Lemma 22.21.3. Let $(A, \text{d})$ be a differential graded algebra. Let $M$ be a differential graded $A$-module. There exists a homomorphism $M \to I$ of differential graded $A$-modules with the following properties

1. $M \to I$ is injective,

2. $\mathop{\mathrm{Coker}}(\text{d}_ M) \to \mathop{\mathrm{Coker}}(\text{d}_ I)$ is injective, and

3. $I$ sits in an admissible short exact sequence $0 \to I' \to I \to I'' \to 0$ where $I'$, $I''$ are products of shifts of $A^\vee$.

Proof. We will use the functors $N \mapsto N^\vee$ (from left to right differential graded modules and from right to left differential graded modules) constructed in Section 22.19 and all of their properties. For every $k \in \mathbf{Z}$ let $Q_ k$ be the free left $A$-module with generators $x, y$ in degrees $k$ and $k + 1$. Define the structure of a left differential graded $A$-module on $Q_ k$ by setting $\text{d}(x) = y$ and $\text{d}(y) = 0$. Arguing exactly as in the proof of Lemma 22.20.3 we find a surjection

$\bigoplus \nolimits _{i \in I} Q_{k_ i} \longrightarrow M^\vee$

of left differential graded $A$-modules. Then we can consider the injection

$M \to (M^\vee )^\vee \to (\bigoplus \nolimits _{i \in I} Q_{k_ i})^\vee = \prod \nolimits _{i \in I} I_{k_ i}$

where $I_ k = Q_{-k}^\vee$ is the “dual” right differential graded $A$-module. Further, the short exact sequence $0 \to A[-k - 1] \to Q_ k \to A[-k] \to 0$ produces a short exact sequence $0 \to A^\vee [k] \to I_ k \to A^\vee [k + 1] \to 0$.

The result of the previous paragraph produces $M \to I$ having properties (1) and (3). To obtain property (2), suppose $\overline{m} \in \mathop{\mathrm{Coker}}(\text{d}_ M)$ is a nonzero element of degree $k$. Pick a map $\lambda : M^ k \to \mathbf{Q}/\mathbf{Z}$ which vanishes on $\mathop{\mathrm{Im}}(M^{k - 1} \to M^ k)$ but not on $m$. By Lemma 22.19.3 this corresponds to a homomorphism $M \to A^\vee [k]$ of differential graded $A$-modules which does not vanish on $m$. Hence we can achieve (2) by adding a product of copies of shifts of $A^\vee$. $\square$

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