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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 $$ \cdots \to H^0(U, \mathbb{Z}) \to H^1(V/U, \mathbb{Z}) \to H^1(V, \mathbb{Z})=0 \to H^1(U, \mathbb{Z}) \cdot$$

usr0192
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  • I think that follows from the Universal Coefficient Theorem for cohomology. https://en.wikipedia.org/wiki/Universal_coefficient_theorem#Universal_coefficient_theorem_for_cohomology – T.J. Gaffney Jul 05 '17 at 19:37
  • @Gaffney That possibility crossed my mind, and you may be right. However the universal coefficient theorem statement involves a single space $X$, whereas here we have $X, U$ and $V$. UCT seems to give $H^1(X, \mathbb{Z}) = Hom(H_1(X, \mathbb{Z}), \mathbb{Z})$ – usr0192 Jul 05 '17 at 19:44
  • You're right. I didn't read right. – T.J. Gaffney Jul 05 '17 at 19:56

2 Answers2

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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 $\operatorname{Mor}(\mathbb{Z}^2,\mathbb{Z})\cong\prod_{a\in\mathbb{Z}^2}\mathbb{Z}$.

Kyle Miller
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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. $$H^1(X)\cong \operatorname{Hom}(H_1(X),\mathbb{Z})\cong\operatorname{Hom}(\pi_1(X),\mathbb{Z})\cong\operatorname{Hom}(U,\mathbb{Z}).$$

Kyle Miller
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