Let $f\colon A^\bullet\to I^\bullet$ be a morphism of bounded below complexes in an abelian category. Assume all $I^i$ are injective objects. Assume also that $f$ induces the zero map on cohomology.
Is it true that $f$ is homotopic to 0?
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No. Take a diagram$$\require{AMScd} \begin{CD} 0@>{}>>0 @>{}>> I^{1}@>{}>>0 @>{}>>\cdots\\ @VVV @VVV @VVV^{\!\!\!\operatorname{Id}}@VVV \\ 0@>{}>>I^0 @>{d}>> I^1@>{}>>0@>{}>>\cdots \end{CD}$$with $I^0,I^1$ injective and $d$ a surjective homomorphism which doesn't split. Then $H^0(A^{\bullet})=H^1(I^{\bullet})=0$, so $H^{\bullet}(f)=0$, but $f$ homotopic to $0$ would mean that $d$ splits.
