Topology of a union of facets of a convex polytope 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?
 A: 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.
A: Here is a partial answer.  There are finite acyclic 2-complexes that are not contractible.  Any such 2-complex can be embedded in $\mathbb{R}^5$, and a regular neighbourhood of such a 2-complex will be a compact 5-manifold with boundary that is a subcomplex of (a finite piece of) $\mathbb{R}^5$ and that is acyclic and not contractible.  It thus has the property that the local homology at each point is either that of a 4-ball or a 4-sphere.  Take a larger piece of the given triangulated  $\mathbb{R}^5$ that is homeomorphic to a 5-ball and that contains the given non-contractible piece as a subcomplex.  Now  project the larger ball-shaped piece homeomorphically onto a piece of a 5-sphere.   Presumably (and this is why this is only a partial answer), one can do this in such a way that the given triangulation is part of a simplicial convex polytope structure on $S^5$.  Note that there is a lot of freedom in how to choose the original triangulation of $\mathbb{R}^5$, so I don't see why this last step would fail.
