# Thread: Topology: Fund Grp of wedge of two real projective planes

1. ## Topology: Fund Grp of wedge of two real projective planes

How can I compute the fundamental group of the wedge of two real projective planes? I would guess that it is (a,b | a^2 = 1 = b^2} with a,b each loops in each respective real projective plane, but I'm not seeing a way to prove it. I'd like to do it without using covering spaces if possible, maybe with SVK, but any suggestion is very welcome at this point.

2. SVK will give you that the fundamental group of the wedge product is the free product of the fundamental groups, which agrees with your guess.

3. I think this works but there is still a gap.

$\displaystyle \mathbb{P}^2$ can be considered a closed disk with boundary identified antipodally. In the proof below I have assumes that the wedge point is taken in the open disk from each projective plane. If I wedge on the boundary of $\displaystyle \mathbb{P}^2$ I get confused - not clear how to choose the open sets.

Here's the gap: It must be true that any two wedges of projective planes are homeomorphic, but how do you prove it.

If I assume the above is true, then without losing generality we can wedge two projective planes at a point interior to $\displaystyle \mathbb{D}^2$ - call this wedge space X at the point q. Let $\displaystyle A = \mathbb{P}^2 \vee \mathbb{D}^2$ and $\displaystyle B = \mathbb{D}^2 \vee \mathbb{P}^2$. Then A, B are open, $\displaystyle A \cup B = X$, and $\displaystyle A \cap B = \mathbb{D}^2 \vee \mathbb{D}^2$ which has a trivial fundamental group. So SVK gives that $\displaystyle \pi_1(X,q) = \pi_1(A,q) * \pi_1(B,q) \cong \{ a,b | a^2 = 1 = b^2 \}$ where * denotes the free group.