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Let $X$ be a compact manifold (so that all (co)homologyies have finite rank). The universal coefficient formula ($\mathbb{Z}$-coefficient for simplicity) says that we have the following short exact sequence $ 0\rightarrow Ext^{1}(H_{k-1}(X,\mathbb{Z}),\mathbb{Z})\rightarrow H^{k}(X,\mathbb{Z}) \rightarrow Hom(H_{k}(X,\mathbb{Z}),\mathbb{Z})\rightarrow 0. $ This is a split sequence but the splitting is not natural but it seems to me that this splits quite canonically in case the coefficient is $\mathbb{Z}$. Am I right?

Here is my argument; since $Ext^{1}(H_{k-1}(X,\mathbb{Z}),\mathbb{Z})$ is torsion and $Hom(H_{k}(X,\mathbb{Z}),\mathbb{Z})$ is free. Thus it splits canonically as a direct sum of the torsion part and the free part $ H^{k}(X,\mathbb{Z})=Ext^{1}(H_{k-1}(X,\mathbb{Z}),\mathbb{Z})\oplus Hom(H_{k}(X,\mathbb{Z}),\mathbb{Z}). $

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    You are right. There is no way to canonically identify $H^k(X,\mathbb{Z})/Ext(H_{k-1}(X,\mathbb{Z}),\mathbb{Z})$ with $Hom(H_{k}(X,\mathbb{Z}),\mathbb{Z})$. Thanks.2012-08-12

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