derivation of cohomology of a torus Let $V$ be an $n$-dimensional complex vector space and $U \subseteq V$ a full dimensional lattice (i.e. $U \cong \mathbb{Z}^{2n}$) and let $X=V/U$.
Something I'm reading says "since $V$ is contractible, $H^1(X, \mathbb{Z}) = Hom(U, \mathbb{Z})$". Why? I don't follow. 
Perhaps there is a long exact sequence like 
$$ \cdots \to H^0(U, \mathbb{Z}) \to H^1(V/U, \mathbb{Z}) \to H^1(V, \mathbb{Z})=0 \to H^1(U, \mathbb{Z}) \cdot$$
 A: Edit: The question is surely about the group quotient $V/U$ instead of a topological quotient $V/U$.  This answer is for what happens if it's a topological quotient; see my other answer for the group quotient.
There is a CW decomposition of $V$ such that $U$ is a subcomplex, hence $(V,U)$ is pair with a long exact sequence of reduced cohomology groups
$$\widetilde{H}^0(X) \to \widetilde{H}^0(V) \to \widetilde{H}^0(U) \to \widetilde{H}^1(X)\to \widetilde{H}^1(V)\to\widetilde{H}^1(U).$$
Since $V$ is contractible, both of the reduced cohomology groups associated to $V$ are $0$, so we have an isomorphism
$$\widetilde{H}^0(U)\cong\widetilde{H}^1(X).$$
In general, $\widetilde{H}^1(X)\cong H^1(X)$.  The space $U$ is discrete, so $\widetilde{H}^0(U)$ should be $\operatorname{Mor}(\mathbb{Z}^2,\mathbb{Z})/\mathbb{Z}$, which is isomorphic as a $\mathbb{Z}$-module to $\operatorname{Mor}(\mathbb{Z}^2,\mathbb{Z})$.  Here, morphism sets are sets of $\mathbb{Z}$-valued functions on the space $\mathbb{Z}^2$, and in particular $\operatorname{Mor}(\mathbb{Z}^2,\mathbb{Z})\cong\prod_{a\in\mathbb{Z}^2}\mathbb{Z}$.
A: The space $X$ is a space with a contractible cover $V$, with $U$ the group of deck transformations.  It follows that $\pi_1(X)\cong U$.  Using your favorite method, you can then deduce $H^1(X)\cong \operatorname{Hom}(U,\mathbb{Z})$.
One method is that $H^1(X,\mathbb{Z})$ is isomorphic to $[X,S^1]$ (since $S^1$ is a $K(\mathbb{Z},1)$).  Then, since $\pi_1(X)=U$, $[X,S^1]$ is $\operatorname{Hom}(U,\mathbb{Z})$.
Another method is the universal coefficient theorem.  $$H^1(X)\cong \operatorname{Hom}(H_1(X),\mathbb{Z})\cong\operatorname{Hom}(\pi_1(X),\mathbb{Z})\cong\operatorname{Hom}(U,\mathbb{Z}).$$
