Showing that $\Bbb R^2/\Bbb Z^2$ is homeomorphic to $S^1 × S^1.$

Question

Show that $\Bbb R^2/\Bbb Z^2$ is homeomorphic to $S^1 × S^1.$

$\textbf {My Attempt} :$ Consider the map $p : \mathbb R^2 \longrightarrow S^1 \times S^1$ defined by $p(x,y) = (e ^{2 \pi i x}, e^{2 \pi i y}),$ $(x,y) \in \mathbb R^2.$ Then it is easy to see that $p$ is continuous and surjective. Also $$p^{-1} \{ (e^{2 \pi i x}, e^{2 \pi i y}) \} = \{(x + m, y + n)\ |\ m, n \in \mathbb Z \}$$ which is the orbit of $(x,y)$ under the action of $\mathbb Z^2$ on $\mathbb R^2.$ Hence by universal property of quotient topology $p$ will induce a bijective continuous map $q : \mathbb R^2 / \mathbb Z^2 \longrightarrow S^1 \times S^1.$ Since $[x,y] = [x - [x], y - [y]]$ it follows that $\mathbb R^2 / \mathbb Z^2 = [0,1]^2/\mathbb Z^2.$ Since $[0,1]^2$ is compact it follows that $\mathbb R^2/ \mathbb Z^2$ is also compact. Hence $q$ is a homeomorphism.

Is my attempt correct? I think I need also to show that the map from $\mathbb R^2$ to $\mathbb R^2 / \mathbb Z^2$ is a quotient map. Then we can take the composition $$[0,1]^2 \hookrightarrow \mathbb R^2 \xrightarrow{\text {quotient map}} \mathbb R^2 / \mathbb Z^2$$ Isn't it so? Could anybody give me some suggestion regarding this?

Thanks for reading.

Answer 1

This is the same question as Prove that $\Bbb{R}^2/\Bbb{Z}^2\approx S^1\times S^1$ .

Your approach is correct, but it needs a little clarification. First observe that writing $\Bbb R^2/\Bbb Z^2$ is somewhat ambiguous: It could denote

1. the quotient set from $\Bbb R^2$ by collapsing the subspace $\Bbb Z^2$ to a point.

2. the quotient group whose elements are the cosets $(x,y) + \Bbb Z^2$.

Here we definitely need interpretation 2. Via the quotient homomorphism $p : \Bbb R^2 \to \Bbb R^2/\Bbb Z^2$ this set is given the quotient topology. You have correctly shown that you get a continuous bijection $q : \mathbb R^2 / \mathbb Z^2 \longrightarrow S^1 \times S^1$. This is even an isomorphism of groups, but that is irrelevant.

I think the equation $\mathbb R^2 / \mathbb Z^2 = [0,1]^2/\mathbb Z^2$ is misleading. What you mean is certainly $p([0,1]^2) = \mathbb R^2 / \mathbb Z^2$. This is true and proves that $\mathbb R^2 / \mathbb Z^2$ is compact.