TITLE: How to recover the cohomology of a torus from its description of a quotient
QUESTION [5 upvotes]: 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...

REPLY [3 votes]: 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, 
$$
{ n \choose k} = \dim_{\mathbb{R}} \mathcal{H}^k(T) = \dim_{\mathbb{R}} H^k_{dR}(T,\mathbb{R}).
$$