## 22.20 P-resolutions

This section is the analogue of Derived Categories, Section 13.29.

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

$0 = F_{-1}P \subset F_0P \subset F_1P \subset \ldots \subset P$

by differential graded submodules such that

1. $P = \bigcup F_ pP$,

2. the inclusions $F_ iP \to F_{i + 1}P$ are admissible monomorphisms,

3. the quotients $F_{i + 1}P/F_ iP$ are isomorphic as differential graded $A$-modules to a direct sum of $A[k]$.

In fact, condition (2) is a consequence of condition (3), see Lemma 22.16.1. Moreover, the reader can verify that as a graded $A$-module $P$ will be isomorphic to a direct sum of shifts of $A$.

Lemma 22.20.1. Let $(A, \text{d})$ be a differential graded algebra. Let $P$ be a differential graded $A$-module. If $F_\bullet$ is a filtration as in property (P), then we obtain an admissible short exact sequence

$0 \to \bigoplus \nolimits F_ iP \to \bigoplus \nolimits F_ iP \to P \to 0$

of differential graded $A$-modules.

Proof. The second map is the direct sum of the inclusion maps. The first map on the summand $F_ iP$ of the source is the sum of the identity $F_ iP \to F_ iP$ and the negative of the inclusion map $F_ iP \to F_{i + 1}P$. Choose homomorphisms $s_ i : F_{i + 1}P \to F_ iP$ of graded $A$-modules which are left inverse to the inclusion maps. Composing gives maps $s_{j, i} : F_ jP \to F_ iP$ for all $j > i$. Then a left inverse of the first arrow maps $x \in F_ jP$ to $(s_{j, 0}(x), s_{j, 1}(x), \ldots , s_{j, j - 1}(x), 0, \ldots )$ in $\bigoplus F_ iP$. $\square$

The following lemma shows that differential graded modules with property (P) are the dual notion to K-injective modules (i.e., they are K-projective in some sense). See Derived Categories, Definition 13.31.1.

Lemma 22.20.2. Let $(A, \text{d})$ be a differential graded algebra. Let $P$ be a differential graded $A$-module with property (P). Then

$\mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(P, N) = 0$

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 $P$ as in property (P). The short exact sequence of Lemma 22.20.1 produces a distinguished triangle. Hence by Derived Categories, Lemma 13.4.2 it suffices to show that

$\mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(F_ iP, N) = 0$

for all acyclic differential graded $A$-modules $N$ and all $i$. Each of the differential graded modules $F_ iP$ has a finite filtration by admissible monomorphisms, whose graded pieces are direct sums of shifts $A[k]$. Thus it suffices to prove that

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

for all acyclic differential graded $A$-modules $N$ and all $k$. This follows from Lemma 22.16.2. $\square$

Lemma 22.20.3. Let $(A, \text{d})$ be a differential graded algebra. Let $M$ be a differential graded $A$-module. There exists a homomorphism $P \to M$ of differential graded $A$-modules with the following properties

1. $P \to M$ is surjective,

2. $\mathop{\mathrm{Ker}}(\text{d}_ P) \to \mathop{\mathrm{Ker}}(\text{d}_ M)$ is surjective, and

3. $P$ sits in an admissible short exact sequence $0 \to P' \to P \to P'' \to 0$ where $P'$, $P''$ are direct sums of shifts of $A$.

Proof. Let $P_ k$ be the free $A$-module with generators $x, y$ in degrees $k$ and $k + 1$. Define the structure of a differential graded $A$-module on $P_ k$ by setting $\text{d}(x) = y$ and $\text{d}(y) = 0$. For every element $m \in M^ k$ there is a homomorphism $P_ k \to M$ sending $x$ to $m$ and $y$ to $\text{d}(m)$. Thus we see that there is a surjection from a direct sum of copies of $P_ k$ to $M$. This clearly produces $P \to M$ having properties (1) and (3). To obtain property (2) note that if $m \in \mathop{\mathrm{Ker}}(\text{d}_ M)$ has degree $k$, then there is a map $A[k] \to M$ mapping $1$ to $m$. Hence we can achieve (2) by adding a direct sum of copies of shifts of $A$. $\square$

Lemma 22.20.4. Let $(A, \text{d})$ be a differential graded algebra. Let $M$ be a differential graded $A$-module. There exists a homomorphism $P \to M$ of differential graded $A$-modules such that

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

2. $P$ has property (P).

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

$0 \to M_{i + 1} \to P_ i \to M_ i \to 0$

where the maps $P_ i \to M_ i$ are chosen as in Lemma 22.20.3. This gives a “resolution”

$\ldots \to P_2 \xrightarrow {f_2} P_1 \xrightarrow {f_1} P_0 \to M \to 0$

Then we set

$P = \bigoplus \nolimits _{i \geq 0} P_ i$

as an $A$-module with grading given by $P^ n = \bigoplus _{a + b = n} P_{-a}^ b$ and differential (as in the construction of the total complex associated to a double complex) by

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

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

$F_{2i}P = \bigoplus \nolimits _{i \geq j \geq 0} P_ j \subset \bigoplus \nolimits _{i \geq 0} P_ i = P$

and we add $P'_{i + 1}$ to $F_{2i}P$ to get $F_{2i + 1}$. These are differential graded submodules and the successive quotients are direct sums of shifts of $A$. By Lemma 22.16.1 we see that the inclusions $F_ iP \to F_{i + 1}P$ are admissible monomorphisms. Finally, we have to show that the map $P \to M$ (given by the augmentation $P_0 \to M$) is a quasi-isomorphism. This follows from Homology, Lemma 12.26.2. $\square$

Comment #2157 by shom on

Minor Typo: In the proof of Lemma 22.13.1 above the second line should read "...the negative of the inclusion map $F_{i}P \rightarrow F_{i+1}P$... " instead of "......the negative of the inclusion map $F_{i}P \rightarrow P_{i+1}P$...

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).