CAT($0$) cube complex and homeomorphism

Question

If $X$ is a CAT($0$) cube complex and $Y$ is homeomorphic to $X$, I think homeomorphism preserves the cube complex structure, simply-connected and the link condition. So can I say $Y$ is also a CAT($0$) cube complex? i.e. is the CAT($0$) cube complex structure preserved under homeomorphism?

Answer 1

Your question is somewhat unclear, because you said that $X$ is a $\text{CAT}(0)$ cube complex, but you did not say what $Y$ is.

---

However, if your intention was that $Y$ is also a $\text{CAT}(0)$ cube complex, the answer is that it is possible that $X$ and $Y$ are homeomorphic but that no homeomorphism preserves the $\text{CAT}(0)$ cube complex structure.

For example, take $X$ to be the four Euclidean squares fitting together cyclically around a vertex, and then take $Y$ to be five Euclidean squares fitting together cyclically around a vertex. These are both $\text{CAT}(0)$ cube complexes, and they are homeomorphic to each other because both are homeomorphic to the closed unit disc $D^2 \subset \mathbb R^2$. But obviously no homeomorphism preserves the $\text{CAT}(0)$ cube structure, because you can't take four squares to five squares by a bijection.

---

From your comment it seems, on the other hand, that your intention was instead to ask whether, given a $\text{CAT}(0)$ cube complex $X$, a topological space $Y$, and a homeomorphism $f : X \to Y$, one can use $f$ to induce a $\text{CAT}(0)$ cube complex structure on $Y$. The answer is yes: the $n$-cubes of $Y$ are simply the images $f(Q)$ of the $n$-cubes of $X$.