Prove that $\Bbb{R}^2/\Bbb{Z}^2\approx S^1\times S^1$

Question

Here $\Bbb{R}^2/\Bbb{Z}^2$ is the quotient space obatined from $\Bbb{R}^2$ by identifying points of $\Bbb{Z}^2$ i.e. $(x,y)\sim (x',y')\iff (x,y),(x',y')\in\Bbb{Z}^2$.

$S^1\times S^1:=\{(z,w)|\ z,w\in S^1\}$

I define $f:\Bbb{R}^2\to {S^1}\times S^1$ by $f(x,y)=(e^{2\pi i x},e^{2\pi i y})$. $f$ is continuous and onto.

As $f(n,m)=(1,1)\ \forall (n,m)\in \Bbb{Z}^2$ i.e. $f$ agrees on $\Bbb{Z}^2$. By the property of quotient space, $f$ induces a continuous map $\tilde{f}:\mathbb{R}^2/\Bbb{Z}^2\to S^1\times S^1$ such that $\tilde{f}([x,y])=f(x,y)$. This map is onto as well. But this map is not injective. I couldn't move forward from here.

Although I have observed one thing- instead of only identifying the points of $\Bbb{Z}^2$ if we identify the points as follows- $$(x,y)\sim (x',y')\iff (x-x',y-y')\in\Bbb{Z}^2$$ Then we would have $\Bbb{R}^2/\sim\approx S^1\times S^1$, the same $f$ will give rise to this homomorphism.

Can anyone help me to solve the problem? Thanks for help in advance.

Answer 1

What you are observing is an ambiguity in the "slash" notation $/$ for quotient spaces.

In the context of $\mathbb R^2 / \mathbb Z^2$, you should think of the group $\mathbb Z^2$ acting on the space $\mathbb R^2$, where $(m,n) \in \mathbb Z^2$ takes the point $(x,y) \in \mathbb R^2$ to the point $(x',y') = (x+m,y+n) \in \mathbb R^2$. This action defines a decomposition of $\mathbb R^2$ into orbits: the orbit of $(x,y) \in \mathbb R^2$ is just the set of all $(x',y') \in \mathbb R^2$ such that $x-x',y-y' \in \mathbb Z^2$. And the meaning of $\mathbb R^2 / \mathbb Z^2$ in this context is that it is the orbit space of the action, namely the quotient space with respect to the decomposition of $\mathbb R^2$ into orbits.

As your question indicates, the slash notation is also used for a different kind of quotient: namely when one is given a topological space $X$ and a subset $A \subset X$, one forms the quotient $X/A$ whose decomposition elements are $A$ itself and all singleton subsets $\{x\} \subset X$ for which $x \in X-A$.

Answer 2

You are right, there is probably a typo in your original question. It should be $(x,y)\sim (x',y')\iff (x,y)-(x',y')\in\Bbb{Z}^2$

Answer 3

Your map $f$ is a surjective group homomorphism between the abelian groups $\mathbb R^2 = \mathbb R \times \mathbb R$ (with componentwise addition) and $S^1 \times S^1$ (with componentwise complex multiplication). We have $\ker f = \mathbb Z^2$, thus $f$ induces a group isomorphism $$\bar f : \mathbb R^2/\mathbb Z^2 \to S^1 \times S^1 .$$ Since we give $\mathbb R^2/\mathbb Z^2$ the quotient topology with respect to the map $p : \mathbb R^2 \to \mathbb R^2/\mathbb Z^2$, $\bar f$ is continuous. It remains to show that $\bar f$ is a homeomorphism. To do so, it suffices to show that $f$ itself is a quotient map (which means that $V \subset S^1 \times S^1$ is open iff $f^{-1}(V) \subset \mathbb R^2$ is open). We shall prove that $f$ is an open map; this implies that it is quotient map. So let $U \subset \mathbb R^2$ be open. It is well-known that $g : \mathbb R \to S^1, g(t) = e^{2\pi i t}$, is an open map (see e.g. Open sets on the unit circle $S^1$). Thus $f = g \times g$ is an open map.

If you identify $\mathbb Z^2$ to a point, then you do not consider the quotient group $\mathbb R^2/_{group}\mathbb Z^2$, but the topological quotient space $\mathbb R^2/_{top}\mathbb Z^2$.