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.
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.
I think this works but there is still a gap.
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 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 - call this wedge space X at the point q. Let and . Then A, B are open, , and which has a trivial fundamental group. So SVK gives that where * denotes the free group.