The following question arose from a survey paper I am writing on
combinatorial reciprocity. Let $\mathcal{P}$ be a $d$-dimensional
convex polytope. Let $\mathcal{Q}$ be a union of facets (codimension
one faces) of $\mathcal{P}$. Suppose that $\mathcal{Q}$ has Euler
characterisic $1$, and that the local Euler characteristic about any
point $p$ of $\mathcal{Q}$ is either $1$ or $1+(-1)^d$. Does it follow
that $\mathcal{Q}$ is homeomorphic to a ball?


If the answer is negative, what if we assume instead that $\mathcal{Q}$
is acyclic and the local homology at each point is the same as that of
a $(d-2)$-ball or $(d-2)$-sphere?



Topology of a union of facets of a convex polytope
Here's an $(n+1)$-dimensional polytope that one can use to construct countereaxamples:  

The convex hull of the points
$$
\left(x\_1,x\_2,\ldots,x\_n,\sum\_{1=1}^nx\_i^2\right)
$$
where $x\_i\in \{-N,\ldots,N-2,N-1,N\}$, for some fixed $N\gg1$.


This polytope's boundary contains an embedded copy of a large chunk of $\mathbb R^n$ with its standard cubulation.


Now fix any cell complex with Euler characteristic $1$ that embeds in $\mathbb R^n$. Thicken it so as to make another space that deformation retracts to it, and which is a union of unit cubes. If the thickening is an $n$-dimensional manifold with boundary, then you've got your counterexample.


