretty nice cube structure on the surface, but it is not locally CAT(0). Is that a problem? Here is the example - Let $C$ be the three-torus with its usual cube structure (one vertex, three edges, three faces, one three-cell) and let $S$ be the two-torus, mapped in linearly to be orthogonal to the vector $(1, 1, 1)$. Then we can make $S$ dual to the cubes (and get triangles and hexagons) or we can homotope $S$ into the two-skeleton (and get a pretty lozenge tiling). In neither case do we get a locally CAT(0) cube structure on $S$. $\endgroup$– Sam NeadCommented Feb 16, 2023 at 12:23
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1$\begingroup$ @SamNead: I definitely don't expect the cube complex structure on the surface to be locally CAT(0). $\endgroup$– UrsulaCommented Feb 16, 2023 at 15:07
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This is not an answer but instead is a "story" to show that the answer is perhaps "no".
Suppose that $C$ is the given locally CAT(0) cube complex. Suppose that $S$ is the given surface of genus $g \geq 1$. Suppose that $f \colon S \to C$ is the given continuous map.
Suppose, as a very special case, that $f_* \colon \pi_1(S) \to \pi_1(C)$ is injective. Then we homotope $f$ to be transverse to the cell structure of $C$, use Stallings-like techniques to homotope $f$ into good position, and pull-back a good structure on $S$. (There are of course many details to fill in here. The most troubling is that Stallings wants $f$ to be transverse, while the original question wants the image of $f$ to lie in the two-skeleton.)
Suppose more generally that $f_*$ is not injective. If there is a simple closed curve in the kernel, then we can crush that, send it someplace standard, and then induct.
That is, this sort of inductive plan of attack seems (?) to need a version of the simple loop conjecture, where the target is a locally CAT(0) cube complex. Unfortunately the simple loop conjecture is surely (?) false in high dimensions, even when the target is a very nice space (say a hyperbolic four-manifold).
As HJRW points out, we do not need high dimensionality to reach a possibly concerning place. A lovely example of the simple loop conjecture, when $S$ is a genus $g$ surface and $C$ is the product of a pair of rank $g$ graphs, appears on pages 85 and 86 of Stallings paper How not to prove the Poincaré conjecture. This connection deserves to be better known!
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2$\begingroup$ Even more worryingly, the “simple loop conjecture” when the target is a product of two graphs implies the Poincaré conjecture! $\endgroup$– HJRWCommented Feb 16, 2023 at 17:57
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$\begingroup$ @HJRW - Ah - is this "How not to prove the Poincare conjecture"? $\endgroup$– Sam NeadCommented Feb 17, 2023 at 9:00
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2$\begingroup$ Exactly! Indeed, on that subject, the recent preprint of Sadanand is surely relevant to this question. arxiv.org/abs/2301.09527v1 $\endgroup$– HJRWCommented Feb 17, 2023 at 12:42