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Let $P$ be a polytope of dimension $n$, and let $\mathcal{C}$ be a polyhedral subdivision of $P$. That means that $\mathcal{C}$ is a finite collection of polytopes whose union is $P$, such that the intersection of any two elements of $\mathcal{C}$ is a common face of both and is itself contained in $\mathcal{C}$.

Let $U$ be a convex open set that is contained in the interior of $P$, and let $A$ be the union of the relative interiors of all elements of $\mathcal{C}$ that touch $U$. That is, $$A := \bigcup_{\substack{Q\in \mathcal{C}\\ Q\cap U\neq \emptyset}} \operatorname{Int}(Q).$$ Note that $A$ is an open subset of the interior of $P$.

I would like to prove that $A$ is homeomorphic to a ball. I would be just as happy to prove the following slightly weaker statement: $$H_k(P,P\setminus A; \mathbb{Q}) \cong H_k(S^n; \mathbb{Q})$$ for all $k$.

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  • $\begingroup$ Sorry this is a vague comment but I think discrete Morse theory (en.wikipedia.org/wiki/Discrete_Morse_theory) is a powerful tool for thinking about this kind of thing $\endgroup$ Commented Apr 14, 2022 at 20:01
  • $\begingroup$ Let $\mathcal{D}$ be the set of $Q$ you are taking union over to define $A$ (i.e., the $Q\in \mathcal{C}$ with $Q\cap U\neq \varnothing$). It seems that two important properties of $\mathcal{D}$ are: if $Q \in \mathcal{D}$ then $Q' \in \mathcal{D}$ for all $Q' \supset Q$; and if $Q$ is a face not on the boundary of $P$ with $Q' \in \mathcal{D}$ for all $Q' \supset Q$, then $Q \in \mathcal{D}$. Is it possible that these properties together already force your $A$ to be an open ball? $\endgroup$ Commented Apr 16, 2022 at 2:47
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    $\begingroup$ @SamHopkins I agree with your first statement; this is how I know that $A$ is open. I don't think that the second statement is true, though. For example, suppose that $P$ is a triangle, and $\mathcal{C}$ is a decomposition of $P$ into three triangles meeting at a single vertex in the middle (so the 1-skeleton is $K_4$). Then it is possible to choose $U$ in such a say that $A$ consists of all three (open) triangles along with two of the three interior edges, but not the third. $\endgroup$ Commented Apr 16, 2022 at 4:18
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    $\begingroup$ I believe that the examples that I really care about do have the property that you describe, but I don't think this property follows from the hypotheses that I wrote in the question. $\endgroup$ Commented Apr 16, 2022 at 4:18
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    $\begingroup$ @TimothyChow: maybe what's tripping you up is that the polyhedral complex $\mathcal{C}$ consists of faces (polytopes) of all dimensions. So the $\mathrm{Int}(Q)$ are relatively open sets (in $Q$), but their union forms a connected, open set. $\endgroup$ Commented Apr 16, 2022 at 13:10

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