$S^2$ without $n$ points is homeomorphic to $S^2$ without $m$ points if and only if $n = m$

Question

Consider the unit sphere $S^2$ with the subspace topology of $\mathbb{R}^3$. Now let $n,m$ be positive integers. Prove that $S^2$ with $n$ different points removed from it is homeomorphic to $S^2$ with $m$ different points removed from it if and only if $n = m$.

Let's assume that the $n$ points removed are $A = \{a_1, \cdots, a_n \} \subset S^2$ and that the $m$ points removed are $B = \{b_1, \cdots, b_m \} \subset S^2$.

First, assume that $n = m$. Using the stereographic projections, $S^2 \setminus A$ is homeomorphic to $\mathbb{R}^3$ without $n-1$ points and $S^2 \setminus B$ is homeomorphic to $\mathbb{R}^3$ without $m-1$ points. Since $n = m$, I assume that $\mathbb{R}^3$ without $n-1$ points is homeomorphic to $\mathbb{R}^3$ without $n-1$ (possible different) points, but I don't know how to explicetly prove this.

If $S^2 \setminus A$ is homeomorphic to $S^2 \setminus B$, then, using again the stereographic projection, if $n \neq m$, then I assume that $\mathbb{R}^3$ without $n-1$ points is not homeomorphic to $\mathbb{R}^3$ without $m-1$ points, but again, I don't know how to prove this.

Answer 1

By this answer, there is a homeomorphism $f:S^2\to S^2$ which maps any $n$ distinct points to any other $n$ distinct points. Restricting this map to the complement of the $n$ points proves the reverse direction of your question.

The forward direction is easy with a bit of algebraic topology--for instance, the Euler characteristic of a sphere with $n$ points removed is $2-n$. In the comments you mention expecting a proof without algebraic topology, though, so here is one. Let $X_n$ denote $S^2$ with $n$ points removed. Note that for any compact $A\subseteq X_n$, there exists a compact $B$ such that $A\subseteq B\subseteq X_n$ and $X_n\setminus B$ has exactly $n$ connected components. Namely, let $B$ be the complement of a union of small balls around each of the $n$ points. On the other hand, if $m>n$ then $X_m$ does not have this property: if $A\subseteq X_m$ is the complement of a union of small balls around each of the $m$ points, then any for any compact $B$ with $A\subseteq B\subseteq X_m$, $X_m\setminus B$ must have at least $m$ connected components, since there must be at least one contained in each component of $X_m\setminus A$. Thus $X_m$ and $X_n$ are not homeomorphic for $m>n$, or for $m<n$ by swapping $m$ and $n$.

(In more fancy language, this argument essentially shows that $X_n$ has exactly $n$ ends and thus $X_n\cong X_m$ implies $n=m$.)

Answer 2

Not sure why algebraic topology seems to be eschewed here, but for the harder direction I would guess that that $H_1(X)\cong\Bbb Z^{n-1}$, whereas $H_1(Y)\cong\Bbb Z^{m-1}$.

For this one can use that $X$ deformation retracts onto a "rose with $n-1$ petals".