Fundamental group of RP^n by recurrence?

[quasar987]

Fundamental group of RP^n by recurrence!?

Homework Statement

That's it. Find the fundamental group of RP^n by recurrence.

The Attempt at a Solution

It's just obvious to me that it's Z/2 no matter n but what is this recurrence argument that I'm supposed to use?

 

[matt grime]

Can you write RP^n as the union of copies of RP^m for m<n? Then you can use the Seifert Van Kampen theorem.

EG you can show the fundamental group of R^n is trivial by saying R^n= R^{n-1}uR^{n-1}, picking the first and last n-1 coordinates, the intersection being R^{n-2} thus by induction the fundamental group is e amalgam e over e.

Or more interestingly, you can show that the fundamental group of the bouquet of n copies of S^1 is F_n the free group on n generators.You can probably do the torus with n holes in it if you wanted to, though it requires a bit more thought.

 

[quasar987]

I found $$\mathbb{R}P^n= \mathbb{R}P^{n-1}\cup B_n$$ (where B_n is the unit open ball) but this is only true as sets. The topology is wrong because in $$\mathbb{R}P^{n-1}\cup B_n$$, $$\mathbb{R}P^{n-1}$$ is open, but it's not supposed to be in $$\mathbb{R}P^n$$. Plus, the intersection is void.

Anyway, I got something else though. $$\mathbb{R}P^n\approx \mathbb{R}P^{n-1}\times \overline{B}_{n-1}$$. And this, we've seen in class, implies that
$$\pi_1(\mathbb{R}P^n)=\pi_1(\mathbb{R}P^{n-1})\times \pi_1(\overline{B}_{n-1})=\pi_1(\mathbb{R}P^{n-1})\times \{e\}=\pi_1(\mathbb{R}P^{n-1})$$

Tadam!