# Klein bottle covered by the torus

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?

• Nov 14, 2013 at 21:36
• @WillJagy If you read the question , you will se that I've already seen these answers, however the problem persists. Nov 14, 2013 at 21:57
• The ususal behavior in your case is to include at least one link to a previous instance of the question, as you refer to those. You failed to do that. Nov 14, 2013 at 22:05
• I don't understand. Are you claiming that the double cover of the Klein bottle by the torus illustrated by subdividing the usual rectangular quotient to the torus, along a line connecting the centers of two opposing edges of the square, is not a true covering map of the Klein bottle? Nov 16, 2013 at 15:18
• @DanielRust Actually I'm claiming that the cover is not a torus, because if one identifies each end of a rectangle with the line passing through the center, then one gets a bouquet of 3 circles glued with a 2-cell (the pullback). Nov 18, 2013 at 4:09

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.
• Yes, but you can label it the way I did above, $abab^{-1}$ and $a'^{-1}b' a'^{-1} b'^{-1}$ with $a = a'$ (an hexagon with the labels identified properly), and it's equivalent to yours. However it's not a surface but apparently it's equivalent to a torus. I know that I'm missing something, but i don't know what it is. Nov 21, 2013 at 0:11