Why does this exact sequence exist?

Question

I'm reading a proof and don't understand a certain part. Let $A^\bullet$ be a (cochain) complex of abelian groups. Let $I^\bullet$ be an injective resolution of an abelian group $B$. Then there is a short exact sequence \begin{equation} 0\to \mathrm{Ext}^1_{\mathrm{Ab}}(h^{n+1}(A^\bullet), B)\to h^{-n}(\mathrm{Hom}^\bullet(A^\bullet, I^\bullet))\to \mathrm{Hom}(h^n(A^\bullet), B)\to 0. \end{equation}Moreover, the sequence splits. I don't see where this sequence is coming from. It is given without further attention or reference, so it should be something that is standard or not that hard to see. It does remind me of the Universal Coefficient Theorem, but I am not able to make the link... Any help, proof or reference is appreciated.

Answer 1

I saw this old question when viewing my own profile, and thought it might be a good idea to answer it. The exact sequence is indeed just a form of the UCT. The confusion is due to the conventions of chain vs cochain. Very explicit:

For a chain complex $A_\bullet$, write $\phi(A_\bullet)$ for the associated cochain complex. Then $h^{-n}(\phi(A_\bullet)) = h_{n}(A_\bullet)$.

We have $\phi(\mathrm{Hom}(A^\bullet, B)) = \mathrm{Hom}^\bullet(A^\bullet, B[0])$, where $B[0]$ is the cochain complex with $B$ in degree zero and $0$ everywhere else. As $I^\bullet$ is an injective resolution of $B$ we have \begin{equation*} (\star) \ \ \ \ \ \ h_n(\mathrm{Hom}(A^\bullet, B)) = h^{-n}(\mathrm{Hom}^\bullet(A^\bullet, B[0])) = h^{-n}(\mathrm{Hom}^\bullet(A^\bullet, I^\bullet)). \end{equation*}

If $A^\bullet$ is projective, then the UCT gives a split exact sequence \begin{equation*} 0\to \mathrm{Ext}^1(h^{n+1}(A^\bullet), B)\to h_n(\mathrm{Hom}(A^\bullet, B))\to \mathrm{Hom}(h^n(A^\bullet), B)\to 0. \end{equation*}

We now conclude by ($\star$).