This should be quite standard but the closest I found was by AreaMan, where full proof is not provided.
My definition of $n$dimensional torus: $\mathbb{T}^n := \mathbb{R}^n /\mathbb{Z}^n$. Problem: Prove it is compact and hausdorff.
My proof:
- The map $f: \mathbb{R}^n \rightarrow S^1 \times \cdots \times S^1$ where $(x_i) \mapsto (e^{2\pi i x_i })$, is a quotient map:
It is surjective and open: let $x=(x_i) \in \prod (a_i, b_i)$, and consider $f(x)$, then $\prod \{ e^{2 \pi i \theta} \, : \, \theta \in (x_i- \varepsilon, x_i+\varepsilon) \}$ is an open nhood in image containing $f(x)$.
- If 1. is correct. Then we have an induced homeomorphism $\mathbb{T}^n \cong S^1 \times \cdots \times S^1$.
using the more general result
Let $f:X \rightarrow Y$ be a continuous surjection. Then $f$ is a quotient map if and only if, for all spaces $Z$ and all functions $g: Y \rightarrow Z$ one has $g$ continuous iff $gf$ is continuous.
The homeomorphism then follows as the quotient maps $\pi, f: \mathbb{R}^n \rightarrow \mathbb{R}^n/\mathbb{Z}^n, S^1 \times \cdots \times S^1$ have the same identification.
I would happy to know if my proof is correct, particularly step 1.
The first part is correct, but you only claimed, without proof, that $f$ is open. It is clear that $f$ induces a bijection from $\mathbb{R}^n/\mathbb{Z}^n$ onto $(S^1)^n$. Since it is open, $f^{-1}$ is continuous. And since the domain of $f^{-1}$ is compact and $f^{-1}$ is continuous, $f^{-1}$ is a homeomorphism. Therefore, $f$ is a homeomorphism.