Proof of the comparison theorem (homological algebra) for injective resolutions I am trying to prove the existence part of the comparison theorem for injective resolutions. Every reference I find says that it is just the dualization of the proof for projective resolutions, but I am not seeing how to dualize it. The issue is that in the proof of the projective case, they assume inductively that a chain map has been constructed at degree $n$. Then they use the commutativity of the square to show that $n$-cycles get taken to $n$-cycles. 
My idea to dualize this was to use the fact that boundaries get taken to boundaries. However, there's no way to do this because in the dual case, you don't yet have the chain map $f_{n+1}$ that can take those boundaries to boundaries. 
Has anyone got a link to a proof for the injective case, or is someone able to tell me how to dualize the standard proof for the projective case? 

Suppose that $f : M \to N$ is a map of modules, and that $M\to I^*$ and $N\to J^*$ are injective resolutions of $M$ and $N$ respectively. Begin by noting that we have an incomplete diagram of the form
$$
\require{AMScd}
\begin{CD}
M @>>> I^0\\
@VfVV   \\
N @>>> J^0
\end{CD}$$
where both horizontal maps are injective. Now consider the map $M\to J^0$ obtained by composition. Since $J^0$ is injective, this can be extended to a map $I^0\to J^0$, rendering the diagram commutative. This starts the induction. In the projective case you would now consider the following diagram
$$
\require{AMScd}
\begin{CD}
 &&   P^k\\
&&@VVV   \\
Q^k @>>> \operatorname{im} d@>>> 0
\end{CD}$$
to complete the induction. Can you imagine what to do in this dual case? You should try to obtain a diagram of the form
$$\require{AMScd}
\begin{CD}
0 @>>> ?? @>>>    I^k\\
&&@VVV   \\
&& J^k 
\end{CD}$$
and use that $J^k$ is injective. 
