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 $H_n(\mathbb{R}^m) = H_n(\{ \ast \})$.
Best Answer
$\mathbb{R}^n$ is contractible therefore homotopy equivalent to a point and so $H_n(\mathbb{R}^m) = H_n(\{ \ast \})$.
$$ H_n(\{ \ast \}) = 0 , n > 0$$ $$ H_n(\{ \ast \}) = \mathbb{Z} , n = 0$$