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When $K$ is a simplicial complex, the dual complex $C^*(K)$ to the chain complex $C_*(K)$ has a concrete interpretation: an element in $C^n(K)$ is given by assigning an integer to every oriented $n$-simplex in $K$. We define the $n$^{th} cohomology group of this cochain complex as $H^n(K)$ ($H_n(K)$ for homology).

As noted on p.190 in Hatcher's Algebraic Topology, there is a canonical homomorphism $h$ from $H^n(K)$ to Hom$(H_n(K), \mathbb{Z})$, with $h$ surjective.

Suppose $H_n (K)$ is free abelian for each $n \in \mathbb{N}$.

My question: How would I use this fact to verify that $h$ is injective as well, and hence $h$ is an isomorphism? Any help would be appreciated.

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Take a look at the Universal Coefficient Theorem:… Are you sure you don't need to assume $H_{n-1}(K)$ is free abelian? – you Mar 11 '12 at 0:46
Dear Vulcan, The homomorphism $h$ is not injective in general, as @you is (slightly indirectly) noting in their comment. Regards, – Matt E Mar 11 '12 at 1:48
@MattE: Speaking to your remark, are you aware of any counterexamples? – Vulcan Mar 11 '12 at 19:37
@Vulcan: Dear Vulcan, There are counterexamples in those cases where the homology is not torsion free, as the universal coefficients theorem shows. Regards, – Matt E Mar 11 '12 at 23:27

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