3
$\begingroup$

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 $f$?

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.