Express the quotient of $\mathbb{R}^2/\mathbb{Z}^2$ by the relation $(x,y) \equiv (-x,-y)$ as the quotient of a polygon

Question

This is another old qualifying exam question, and I'm kinda grasping at straws for this one.

Let $\mathbb{T}^2$ denote the quotient space $\mathbb{R}^2/\mathbb{Z}^2$, and let $M$ denote the quotient of $\mathbb{T}^2$ by the relation $(x,y) \equiv (-x,-y)$.

• Express $M$ as a quotient of a polygon with sides defined.
• What is the fundamental group of $M$?

So far, what I did was divide the standard quadrilateral expression for $\mathbb{T}^2$ by taking the center to be origin of my coordinate system and then adjusting the sides to get what should be the new quotient on the sides that correspond to $M$ as shown below:

After that, I realized that the relation for $M$ would "rotate" the left half of this image into the right half which, after the relations for $M$ are then added to the centerline, would result in the polygon given by

but at this point, I get really stuck. I'm not even sure if what I've done is the correct way to approach this, so any help would be greatly appreciated. Thank you!

Answer 1

Your second diagram looks correct to me.

And then one can see that the quotient is homeomorphic to $S^2$, by simply gluing side pairs of the second diagram one-pair-at-a-time.