What shapes, with boundary collapsed to a point, are homeomorphic to $S^n$? Consider the following construction:

Given a set $A\subseteq\Bbb R^n$, form the quotient space $A/\sim$ which identifies all the points on the boundary $\partial A$ (w.r.t $\Bbb R^n$).

For which sets $A$ is the resulting topological space homeomorphic to $S^n$? Obviously this works for $D^n$ (the unit ball in $\Bbb R^n$), as well as $[0,1]^n$ and $\Delta^n$ (the simplex). But I think it will also work on much more complicated sets, like the Koch curve (with interior). In 2D I believe the Riemann mapping theorem will help to construct a homeomorphism, but I don't know whether that generalizes to $\Bbb R^n$.
Some necessary conditions:


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*$A^\circ$ must be path-connected, because there is a path on $S^n$ connecting any two points and avoiding the pole. More generally, $A^\circ$ must be homeomorphic to $\Bbb R^n$, so it must in fact be simply connected.

*For similar reasons, $A$ cannot be nowhere dense.

*$A^\circ$ must be bounded. If not, take some sequence $(x_n)\in A^\circ$ that diverges to infinity (and satisfies $d(x_m,x_n)\ge1$ for $m\ne n$), and take a subsequence that converges in $S^n$ (necessarily to the pole). Then $A\setminus\bigcup_n\bar B(x_n,\frac12\min(d(x_n,\partial A),1))$ is open in $A$, because all the closed sets in the union are separated from each other, and it contains $\partial A$, hence the image is an open set containing the pole and missing all the $x_n$'s, a contradiction.

 A: Necessary and sufficient conditions are that $A\supseteq\overline{A^\circ}$, $\overline{A^\circ}$ is compact, and $A^\circ$ is homeomorphic to $\mathbb{R}^n$.  First, suppose these conditions hold.  Note that $A/{\sim}\cong\overline{A^\circ}/{\sim}$, so we may assume $A=\overline{A^\circ}$ and in particular that $A$ is compact.  Then since $\sim$ is a closed equivalence relation on a compact Hausdorff space, the quotient $A/{\sim}$ is also compact Hausdorff.  It is also easy to see that the quotient map restricts to a homeomorphism (onto its image) on $A^\circ$.  Thus $A/{\sim}$ is a one-point compactification of a space homeomorphic to $\mathbb{R}^n$, and so $A/{\sim}$ is homeomorphic to $S^n$.
Conversely, suppose $A/{\sim}\cong S^n$.  You have already noted that $A^\circ\cong\mathbb{R}^n$.  If $\overline{A^\circ}$ is not compact or is not contained in $A$, we can choose a sequence $(x_n)$ in $A^\circ$ that does not accumulate at any point of $A$.  The set $X=\{x_n\}$ is then closed in $A$ and saturated under $\sim$, so its image in $A/{\sim}$ is closed.  But the same reasoning shows that the image of any subset of $X$ is also closed.  Thus the image of $X$ is an infinite closed discrete subset of $A/{\sim}$.  This is a contradiction, since $A/{\sim}\cong S^n$ is compact.
