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In the universal coefficient theorem for cohomology with a chain complex $C$ of free abelian groups having homology groups $H_n(C)$, \begin{eqnarray} 0\rightarrow\text{Ext}(H_{n-1}(C),G)\rightarrow H^n(C,G)\xrightarrow{h}\text{Hom}(H_n(C),G)\rightarrow0, \end{eqnarray} $\text{Ext}(H_{n-1}(C),G)$ can be identified with $\text{Coker}i^*_{n-1}$ in the dualized sequence \begin{eqnarray} 0\leftarrow B_{n-1}^*\xleftarrow{i^*_{n-1}}Z_{n-1}^*\leftarrow H_{n-1}(C)^*\leftarrow0 \end{eqnarray} where $i^*_{n-1}$ is dualization of the inclusion $B_{n-1}\xrightarrow{i_{n-1}}Z_{n-1}$. However, I cannot see why $\text{Coker}i^*_{n-1}$ (equivalently $\text{Ext}(H_{n-1}(C),G)$) can be nontrivial in the above case because any homomorphism Hom$(B_{n-1},G)$ seems to be able to be constructed from the restriction of Hom$(Z_{n-1},G)$ on $B_{n-1}$, namely $i^*_{n-1}$ being surjective.

Could you tell me how to see the obstruction in construction of Hom$(B_{n-1},G)$ from the restriction of Hom$(Z_{n-1},G)$ on $B_{n-1}$.

Thanks very much in advance!

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We typically get examples with non-trivial cokernels where there is torsion around. For a simple example, let $C_0$ and $C_1$ both equal $\mathbb{Z}$ with all other groups $C_n$ being zero. Let $d:C_1\to C_0$ be the map taking $m$ to $2m$. Take $G$ to be $\mathbb{Z}$. In this case $Z_0=\mathbb{Z}$ and $B_0=2\mathbb{Z}$. Not every element in $B_0^*=\textrm{Hom}(2\mathbb{Z},\mathbb{Z})$ is the restriction of an element of something in $Z_0^*=\textrm{Hom}(\mathbb{Z},\mathbb{Z})$; for instance the map $2m\mapsto m$ isn't.

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  • $\begingroup$ Thanks very much for your brief but helpful answer and I have got your point! $\endgroup$
    – Smart Yao
    Commented Apr 14, 2017 at 12:57

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