How to recover the cohomology of a torus from its description of a quotient Note: here, "cohomology" means "De Rham cohomology".
I know how to compute the De Rham Cohomology of a torus $T=\left(S^1\right)^n$ using Kunneth formula. 
But a torus can also be obtain as a quotient $T_\Gamma$ of $\mathbb R^n$ by a discrete subgroup $\Gamma$ of rank $n$ of $\mathbb R^n$. Is it possible to compute the cohomology of the torus from that information? 
My only idea is to use the canonical surjection $\pi:\mathbb R^n\to T_\Gamma$ to induce an application $\pi^\ast$ from $H^p(T_\Gamma)$ into $H^p(\mathbb R^n)$, but since $H^p(\mathbb R^n)=0$ for $p>0$, I don't see how it can be useful...
 A: I'll explain an analytic approach to computing the de Rham cohomology, to complement Qiaochu's algebraic one. Put any Riemannian metric on the torus, then there is a natural map from the space $\mathcal{H}^k(T^n)$ of harmonic $k$-forms to the de Rham cohomology $H^k_{dR}(T^n,\mathbb{R})$, where we send a harmonic form to its de Rham class. In fact, the Hodge theorem stipulates that this map is an isomorphism; that is, each de Rham class has a unique harmonic representative. 
If we put the flat metric on the torus (the metric induced by the quotient map $\mathbb{R}^n \to \mathbb{R}^n/\Gamma = T^n$), then the only global harmonic functions are the constants. The harmonic $k$-forms on $\mathbb{R}^n$ (and hence on $T^n$) are precisely those with harmonic coefficients, the harmonic $k$-forms on $T^n$ are those with constant coefficients. It follows that an $\mathbb{R}$-basis of $k$-forms on $T^n$ is
$$
\{ {dx}_{i_1} \wedge \ldots \wedge {dx}_{i_k} \colon 0 \leq i_1 < \ldots < i_k \leq n \},
$$
where $x_1,\ldots,x_n$ are the coordinates on $T^n$ inherited from $\mathbb{R}^n$.
Therefore, 
$$
{ n \choose k} = \dim_{\mathbb{R}} \mathcal{H}^k(T) = \dim_{\mathbb{R}} H^k_{dR}(T,\mathbb{R}).
$$
A: $\mathbb{R}^n$ is contractible, so this description tells you that the torus is a classifying space $B \mathbb{Z}^n$, or equivalently an Eilenberg-MacLane space $K(\mathbb{Z}^n, 1)$, and hence its cohomology can be identified with the group cohomology of $\mathbb{Z}^n$. You can also compute this using the Kunneth formula, but there are other ways. 
The really nice general statement is about quotients by free actions of finite groups, but infinite groups are much trickier. For the case of finite groups $G$, if $X \to X/G$ is such a quotient then $H^{\bullet}(X/G, \mathbb{R}) \to H^{\bullet}(X, \mathbb{R})$ is an injection (note that this fails for the torus) with image precisely the $G$-invariant subalgebra $H^{\bullet}(X, \mathbb{R})^G$. The way you prove this in de Rham cohomology is by averaging forms over $G$, which you can't do if $G$ is infinite. 
