Let $A,B$ be two bounded below complexes in module category, and $A \longrightarrow I$ (resp. $B \longrightarrow J$) a injective resolution. If $f: A \longrightarrow B$ is a morphism of complexes.


My question is: how to construct a morphism from $I$ to $J$ which induced by $f$?



morphism of injective objects
Note that there is in general *no* morphism $g: I \to J$ such that the following diagram commutes:
$$\begin{array}{ccc}
I & \xrightarrow{g} & J \newline 
i\uparrow & & \uparrow j \newline
A & \xrightarrow[f]{} & B
\end{array}\tag{$\ast$}$$
For, the commutativity of the diagram forces $\ker(i) \subseteq \ker(j\circ f)$, but the three maps $i,j,f$ are completely independent from each other. *However* $g$ can be choosen such that the diagram commutes *up to homotopy*. Moreover, $g$ is unique up to homotopy. The proof uses the 



> 
> **Comparison Theorem:** If $I$ is a bounded below complex of injectives and $\alpha: X \to Y$ is a weak equivalence of two arbitrary complexes, then 
>  $$[Y,I] \to [X,I],\; [h] \mapsto [h \circ \alpha]$$
>  is an isomorphism of homotopy classes. 
> 
> 
> 


A reference for (the projective version of) the comparison theorem is Brown: Cohomology of Groups, Theorem I.8.5. 


We are now ready to construct $g$ above: $i: A \to I$ is a weak equivalence and $J$ is injective, bounded below. Hence $[I,J] \to [A,J],\; [h] \mapsto [h \circ i]$ is an isomorphism. Because of $[j \circ f] \in [A,J]$ there is consequently $g: I \to J$ with $[g \circ i]=[j \circ f]$, i.e. $g\circ i \simeq j \circ f$. The uniqueness of $g$ is shown in the same way using the injectivity of the comparison map. 


