# Choosing an injective resolution of a short exact sequence of complexes

Lemma: Given a short exact sequence of cochain complexes in an abelian category $\mathcal{C}$ with enough injectives, $$0 \to P^\bullet \xrightarrow{f} Q^\bullet \xrightarrow{g} R^\bullet \to 0,$$ There exist complexes $I^\bullet, J^\bullet, K^\bullet$ of injective objects, and a diagram commmutative up to homotopy $$\begin{array}{} 0 & \rightarrow & P^\bullet & \overset{f}{\rightarrow} & Q^\bullet & \overset{g}{\rightarrow} & R^\bullet & \rightarrow & 0 \\ && \!\!\!\! \sim \downarrow a&& \!\!\!\!\sim \downarrow b && \!\!\!\!\sim \downarrow c \\ 0 & \rightarrow & I^\bullet & \overset{u}{\rightarrow} & J^\bullet & \overset{v}{\rightarrow} & K^\bullet & \rightarrow & 0 \end{array}$$ where the rows are exact, and the vertical arrows are homotopy equivalences.

In the proof of this theorem in Iversen's book, he begins by choosing an injective resolution of $Q^\bullet$ and $R^\bullet$, and a map $v$ (unique up to homotopy) between them: $$\begin{array}{} 0 & \rightarrow & P^\bullet & \overset{f}{\rightarrow} & Q^\bullet & \overset{g}{\rightarrow} & R^\bullet & \rightarrow & 0 \\ &&&& \!\!\!\!\sim \downarrow b && \!\!\!\!\sim \downarrow c \\ & & & & J^\bullet & \overset{v}{\rightarrow} & K^\bullet & \rightarrow & 0 \end{array}$$ Since the diagram is homotopy commutative, he selects a homotopy $Q^\bullet \xrightarrow{s} K^\bullet[-1]$ between $c \circ g \Rightarrow v \circ b$.

So far, so good. Then he says

Choose a map $t:Q^\bullet \xrightarrow{} J^\bullet[-1]$ such that $s=v \circ t$.

This is where I fall off. Why does $s$ necessarily factor through $v[-1]$?

• Do you know the proof of the dual statement? – Pedro Tamaroff Jun 25 '16 at 23:33
• Note you are changing the names of your objects and arrows. – Pedro Tamaroff Jun 25 '16 at 23:38
• @PedroTamaroff Whoops, just fixed it. No, I don't know the proof of the dual. Only for a projective resolution of a short exact sequence of objects (via the "horseshoe lemma"). Not for complexes. – Eric Auld Jun 25 '16 at 23:42

He means choose a map of graded modules, not a chain map. This is possible because $v$ admits a section as a map of graded modules (not as chain complexes).