Fundamental Group of the projective plane after we remove n points?

[Geometrick]

So I have been wondering, what is the fundamental group of a projective plane after we remove n points?

I tried doing this using Van Kampens Theorem, maybe I am applying in incorrectly, I am getting that it is the Free group on n generators.

However, when I think of RP^2 as a quotient of the sphere, it's the same thing as a sphere with 2n points removed, which is the same thing as the Euclidean plane with 2n-1 points removed by stereographic projection, which has Fundamental Group F(2n-1), the free group on 2n-1 generators.

Which is correct?

[Office_Shredder]

Not an expert in algebraic topology by any means, but keep in mind that each of the pair of points that you remove is equivalent on the sphere so you only have n distinct points. You really only map the top half of the sphere to Euclidean space and only half the points will be on that half of the sphere (you might have some trouble with boundary points but you can pick an equator that doesn't pass through any of your points).

Hope this helps (sorry if it doesn't)

[Geometrick]

Not an expert in algebraic topology by any means, but keep in mind that each of the pair of points that you remove is equivalent on the sphere so you only have n distinct points. You really only map the top half of the sphere to Euclidean space and only half the points will be on that half of the sphere (you might have some trouble with boundary points but you can pick an equator that doesn't pass through any of your points).

Hope this helps (sorry if it doesn't)

I don't know if that's true; a point removed on the projective plane corresponds to a point, antipodal point pair being removed on the sphere, i.e. the sphere is a 2-1 covering of the projective plane. So removing a point on the projective plane corresponds to removing 2 points on the sphere.