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It is well known that if you start with a two-dimensional square and could glue — in the most straightforward way — the top and bottom edges to each other, and likewise the left and right edges to each other, then the resulting surface is a torus.

Let the square be the points (x,y) of the plane with 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.

Suppose instead that we glue the point (x,0) to the point (x',1) where x' = x + 1/2 (modulo 1), for all x with 0 ≤ x ≤ 1. And likewise, glue the point (0,y) to the point (1,y'), where y' = y + 1/2 (modulo 1), for all y with 0 ≤ y ≤ 1.

What surface is the result, topologically?

(Note that for any number t with 0 ≤ t ≤ 1, the expression

t + 1/2 (modulo 1)

means t + 1/2 if t < 1/2, and it means t - 1/2 if t ≥ 1/2.)

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This is equivalent to gluing opposite sides of an octagon in an orientable manner. The result is a double torus (genus 2), as shown in the picture below taken from this MSE question: All segments meet at one point on the central handle

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