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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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Why is the split into torsion part and free part canonical? –  Cocopuffs Aug 11 '12 at 9:35
I think that the torsion part of an Abelian group $H^K(X.\mathbb{Z})$ is intrinsic. It consists of elements $x\in H^K(X.\mathbb{Z})$ such that $nx=0$ for some $n\in \mathbb{N}$. The free part is also intrinsic. –  Michel Aug 11 '12 at 15:23
Identifying the quotient $M / Tor(M)$ with the free part of $M$ usually involves choosing a basis somewhere ($M$ = Z-module = abelian group). I don't think you can do this canonically –  Cocopuffs Aug 11 '12 at 18:24
Really? Could you give me an example? –  Michel Aug 11 '12 at 21:19
math.ku.dk/~gelvin/Modules.pdf, see proposition 4.8 –  Cocopuffs Aug 12 '12 at 6:15
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