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The projective plane $\mathbb{R} P^2$ is constructed via identifying antipodal boundary points of a disk $D^2$, I want to use van-kampen theorem to tackle this problem, first I remove a disk in the polygonal region that represent the real projective plane and then I prove that the fundamental group of a disk minuses finite points is $Z*Z....*Z$, and then I don't know what should I do.

Arctic Char
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YuerCauchy
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  • Based on extending the argument from this answer, it seems like the fundamental group of $\mathbb{R}P^2$ minus $n \geq 1$ points is the free product of $n$ instances of $\mathbb{Z}$ – The Zach Man Jun 09 '21 at 05:33

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You're on the right track! But you don't even need van kampen here. Just working with a simple deformation retract will do. We'll work with $4$ punctures for concreteness, but you'll see the argument is the same with any (finite) number.

We work with this picture:

RP2 with punctures

Notice it deformation retracts

deformation retract

to two antipodal points (which get identified) with one edge $a$ (from outside the punctures) and three edges $b_1$, $b_2$, and $b_3$ (from between the punctures).

Making the necessary identifications gives

a bouquet of 4 circles

a bouquet of $4$ circles. Which just so happens to be the number of punctures we started with...

Of course, it's a well known exercise to show that this has fundamental group $F_4$, or if you prefer, $\mathbb{Z} \ast \mathbb{Z} \ast \mathbb{Z} \ast \mathbb{Z}$.

Can you show that this works for an arbitrary number of points?

I hope this helps ^_^

HallaSurvivor
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