Complement of an n-simplex in a triangulated n-sphere

Question

Let $\Sigma^n$ be a finite simplicial complex homeomorphis to $S^n$. Is it true that, after removing one $n$-simplex, the rest is a triangulated disc? (By that I mean a~simplicial complex homeomorphic to $D^n$)

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My attempts: First I thought it is obvious; then I realised that an analogy is not true in the continuous category (the complement of the Alexander horned 3-disc in $S^3$ is wild). But still, if $\sigma$ is an $n$-simplex in a finite triangulation $\Sigma^n$, then $(\Sigma^n, \sigma)$ has the homotopy extension property. Note sure how to continue..

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After edit: The question is partially motivated by this:

Is "recognising a triangulated disc" a decidable problem?

If the above conjecture is true, then the undecidability of sphere-recognition in dimension $\geq 5$ implies the undecidability of disc recognition.

Answer 1

If you remove a $n$-simplex, the result is a deformation retracts of $S^n$-point which is homeomorphic to $\mathbb{R}^n$ via the stereographic projection.

Answer 2

There are several questions one can ask (and answer) along the lines of your question. The more natural one is:

Let $M$ be a simplicial complex homeomorphic to $S^n$, let $\Delta$ be an open $n$-simplex in $M$. Is it true that $M- \Delta$ is PL homeomorphic to $D^n$ (the standard closed $n$-disk with the standard PL structure).

One can also ask a weaker question (and this is what you seem to be asking, except you are not making it clear if your simplex is open or closed), namely, if $M- \Delta$ is homeomorphic to the closed $n$-dimensional ball $D^n$.

I will answer the second question, but with a bit more care, the same proof will go through for the first one. A good reference for all this is

[RS] Rourke and Sanderson "Introduction to Piecewise-Linear Topology".

I will try to consistently use $\Delta$ for an open n-dimensional simplex and $\bar\Delta$ for its closure. I will use the notation $$ A=\{x\in R^n : 1/2\le |x|\le 1\} $$ for the $n$-dimensional annulus and $D^n$ for a closed unit ball in $R^n$ (centered at the origin).

1. First of all, if $N$ is a regular neighborhood of a simplex $\bar\Delta$ in a triangulated manifold $M$, then $N$ is PL homeomorphic to $D^n$. In order to prove this, first verify that $N - \Delta\cong A$ (Corollary 3.18 of [RS]). Then construct a PL homeomorphism $N\to D^n$ by sending $\bar\Delta$ to the ball of radius $1/2$ and then extend the boundary map to $N - \Delta$ sending it to the annulus $A$.

2. Given this, the problem reduces to the following: Let $\Delta_2$ be an open facet of a triangulated manifold $M$ homeomorphic to the $n$-dimensional sphere $S^n= R^n \cup \{\infty\}$, let $\bar\Delta_1 \subset \Delta_2$ be a closed simplex. We need to show that $M- \Delta_1$ is homeomorphic to $D^n$. To prove this, take a homeomorphism which sends $\bar\Delta_2$ to a unit ball $D^n\subset R^n$ and takes $\bar\Delta_1$ to a concentric ball $\frac{1}{2}D^n$ of half the radius (see Part 1). Now, I leave it to you to check that $$R^{n}\cup\{\infty\} - int(\frac{1}{2}D^n)$$ is homeomorphic to $D^n$. (Hint: Use the inversion.)