Lets say I take an arbitrary closed and smooth $d$-manifolds $\mathcal{M}$. Now, it is a well-known fact that whenever I take two (sufficiently nice embedded) closed $d$-balls $B_{1}$ and $B_{2}$ in $\mathcal{M}$, then the manifold obtained by cutting out the interior of $B_{1}$ is homeomorphic to the manifold obtained by cutting out the interior of $B_{2}$ from $\mathcal{M}$. In other words, for a closed manifold, it does not depend on where I cut out a ball (up to homeomorphism). This is basically a consequence of the famous and highly non-trivial annulus theorem. For dimension 2 and 3, it follows from the famous triangulation theorems by Radó (1924) and Moise-Bing (1952,1959), but for higher dimension it was only proven in 1969 by Kirby ($d>4$) and in 1982 by Quinn (for d=4).

I am wondering if some similar statement is true for pseudomanifolds. By a pseudomanifold, I mean the following:

Let $\Delta$ be a finite abstract $d$-dimensional simplicial complex. Its geometric realization $\vert\Delta\vert$ is $d$-dimensional

pseudomanifold(without boundary), if the following conditions are fulfilled:

- It is "pure", i.e. every simplex $\sigma\in\Delta$ of dimension $<d$ is the face of some $d$-simplex.
- It is "non-branching", i.e. every $(d-1)$-simplex is face of exactly two $d$-simplices.
- It is "strongly connected", i.e. for every two $d$-simplices $\sigma,\tau\in\Delta_{d}$, there is a sequence of $d$-simplices $\sigma=\sigma_{1},\sigma_{2},\dots,\sigma_{k}=\tau$ such that $\sigma_{l}\cap\sigma_{l+1}$ is a $(d-1)$-simplex $\forall l$.

Obviously, every PL-manifold is a pseudomanifold, but not vice versa. A famous example is the pinched torus, which is obtained by identifying two distinct points of the $2$-sphere.

So, **my question is**,

when I remove a closed $3$-ball inside a closed pseudomanifold (for example, by deleting the interior of a single $3$-simplex or any subcomplex PL-homeomorphic to it), does the result depend on where I cut the ball (up to PL-homeomorphism)?

I am mostly interested in the $3$-dimensional case, which is usually a little bit easier than the higher-dimensional cases...