TITLE: Homology of punctured projective space
QUESTION [13 upvotes]: 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

REPLY [3 votes]: 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 $$ H_k(X)=H_k(\mathbb{RP}^{n-1}) \operatorname {for all} k\geq 0$$