Why does this exact sequence exist?

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.

• It is related to the universal coefficients theorem, actually some authors even consider this part of the universal coefficients theorem. However it seems you're missing some hypotheses on $A$, are you sure there are not more ? – Max May 16 at 16:21
• In the context I encountered this, it was applied to a specific complex $A^\bullet$. It could well be that this specific complex satisfies the extra hypotheses on $A^\bullet$ you are talking about. Could you tell me what the missing hypotheses are? Do you happen to know a reference for this formulation of Universal Coefficients? – Yo. May 16 at 19:05
• Usually $A$ is assumed to be a chain complex of free abelian groups or something along those lines. I don't have a reference just right now but I'll come back if this hasn't been answered when I have the time – Max May 16 at 19:16
• What means $h^n$? – san Jun 22 at 20:23
• @san $h^n(A^\bullet)$ is the $n$-th cohomology of the complex $A^\bullet$ – Yo. Jun 26 at 12:20