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I've been computing some singular homology groups of different spaces. In particular, I know how to compute the homology of a cell complex. Now I'm wondering how to compute the homology of $\mathbb{R}^m$. Since homology is a way of counting holes and $\mathbb{R}^m$ doesn't have any I guess $H_n(\mathbb{R}^m) = 0$ for all $n,m$.

But how do I rigorously compute this? Thanks for your help.

Edit: I think I can use that $\mathbb{R}^n$ is contractible and then $H_n(\mathbb{R}^m) = H_n(\{ \ast \})$.

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    I wonder if the Jordan curve theorem could be used here; I know it is like using a tank to kill a fly, but I wonder if JCT says that every cycle (simple closed curve; given an orientation) bounds.2011-08-13

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$\mathbb{R}^n$ is contractible therefore homotopy equivalent to a point and so $H_n(\mathbb{R}^m) = H_n(\{ \ast \})$.

$ H_n(\{ \ast \}) = 0 , n > 0$ $ H_n(\{ \ast \}) = \mathbb{Z} , n = 0$

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    Yes, I do, thanks. Sorry, I used $n$ here and $k$ on paper : P2011-08-14
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I think this may be an alternative proof; maybe overkill, but I think it may work:

I think cycles C on X can be geenralized as injective continuous , injective maps

$S^n\rightarrow X$. JCT states that (the image of ) C separates the plane into two

connected regions with C as the boundary; the interior of C can then be seen as being

bounded by the cycle, C, so that we can then say that every cycle C in $\mathbb R^n$

bounds, and so the homology of $\mathbb R^n $is trivial.

Edit: as Grigory pointed out, my idea was too simple and too good to be true ; e.g.,genus-g surfaces cannot be represented as images of spheres. Maybe my argument can be shown that $\pi_n(\mathbb R^n)=0$, but I am not betting on it.

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    @gary: thanks anyway, for trying!2011-08-14