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I'm trying to prove that $\mathbb{R^2}$ punctured n times is homeomorphic to $\mathbb{R^2}$ punctured n times, where the two spaces have different punctures, say $\{a_1, a_2, ... , a_n\}$ and $\{b_1, b_2, ... , b_n\}$. My attempt goes by induction; the case n=1 is straightforward, and then assuming the hypothesis holds for n-1, given aforementioned punctures we have a homeomorphism $F:\mathbb{R^2}-\{a_1, a_2, ... , a_{n-1}\}\rightarrow\mathbb{R^2}-\{b_1, b_2, ... , b_{n-1}\}$. We assume $F$ doesn't map $a_n$ to $b_n$ otherwise we're done. If I can show $\mathbb{R^2}-\{a_1, a_2, ... , a_n\}$ is homeomorphic to $\mathbb{R^2}-\{a_1, a_2, ... , F^{-1}(b_n)\}$ then we're not very far away from a complete proof, but I'm having trouble constructing such a homeomorphism that fixes the other n-1 points.

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    $\begingroup$ Pick an arc from $a_n$ to $b_n$ that touches none of the other points. Pick a small enough neighborhood of it that none of the other points are included in its closure. Now you have those two points in a closed $k$-ball, $D^k$. It suffices to show that, given two points $a,b$ in the interior of $D^k$, there is a homeomorphism $f$ of $D^k$ such that $f(a)=b$ and $f$ is the identity on the boundary. Then you can extend $f$ to the rest of $\Bbb R^k$ by the identity map. $\endgroup$ – user98602 Sep 9 '15 at 1:47
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    $\begingroup$ I wrote up a comment that was almost the same as @MikeMiller’s, but delayed posting it while I actually constructed a map of the sort he says exists. So that is definitely the way to do it. $\endgroup$ – Lubin Sep 9 '15 at 1:59
  • $\begingroup$ @Lubin: If you have written down the homeomorphism I hope you'll post it as an answer; as you say it is the canonical answer. $\endgroup$ – user98602 Sep 9 '15 at 1:59
  • $\begingroup$ Thanks very much to you both! $\endgroup$ – Ssarutto Sep 9 '15 at 3:37
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In response to @MikeMiller’s request, here’s just one way of constructing a self-homeomorphism of the disk that moves one point to another while leaving points on the boundary fixed. Let your disk be the unit disk in the plane, and first map the interior to all of the complex plane, by $\varphi:(r,\theta)\mapsto \frac r{1-r}(\cos\theta+i\sin\theta)$. Here, of course, $0\le r<1$. Notice what happens to points on the boundary, $r=1$: they go to what you might call “vanishing points on the horizon”, one for each direction $\theta$, but at a distance $+\infty$. Now just take your two points in the disk, call them $a$ and $b$, and after $\varphi$, apply $\tau:z\mapsto\varphi(b)-\varphi(a)+z$. This is a parallel translation of the whole complex plane, and leaves the “vanishing points” fixed. Now apply $\varphi^{-1}$.

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