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My question is how to prove that for $m \neq n$: $\mathbb{R}^2 $ without $n$ points and $\mathbb{R}^2 $ without $m$ points are not homotopy equivalent.

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Look at their fundamental groups! –  andybenji Jan 19 '13 at 20:34

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Observe that $R_n = \mathbb{R}^2 - \{n\mbox{ points}\}$ retracts onto a wedge of $n$ circles, so that its fundamental group is free on $n$ generators. So if $m\neq n$, the two spaces $R_m$ and $R_n$ cannot be homotopy-equivalent (because, as Zhen Lin points out below, two free groups of different rank are non-isomorphic).

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(One should show that the free group on $m$ generators is not isomorphic to the free group on $n$ generators if $m \ne n$...) –  Zhen Lin Jan 19 '13 at 20:39
    
@ZhenLin Indeed, and thank you. I have added a point mentioning this. –  Neal Jan 19 '13 at 21:57
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An easy way to see this is to abelianize. –  Grumpy Parsnip Jan 19 '13 at 22:01
    
And how do I know that the fundamental group of a wedge of $n$ circles is $\mathbb{Z}_n$ ? –  Anne Jan 19 '13 at 22:11
    
@Anne, do you know van Kampen's theorem? –  user27126 Jan 19 '13 at 22:36

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