# Prove that the quotient space $\mathbb{R^{2}} \sim$ with is homeomorphic to the Torus

Define $\sim$ to be a relation in $\mathbb{R^{2}}$ :$(x,y) \sim (x',y') \Leftrightarrow x-x'=2nπ$ and $y-y'=2nπ$ Prove that the Quotient space $\mathbb{R^{2}} \sim$ is homeomorphic to the torus .

Now i know i have to find a surjective map continous map fro $\mathbb{R^{2}}$ to the Torus the along with the quotient map create a composition that will give me the homeomorfism. But im stuck as to find the map and then prove that it is the right onethat induces a homeomorphism. Also im trying what the quotient space is geometrically maybe that will help me as to understant what map to look for. But what i understant is that the element of the quotient space are classes of lines? maybe im wrong.

Hint. Let define $f\colon\mathbb{R}^2\rightarrow\mathbb{T}^2$ by the following formula: $$f(x,y)=(e^{2i\pi x},e^{2i\pi y}).$$ Prove that this map is continuous surjective and goes through the quotient to an injective map.
Conclude using the universal property of quotient topology and the compactness of $\mathbb{R}^2/\sim=[0,2\pi]^2/\sim$.
Geometrically, $\sim$ consists in glueing the opposite edges of each square of length $2\pi$ without any twisting.