let $I=[0,1]^2$.
Let's provide $\mathbb{Z}^2$ with the group operation: $(k, l) ∗ (k' , l' ) = (k + (−1)^l k' , l + l' )$
Let's then define $\mathbb{Z}^2$ action on $\mathbb{R}^2$ given by:
$((k, l)* (x, y)) → ((−1)^l x + k, y + l)$
This action is free and proper so the quotient $\mathbb{R}^2/\mathbb{Z}^2$ is a differential manifold, that we define as the Klein bottle.
Show that K is hemeomorphic to $I^2/∼$ where $∼$ is the equivalence relashinship $(0, y) ∼ (1, y)$ and $(x, 0) ∼ (1 − x, 1)$.
What I did:
Let's take the injection $i: I^2 \rightarrow \mathbb{R}^2 $.
Let $p: I^2 \rightarrow I^2/ ∼$ and $\pi: \mathbb{R}^2 \rightarrow \mathbb{R}^2/\mathbb{Z}^2 $ be the canonical projections.
By the universal property of the quotient, there exists a unique $\overline{i}: I^2/ ∼ \rightarrow \mathbb{R}^2/\mathbb{Z}^2$
I need to prove that $\overline{i}$ is a homeomorphism. It is obvious that it is $\overline{i}$ is continuous (because $i$ is continuous) and proper(as $I^2/ ∼$ is compact).
I don't see how to prove it is a bijection. In this try, I didn't use the equivalence relationship definition, nor the group action definition. So I wonder if there is an explicit formulation of such a homermorphism. Thank you for any help in the right direction .