
## 22.14 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

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

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

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

In fact, condition (2) is a consequence of condition (3), see Lemma 22.12.1. The reader can verify that as a graded module $I$ will be isomorphic to a product of $A^\vee [k]$.

Lemma 22.14.1. 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 22.13.1. $\square$

The following lemma shows that differential graded modules with property (I) are the analogue of K-injective modules. See Derived Categories, Definition 13.29.1.

Lemma 22.14.2. 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 22.14.1 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 22.12.3 and the fact that $(-)^\vee$ is an exact functor. $\square$

Lemma 22.14.3. 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. 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 $\text{d}(y) = 0$. Let $I_ k = Q_{-k}^\vee$ be the “dual” right differential graded $A$-module, see Section 22.12. The next paragraph shows that we can embed $M$ into a product of copies of $I_ k$ (for varying $k$). The dual statement (that any differential graded module is a quotient of a direct sum of $P_ k$'s) is easy to prove (see proof of Lemma 22.13.3) and using double duals there should be a noncomputational way to deduce what we want. Thus we suggest skipping the next paragraph.

Given a $\mathbf{Z}$-linear map $\lambda : M^ k \to \mathbf{Q}/\mathbf{Z}$ we construct pairings

$\psi _ n : M^ n \times Q_ k^{-n} \longrightarrow \mathbf{Q}/\mathbf{Z}$

by setting

$\psi _ n(m, ax + by) = \lambda (ma + (-1)^{k + 1}\text{d}(mb))$

for $m \in M^ n$, $a \in A^{-n - k}$, and $b \in A^{-n - k - 1}$. We compute

\begin{align*} \psi _{n + 1}(\text{d}(m), ax + by) & = \lambda \left(\text{d}(m)a + (-1)^{k + 1}\text{d}(\text{d}(m)b)\right) \\ & = \lambda \left(\text{d}(m)a + (-1)^{k + n}\text{d}(m)\text{d}(b)\right) \end{align*}

and because $\text{d}(ax + by) = \text{d}(a)x + (-1)^{-n - k}ay + \text{d}(b)y$ we have

\begin{align*} \psi _ n(m, \text{d}(ax + by)) & = \lambda \left( m\text{d}(a) + (-1)^{k + 1}\text{d}(m((-1)^{-n - k}a + \text{d}(b))) \right) \\ & = \lambda \left( m\text{d}(a) + (-1)^{-n + 1}\text{d}(ma) + (-1)^{k + 1}\text{d}(m)\text{d}(b))) \right) \end{align*}

and we see that

$\psi _{n + 1}(\text{d}(m), ax + by) + (-1)^ n\psi _ n(m, \text{d}(ax + by)) = 0$

Thus these pairings define a homomorphism $f_\lambda : M \to I_ k$ by Lemma 22.12.2 such that the composition

$M^ k \xrightarrow {f^ k_\lambda } I_ k^ k = (Q_ k^ k)^\vee \xrightarrow {\text{evaluation at }x} \mathbf{Q}/\mathbf{Z}$

is the given map $\lambda$. It is clear that we can find an embedding into a product of copies of $I_ k$'s by using a map of the form $\prod f_\lambda$ for a suitable choice of the maps $\lambda$.

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 22.12.3 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 22.14.4. 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

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

2. $I$ 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 22.14.3. This gives a “resolution”

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

Then we set

$I = \prod \nolimits _{i \geq 0} I_ i$

where we take the product in the category of graded $A$-modules and differential defined by

$\text{d}_ I(x) = f_ a(x) + (-1)^ a \text{d}_{I_ a}(x)$

for $x \in I_ a^ b$. 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

$F_{2i}P = \prod \nolimits _{j \geq i} I_ j \subset \prod \nolimits _{i \geq 0} I_ i = I$

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 22.12.1 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 12.23.3. $\square$

Comment #536 by m.o. on

There seems to be a kind of typos in Condition 2) in the definition of Property (I): I wonder if Condition 2) in Property (I) is that the quotients $I/F_iI \to I/F_{i-1}I$ are admissible epimorphisms, which corresponds to Condition 2) in Property (P). If $I$ has only finite many nonzero graded pieces, then these conditions are equivalent, but it seems they are not equivalent in general. Also the actual property that is used in this section is that the quotients $I/F_iI \to I/F_{i-1}I$ are admissible epimorphisms.

Comment #537 by m.o. on

I am sorry, but I have made a mistake. Under Condition 1), Condition 2) is equivalent to the condition that the quotients $I/F_iI\to I/F_{i-1}I$ are admissible epimorphisms.

Comment #548 by on

Actually, I did make the change as I really want this to be, as much as possible, dual to the section on P-resolutions. I also try to say it a bit more clearly. You can find the edit here.

In the paper by Keller, the corresponding sections are not exactly dual either, so either. I am not sure why.

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