Topology: Fund Grp of wedge of two real projective planes

• May 12th 2011, 08:04 PM
huram2215
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.
• May 12th 2011, 08:36 PM
Tinyboss
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.
• May 13th 2011, 03:25 AM
huram2215
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.