Let $\Lambda = \bigoplus_{i=1}^n u_i \mathbb{Z}$ be a complete lattice in $\mathbb{R}^n$. This gives rise to the topological space $M:=\mathbb{R}^n/\Lambda$ equipped with the quotient topology; and I would like to prove that $M$ is a differentiable manifold.
If $\pi \colon \mathbb{R}^n \to M$ denotes the projection and if $U_0:=\{\sum_{i=1}^n a_i u_i : \vert a_i \vert < \frac{1}{2}\}$, then $\pi$ induces a homeomorphism $\varphi_0\colon U_0 \cong \pi(U_0)=:V_0$. For arbitrary $x\in \mathbb{R}^n$, we then have homeomorphisms $\varphi_x\colon U_0 \cong \pi(U_0+x)=:V_x$, $u\mapsto \pi(u+x)$. These define charts and I have to prove that the transition maps are differentiable. What exactly do the transition maps look like?
$\varphi_y^{-1} \circ \varphi_x$ sends $a=\sum_i a_i u_i$ to $\varphi_y^{-1}([a+x])$. Now I need to write $[a+x]=[b+y]$ with $b\in U_0$, i.e. $b=a+(x-y)+\sum_i z_i u_i \in U_0$ for some $z_i\in \mathbb{Z}$.
How can I make this more explicit?
Maybe an easier way: $S^1 = \{z \in \mathbb{C} : |z| = 1 \}$ is a differentiable manifold with charts $U_1 = S^1 \setminus \{1\}$ and $U_{-1} = S^1 \setminus \{-1\}$.
Now $\mathbb{R}/\mathbb{Z} \cong S^1$ as topological spaces via $x + \mathbb{Z} \mapsto e^{2 \pi i x}$. Thus you can make $\mathbb{R}/\mathbb{Z}$ into a manifold.
Finally,
$$\mathbb{R}^n/\Gamma \cong \prod\limits_{i=1}^n \mathbb{R}/\mathbb{Z}$$
as topological spaces, and a product of manifolds is naturally a manifold.