I just worked a little bit with triangulations of surfaces. I think the following "triangulation" of the real projective plane is false:
The red (blue) edges are identified in an inverse way. Sorry for this ugly picture.
I know how the right triangulation should look like, but to improve my intuition, I would like to know, why the "triangulation" above is false.
Is it because of the two triangles with vertices A, B and E in the upper left and lower right corner?
The intersection of the two triangles is not a single vertex.
Best Answer
The first problem is that the space that one gets by making the identifications in your picture is not the projective plane. In fact it is not a surface, because in the quotient space the image of the four A vertices has no neighborhood homeomorphic to a disc; it does have a neighborhood homeomorphic to two discs identified at their central point, but that is not allowed in a surface.
A simple relabelling of the vertices will correct this issue. Starting from the upper left corner and going clockwise, instead of labelling the vertices as ABCADEABCADE, replace every other A with an X to get ABCXDEABCXDE.
However, even after doing that correction, your picture still does not define a triangulation in the strict sense, and the reason you state is exactly correct: there are two distinct triangles with vertices ABE.