I'm trying to solve the following:
Let $\mathbb{Z}^n\times \mathbb{R}^n\to \mathbb{R^n}$ be a group action defined as $$(z_1,\dots,z_n)\times (x_1,\dots,x_n)\mapsto (x_1+z_1,\dots,x_n+z_n)$$ Prove that $\mathbb{R}^n/\mathbb{Z}^n$ is an $n$-manifold and show whether it is compact or not.
So what I thought so far:
I have to show that $\mathbb{R}^n/\mathbb{Z}^n$ is $(1)$ second countable, $(2)$ Hausdorff and $(3)$ locally homeomorphic to Euclidean space.
$(2$ Hausdorff $)$ : I need to show that the equivalence classes are closed. So let $x \in[x]$ and $y\in [y]$ and let $d(x,y):=\inf(|x-y|)$ be a metric that gives the smallest distance between two points. Now let $d(x,y)=\epsilon$, then we can find open sets $\bigcup\limits_{i=1}^n B_{\epsilon/2}(x_i)$ and $\bigcup\limits_{i=1}^n B_{\epsilon/2}(y_i)$ such that $$\bigcup\limits_{i=1}^n B_{\epsilon/2}(x_i)\cap \bigcup\limits_{i=1}^n B_{\epsilon/2}(y_i)=\emptyset$$
Then for any vector $z\in\mathbb{Z}^n$ also $$\bigcup\limits_{i=1}^n B_{\epsilon/2}(x_i+z_i)\cap \bigcup\limits_{i=1}^n B_{\epsilon/2}(y_i+z_i)=\emptyset$$
Here I don't know any further
$(3$ locally homeomorphic $)$ : Now chose an open set $U$ such that for $x\in [x]$ we have that $x\in U$. We have to show that $\exists \varphi:\mathbb{R}^n / \mathbb{Z}^n\to \mathbb{R}^n$ such that $\varphi(U)$ is an open subset in $\mathbb{R}^n$.
Now since $\mathbb{R}/\mathbb{Z} \simeq S^1$ we have that $$\mathbb{R}^n/\mathbb{Z}^n = (\mathbb{R}/\mathbb{Z} \times \dots \times \mathbb{R}/\mathbb{Z}) \simeq (S^1\times \dots \times S^1) = S^n$$ so there exist a homoemorphism $g:\mathbb{R}^n/\mathbb{Z}^n \to S^n$.
So let $f^{-1}(y):=e^{iy}$ for $y\in \mathbb{R}^n$ then $f(x)=\frac{1}{i}\ln(x)$ is a continuous map that maps from $S^n\to \mathbb{R}^n$, hence there is a continuous map $\varphi: \mathbb{R}^n / \mathbb{Z}^n \to \mathbb{R}^n$ and it's $\varphi(x):=f(g(x))$ such that $\forall U$ we have that $\varphi(U)\subset\mathbb{R}^n$ is an open set.
$(1)$ This one I don't know. Does it follow from $(2)$?
And I would say that it is a compact, but I'm not how to argue.
If you wish not to deal with all the notation and trying to actually define maps then you can prove the general statement; if $M$ is a manifold and $\Gamma$ is a discrete group s.t $\Gamma$ acts freely, and properly discontinuous on $M$ then $M/\Gamma$ is a manifold. In other words, show that the action $\cdot: \Gamma \times M \to M$ defined by $(z_1,...,z_n) \cdot (x_1,...,x_n) = (x_1+z_1,...,x_n+z_n)$ is free and properly-discointunous.
If you did notice that $M=\mathbb{R}^n/\mathbb{Z}^n = S^1 \times \cdots \times S^1$ then you can use the product of each atlas to define an atlas on $M$.