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I am trying to calculate $H_k(X)$ where $X = \mathbb{R}P^n - \{ x_0 \}$

I started thinking about $k=2$. We can get the projective plane by taking the upper hemisphere with points along the equator identified according to the antipodal map. If we remove a point from the hemisphere, we can then enlarge this such that we are just left with a circle and thus for $k=2$ we just have the homology of the circle.

My geometric intuition starts to fail for $k=3$ and higher spaces. So my questions are:

1) Does the same construction work for higher $k$? (probably not, this seems too easy)

2) If not, what is the nice way to calculate the homology groups (say we know $H_k(\mathbb{R}P^n)$? I guess there is a way to use Mayer-Vietoris, but I just can't see it

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up vote 9 down vote accepted

Your idea should work similarly in general: Take the closed upper hemisphere of $S^n$. Then $\mathbb{R}P^n$ is obtained by identifying antipodal points on the equator, which is an $S^{n-1}$. So removing a point (preferably not on the equator!) would result in something homotopy equivalent to an $S^{n-1}$ with its antipodal points identified- and that's just $\mathbb{R}P^{n-1}$. So if you know the homology of that, you're in business!

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thanks - I guess I am bound to be right every now and then! – Juan S Apr 6 '11 at 7:37

We have a deformation retraction of $X$ to $\mathbb{RP}^{n-1}=\{x_0=0\}\subset X$ given by the homotopy : $$ ([0,1]\times X\to X:(t,[x_0:x_1:\dots:x_n])\mapsto [(1-t)x_0:x_1:\dots:x_n] $$ Since $X$ and $\mathbb{RP}^{n-1}$ are thus homotopically equivalent they have the same homology so that $$ H_k(X)=H_k(\mathbb{RP}^{n-1}) \operatorname {for all} k\geq 0$$

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