Triangulation of the projective plane

Question

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.

Answer 1

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.

Answer 2

How would your surface look in a neighborhood of $A$?

Your diagram has two copies of $BAE$ and nothing at all that connects the angle interval between $AB$ and either $AC$ or $AD$.