Diffeomorphism on the torus Let $S : \mathbb{R}^n → \mathbb{R}^n$ be linear invertible map, then $S$ projects to $\mathbb{T}^n$ diffeomorphism
if and only if $S ∈ GL_n(\mathbb{Z})$.
I can't prove the right to left implication.
 A: I'm assuming that for you $\mathbb T^n = \mathbb R^n / \Gamma$, where $\Gamma$ is a full lattice in $\mathbb R^n$. In this case, the left to right implication is just the fact that if $S \in \text{GL}_n(\mathbb Z)$ then $S(\Gamma) \subset \Gamma$, hence universal property of the quotient concludes.
In fact, being the action of $\Gamma$ over $\mathbb R^n$ properly discontinuous, you can see $\mathbb T^n$ as a quotient in the category of differentiable manifolds. Hence your $S$ factors through $\mathbb T^n$ producing a map $\mathbb T^n \to \mathbb T^n$. In the same way you can factors its inverse, hence you get a diffeomorphism of the torus (you check that the two induced morphisms are inverse each other using again the universal property).
Edit: To do the converse, you don't even need the universal property. Assume in fact that $S$ is a linear invertible map inducing a diffeomorphism of the torus. Apparently I'm not allowed to draw commutative diagrams... So let $p \colon \mathbb R^n \to \mathbb T^n$ be the canonical projection and let $\overline{S} \colon \mathbb T^n \to \mathbb T^n$ the map induced by $S$. Then $p \circ S = \overline{S} \circ p$. This means that $p(S(\Gamma)) = \overline{S}(p(\Gamma)) = \overline{S}([\mathbf 0]) = [\mathbf 0]$, hence $S(\Gamma) \subset \Gamma$. Since $\Gamma = \mathbb Z^n$, and since the vectors $\mathbf e_i = (0,\ldots,1,\ldots,0)$ form a basis both for $\Gamma$ and for $\mathbb R^n$, you can write down the matrix of $S$ explicitly: in the $i$-th column you will have the coefficients of $S(\mathbf e_i)$ with respect to this basis; since $S(\Gamma) \subset \Gamma$ you see that $S(\mathbf e_i) \in \mathbb Z^n$, hence the coefficients must be integers! This implies $S \in M_n(\mathbb Z)$.
Now, your hypothesis is that $\overline{S}$ is a diffeomorphism of $\mathbb T^n$. Let $f \colon \mathbb T^n \to \mathbb T^n$ be its inverse. Since the action of $\mathbb Z^n$ over $\mathbb R^n$ is properly discontinous, then the canonical projection $p$ is a covering; since $\mathbb R^n$ is simply connected, this is the universal covering. Thus $f$ lifts (uniquely!) to a continuous map $T \colon \mathbb R^n \to \mathbb R^n$. Universal property of universal covering implies that $T$ is the inverse of $S$, hence $T \in \text{GL}_n(\mathbb R)$. The same reasoning of above shows $T \in M_n(\mathbb Z)$. Therefore $S, T \in \text{GL}_n(\mathbb Z)$.
A: Just to stress the purely algebraic part of this, I post what follows; I apologize if it is redundant. Since the mapping $\mathbb R^n\to\mathbb T^n$ is a surjective local diffeomorphism, the linear isomorphism $S$ defines a diffeo of $\mathbb T^n$ if and only if it defines an injective mapping of $\mathbb T^n$. We indentify $S$ with its matrix.
(1) To define a mapping of $\mathbb T^n$ means that $x-y\in\mathbb Z^n$ implies $Sx-Sy\in\mathbb Z^n$, or $S$ being linear, $u\in\mathbb Z^n$ implies $Su\in\mathbb Z^n$. Of course this happens if $S$ has integer coefficients. And conversely, as the columns of $S$ are $Se_i$ for $e_i=(0,\dots,1,\dots,0)$. In other words, $S$ defines a mapping of $\mathbb T^n$ if and only if $S$ has integer coefficients. 
(2) The induced mapping is injective exactly when $Sx-Sy\in\mathbb Z^n$ implies $x-y\in\mathbb Z^n$. But this just says that $S^{-1}$ defines a mapping of $\mathbb T^n$, which now we know equivalent to $S^{-1}$ have integer coefficients.
Summing up, $S$ induces a diffeo if and only if both $S$ and $S^{-1}$ have integer coefficiens, or equivalently $S\in GL_n(\mathbb Z)$ (and then of course $\det=\pm1$).
