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Reading through some lecture notes and it says

The torus $T^2$ is the orientation double cover of the Klein bottle $K$, via the covering projection

$p:T^2\to K; [x,y]\mapsto [x,2y]$

Could someone explain this map? Are they taking $[x,y]$ as the equivalence class of the point $(x,y)$ in $I\times I$ with torus identifications ? I don't see how this maps to the Klein bottle.

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    If you know the way of identifying opposite edges of a square to make a Klein bottle or a torus, take a 2x1 rectangle and identify edges to make a torus. Draw a line down the middle of the rectangle dividing it into two squares and with the middle line having opposite orientation to the parallel sides. If you identify both ends with the centre, this does not disturb the identification of the ends with each other (relative orientation is preserved), each square makes a Klein bottle - and you can see your double cover. – Mark Bennet May 03 '12 at 15:51
  • @MarkBennet Does this mean, in some sense, that every map on the torus descends to the klein bottle? I mean, as long as the map coincides on the edges of the cylinder before quotienting to obtain the torus, is not really important what happens in the middle circle since it will trivially be coincident so it is well defined on the covering space. Is this right? – astro Aug 14 '20 at 16:25

1 Answers1

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Posting so this doesn't go unanswered.

If you know the way of identifying opposite edges of a square to make a Klein bottle or a torus, take a 2x1 rectangle and identify edges to make a torus. Draw a line down the middle of the rectangle dividing it into two squares and with the middle line having opposite orientation to the parallel sides. If you identify both ends with the centre, this does not disturb the identification of the ends with each other (relative orientation is preserved), each square makes a Klein bottle - and you can see your double cover. – Mark Bennet May 3 '12 at 15:51

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