This question is a continuation of my previous question on quotient groups. I know that: \begin{eqnarray} \mathbb{Z}/n\mathbb{Z} = \{[0],...,[n-1]\} \tag{1}\label{1} \end{eqnarray} and, from the asnwers I got in my post, it must follow that $\mathbb{Z}^{d}/n\mathbb{Z}^{d} = (\mathbb{Z}/n\mathbb{Z})^{d}$ and $\mathbb{R}^{d}/n\mathbb{Z}^{d} = [0,1)^{d}$. But I don't know how to prove these two equalities.
For the first one, I have a little sketch in my mind that goes like this. Define the following equivalence relation on $\mathbb{Z}^{d}$ $$a \equiv b \hspace{0.1cm} \mbox{(mod $n$)} \iff \mbox{$n|(a_{i}-b_{i})$ for each $i=1,...,d$} $$ where $a= (a_{1},...,a_{d}), b=(b_{1},...,b_{d}) \in \mathbb{Z}^{d}$. We define equivalent classes: $$[a]_{d}:= \{b \in \mathbb{Z}^{d}: \hspace{0.1cm} a \equiv b \hspace{0.1cm} \mbox{(mod $n$)}\}$$ Then, we must have something like: $$[a]_{d} = \{b \in \mathbb{Z}^{d}: \hspace{0.1cm} n|(a_{i}-b_{i}) \hspace{0.1cm} \forall i \in \{1,...,d\}\} = \{b\in \mathbb{Z}^{d}: \hspace{0.1cm} b_{i} \in [a_{i}]\} = [a_{1}]\times\cdots\times [a_{d}]$$ where $[a_{i}]$ is the equivalence class of $a_{i} \in \mathbb{Z}$ in $\mathbb{Z}/n\mathbb{Z}$.Thus, the set of all equivalence classes $[b]_{d}$ should be the set of all cartesian products $[a_{1}]\times\cdots[a_{d}]$ and, by (\ref{1}), we must have $\mathbb{Z}^{d}/n\mathbb{Z}^{d} = (\mathbb{Z}/n\mathbb{Z})^{d}$.But I don't know if my reasoning is correct or if is the best approach to the problem.
Furthermre, I have no idea on how to address the problem $\mathbb{R}^{d}/n\mathbb{Z}^{d} = [0,1)^{d}$. Can someone help me here? Thanks in advance!
Consider the map \begin{align} f :~~~~~~~~\mathbb{Z}^d \to &\left(\mathbb{Z}/n\mathbb{Z}\right)^d\\ \sum_{i=1}^d \alpha_i e_i \mapsto & \left( \overline{\alpha_1},\ldots,\overline{\alpha_d}\right) \end{align} where $\overline{\alpha}$ is the reduction modulo $n$ of $\alpha$, and $\{e_1,\ldots,e_d\}$ stands for the canonical basis of $\mathbb{R}^d$. Show first that it is a group homomorphism (it is easy). Show that it is surjective (easy too). Thus, check its kernel : you can show that $\ker f = \left(n\mathbb{Z}\right)^d$. Hence, $f$ goes to the quotient $\overline{f} : \mathbb{Z}^d / \ker f \to \left(\mathbb{Z}/n\mathbb{Z}\right)^d$ as a group homomorphism, which is surjective (as $f$ is), and injective by construction. It is then an isomorphism.
Careful : $\mathbb{R}/\mathbb{Z}$ is not $[0,1)$ (at least, not written like that). We usually prefer to denote it $\mathbb{S}^1$. The same proof than above show that $\mathbb{R}^d / \mathbb{Z}^d \simeq \left(\mathbb{S}^1\right)^d$.