Prove rigorously that for two points $x, y \in M$, the spaces $M \ \backslash \{x\}$ and $M \ \backslash \{y\}$ are homeomorphic. Let $M$ be a connected topological manifold. Prove rigorously that for two points
$x, y \in M$, the spaces $M \  \backslash \{x\}$ and $M \ \backslash  \{y\}$ are homeomorphic.
I am not sure the best way to start this problem.  I was thinking of using $CW$-complexes, but I didn't get very far.
 A: A manifold locally looks like $\Bbb R^n$, and connectedness implies path connectedness.


*

*Prove, that for an open ball $B\subseteq\Bbb R^n$ around the origin and $x\in B$, there is a homeomorphism $\bar B\to \bar B$ (on the closure of $B$) that is identical on $\partial B$ and maps $0$ to $x$. (Hence, for any $x,y\in B$ there is such a homeomorphism $\bar B\setminus\{x\}\to \bar B\setminus\{y\}$.

*For the given $x,y\in M$, fix a path $\gamma:[0,1]\to M$ that connects them ($\gamma(0)=x,\ \gamma(1)=y$), and consider the open sets $\gamma^{-1}(U)$ for all open $U\subseteq M$ which admits a chart (from the manifold structure) $\varphi:U\to\Bbb R^n$, under which $\varphi(U)$ is an open ball. This is going to be an open cover of $[0,1]$, so finitely many can be chosen that still covers $[0,1]$.

*So, we can point out a finite sequence $0=t_0,\,t_1,\,t_2,\dots,t_N=1$ of moments such that $\gamma(t_i)$ and $\gamma(t_{i+1})$ lie in a same chart ball $U_i$. Apply the homeomorphism given in 1. to $U_i$ and its two given points, and extend this map to be identity on $M\setminus U_i$. This way we will obtain homeomorphisms
$$M\setminus\{\gamma(t_0)\}\ \to \ M\setminus\{\gamma(t_1)\}\ \to \dots \to\ M\setminus\{\gamma(t_N)\} \,.
$$

A: If $f:M\to M$ is a self-homeomorphism satisfying $f(x)=y$, the restriction $$f|_{M\setminus \{x\}}: M\setminus \{x\}\stackrel {\cong}{\to}  M\setminus \{y\}$$ solves  your problem.   
-Ah, you'll say, but is the existence of such a providential self-homeomorphism not a little fishy?
-Well, yes, it is fishy but it is true!
