Maybe this is an idiot question and I'm missing something very trivial. This question question was asked here before, but the answer (which apparently is equal to the one that I created) seems incorrect. Let $p : T \longrightarrow K$ be a double covering of the Klein bottle by defining it as the projection of each Klein bottle inside the torus when you cut it in two pieces say $abab^{-1}$ and $a'^{-1}b' a'^{-1} b'^{-1}$. Then each end will be identified to the line in the center (this means $a = a'$), therefore it will not be the torus, actually it will not be a surface (it will be a bouquet of 3 circles glued with a 2-cell). So what's the covering?
Thanks in advance.
Best Answer
I'm not sure where you get $3$ circles from. The model only has two $1$-cells after identifying - horizontal and vertical edges. This coincides with the usual $1$-skeleton for the Klein bottle given by the wedge of two circles.
The picture you should have drawn is a rectangle with $7$ edges; four are vertical, all labelled $a$ and pointing in the same direction, the last three are horizontal, all labelled $b$ and pointing right, left, right as we go from top to bottom of the rectangle.
There are two $2$-cells, both labelled $A$, the top one oriented clockwise, the bottom one oriented anti-clockwise.
Each half is a fundamental domain equal to the usual model of the Klein bottle, and if you remove the central horizontal $1$-cell, gluing the two $2$-cells together along the edge, and forget about identifying the vertical edges with their diagonal partners (so only identifying top-left with top-right, and bottom-left with bottom-right), then this model is the usual model for the torus.