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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

  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 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

  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

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

  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 22.21.3. 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

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 are admissible epimorphisms, which corresponds to Condition 2) in Property (P). If 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 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 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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