Projective space with 2 points removed

Question

I am interested in finding what $\mathbb{RP}^n$ with two points removed, or in general $n$ point removed is homeomorphic or homotopically equivalent. Here we see once-punctured $\mathbb{RP}^n$ is homotopically equivalent to $\mathbb{RP}^{n-1},$ but for example it does not hold that twice punctured $\mathbb{RP}^2$ is homotopically equivalent to $\mathbb{RP}^0 = {pt},$ here we see: The real projective 2-space $\mathbb{RP}^2$ with $2$ points removed is homotopically equivalent to the wedge of two circles.

So my question: is $n=2$ an exception and $\mathbb{RP}^n$ with two point removed is homotopically equivalent to $\mathbb{RP}^{n-2}$ is true, if not what is it homotopically equivalent?

Answer 1

If $M$ is a (connected) $n$-manifold then $M - \{k \text{ points}\}$ is homotopy equivalent to $(M - \text{point}) \vee \bigvee_{k-1} S^{n-1}$. Here is one way to see this:

First note that because any collection of $k$ points can be connected by disjoint paths, up to homotopy it doesn't matter which $k$ points are removed. Now removing a single point is the same as removing a small disk. Instead, split the disk into $k$ sectors, each homeomorphic to a disk, and remove one point from each sector. Now each of the $k$ punctured sectors can be retracted to its boundary. Note that the subset of the boundary of the disk that belongs to $k - 1$ of the sectors is contractible, since it is $S^n - D^n$. So collapsing this subset is a homotopy equivalence, which results in those $k - 1$ sectors becoming spheres and the rest of the space becoming $M$ with a puncture, all wedged at the collapse point.

Applying this to $\mathbb{RP}^n$, and noting that $\mathbb{RP}^n - \text{point} \simeq \mathbb{RP}^{n-1}$ (one way to see this is to retract $S^n$ punctured at antipodal points to the equator, by an antipode preserving retraction), so $\mathbb{RP}^n - \{k \text{ points}\} \simeq \mathbb{RP}^{n-1} \vee \bigvee_{k-1} S^{n-1}$.