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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.

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  • $\begingroup$ @WillJagy If you read the question , you will se that I've already seen these answers, however the problem persists. $\endgroup$
    – user40276
    Nov 14, 2013 at 21:57
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    $\begingroup$ 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. $\endgroup$
    – Will Jagy
    Nov 14, 2013 at 22:05
  • $\begingroup$ 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? $\endgroup$
    – Dan Rust
    Nov 16, 2013 at 15:18
  • $\begingroup$ @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). $\endgroup$
    – user40276
    Nov 18, 2013 at 4:09

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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.

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  • $\begingroup$ 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. $\endgroup$
    – user40276
    Nov 21, 2013 at 0:11

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