## 22.21 I-resolutions

This section is the dual of the section on P-resolutions.

Let $(A, \text{d})$ be a differential graded algebra. Let $I$ be a differential graded $A$-module. We say $I$ *has property (I)* if it there exists a filtration

\[ I = F_0I \supset F_1I \supset F_2I \supset \ldots \supset 0 \]

by differential graded submodules such that

$I = \mathop{\mathrm{lim}}\nolimits I/F_ pI$,

the maps $I/F_{i + 1}I \to I/F_ iI$ are admissible epimorphisms,

the quotients $F_ iI/F_{i + 1}I$ are isomorphic as differential graded $A$-modules to products of the modules $A^\vee [k]$ constructed in Section 22.19.

In fact, condition (2) is a consequence of condition (3), see Lemma 22.19.1. The reader can verify that as a graded module $I$ will be isomorphic to a product of $A^\vee [k]$.

Lemma 22.21.1. Let $(A, \text{d})$ be a differential graded algebra. Let $I$ be a differential graded $A$-module. If $F_\bullet $ is a filtration as in property (I), then we obtain an admissible short exact sequence

\[ 0 \to I \to \prod \nolimits I/F_ iI \to \prod \nolimits I/F_ iI \to 0 \]

of differential graded $A$-modules.

**Proof.**
Omitted. Hint: This is dual to Lemma 22.20.1.
$\square$

The following lemma shows that differential graded modules with property (I) are the analogue of K-injective modules. See Derived Categories, Definition 13.31.1.

Lemma 22.21.2. Let $(A, \text{d})$ be a differential graded algebra. Let $I$ be a differential graded $A$-module with property (I). Then

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(N, I) = 0 \]

for all acyclic differential graded $A$-modules $N$.

**Proof.**
We will use that $K(\text{Mod}_{(A, \text{d})})$ is a triangulated category (Proposition 22.10.3). Let $F_\bullet $ be a filtration on $I$ as in property (I). The short exact sequence of Lemma 22.21.1 produces a distinguished triangle. Hence by Derived Categories, Lemma 13.4.2 it suffices to show that

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(N, I/F_ iI) = 0 \]

for all acyclic differential graded $A$-modules $N$ and all $i$. Each of the differential graded modules $I/F_ iI$ has a finite filtration by admissible monomorphisms, whose graded pieces are products of $A^\vee [k]$. Thus it suffices to prove that

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(N, A^\vee [k]) = 0 \]

for all acyclic differential graded $A$-modules $N$ and all $k$. This follows from Lemma 22.19.3 and the fact that $(-)^\vee $ is an exact functor.
$\square$

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

$M \to I$ is injective,

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

$I$ 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 $.
$\square$

Lemma 22.21.4. 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 such that

$M \to I$ is a quasi-isomorphism, and

$I$ has property (I).

**Proof.**
Set $M = M_0$. We inductively choose short exact sequences

\[ 0 \to M_ i \to I_ i \to M_{i + 1} \to 0 \]

where the maps $M_ i \to I_ i$ are chosen as in Lemma 22.21.3. This gives a “resolution”

\[ 0 \to M \to I_0 \xrightarrow {f_0} I_1 \xrightarrow {f_1} I_1 \to \ldots \]

Denote $I$ the differential graded $A$-module with graded parts

\[ I^ n = \prod \nolimits _{i \geq 0} I^{n - i}_ i \]

and differential defined by

\[ \text{d}_ I(x) = f_ i(x) + (-1)^ i \text{d}_{I_ i}(x) \]

for $x \in I_ i^{n - i}$. With these conventions $I$ is indeed a differential graded $A$-module. Recalling that each $I_ i$ has a two step filtration $0 \to I_ i' \to I_ i \to I_ i'' \to 0$ we set

\[ F_{2i}I^ n = \prod \nolimits _{j \geq i} I^{n - j}_ j \subset \prod \nolimits _{i \geq 0} I^{n - i}_ i = I^ n \]

and we add a factor $I'_{i + 1}$ to $F_{2i}I$ to get $F_{2i + 1}I$. These are differential graded submodules and the successive quotients are products of shifts of $A^\vee $. By Lemma 22.19.1 we see that the inclusions $F_{i + 1}I \to F_ iI$ are admissible monomorphisms. Finally, we have to show that the map $M \to I$ (given by the augmentation $M \to I_0$) is a quasi-isomorphism. This follows from Homology, Lemma 12.26.3.
$\square$

## Comments (3)

Comment #536 by m.o. on

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