The Stacks project

Lemma 24.24.2. The functors $F, G$ of Lemma 24.24.1 have the following properties. Given a graded $\mathcal{A}$-module $\mathcal{N}$ we have

  1. the counit $\mathcal{N} \to F(G(\mathcal{N}))$ is injective,

  2. the map $\overline{\text{d}} : \mathcal{N} \to \mathop{\mathrm{Coker}}(\mathcal{N} \to F(G(\mathcal{N})))[1]$ is an isomorphism, and

  3. $G(\mathcal{N})$ is an acyclic differential graded $\mathcal{A}$-module.

Proof. We observe that property (3) is a consequence of properties (1) and (2). Namely, if $s$ is a nonzero local section of $F(G(\mathcal{N}))$ with $\text{d}(s) = 0$, then $s$ cannot be in the image of $\mathcal{N} \to F(G(\mathcal{N}))$. Hence we can write the image $\overline{s}$ of $s$ in the cokernel as $\overline{\text{d}}(s')$ for some local section $s'$ of $\mathcal{N}$. Then we see that $s = \text{d}(s')$ because the difference $s - \text{d}(s')$ is still in the kernel of $\text{d}$ and is contained in the image of the counit.

Let us write temporarily $\mathcal{A}_{gr}$, respectively $\mathcal{A}_{dg}$ the sheaf $\mathcal{A}$ viewed as a (right) graded module over itself, respectively as a (right) differential graded module over itself. The most important case of the lemma is to understand what is $G(\mathcal{A}_{gr})$. Of course $G(\mathcal{A}_{gr})$ is the object of $\textit{Mod}(\mathcal{A}, \text{d})$ representing the functor

\[ \mathcal{M} \longmapsto \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}(\mathcal{A})}(\mathcal{A}_{gr}, F(\mathcal{M})) = \Gamma (\mathcal{C}, \mathcal{M}) \]

By Remark 24.22.5 we see that this functor represented by $C[-1]$ where $C$ is the cone on the identity of $\mathcal{A}_{dg}$. We have a short exact sequence

\[ 0 \to \mathcal{A}_{dg}[-1] \to C[-1] \to \mathcal{A}_{dg} \to 0 \]

in $\textit{Mod}(\mathcal{A}, \text{d})$ which is split by the counit $\mathcal{A}_{gr} \to F(C[-1])$ in $\textit{Mod}(\mathcal{A})$. Thus $G(\mathcal{A}_{gr})$ satisfies properties (1) and (2).

Let $U$ be an object of $\mathcal{C}$. Denote $j_ U : \mathcal{C}/U \to \mathcal{C}$ the localization morphism. Denote $\mathcal{A}_ U$ the restriction of $\mathcal{A}$ to $U$. We will use the notation $\mathcal{A}_{U, gr}$ to denote $\mathcal{A}_ U$ viewed as a graded $\mathcal{A}_ U$-module. Denote $F_ U : \textit{Mod}(\mathcal{A}_ U, \text{d}) \to \textit{Mod}(\mathcal{A}_ U)$ the forgetful functor and denote $G_ U$ its adjoint. Then we have the commutative diagrams

\[ \vcenter { \xymatrix{ \textit{Mod}(\mathcal{A}, \text{d}) \ar[d]_{j_ U^*} \ar[r]_ F & \textit{Mod}(\mathcal{A}) \ar[d]^{j_ U^*} \\ \textit{Mod}(\mathcal{A}_ U, \text{d}) \ar[r]^{F_ U} & \textit{Mod}(\mathcal{A}_ U) } } \quad \text{and}\quad \vcenter { \xymatrix{ \textit{Mod}(\mathcal{A}_ U, \text{d}) \ar[r]_{F_ U} \ar[d]_{j_{U!}} & \textit{Mod}(\mathcal{A}_ U) \ar[d]^{j_{U!}} \\ \textit{Mod}(\mathcal{A}, \text{d}) \ar[r]^ F & \textit{Mod}(\mathcal{A}) } } \]

by the construction of $j^*_ U$ and $j_{U!}$ in Sections 24.9, 24.18, 24.10, and 24.19. By uniqueness of adjoints we obtain $j_{U!} \circ G_ U = G \circ j_{U!}$. Since $j_{U!}$ is an exact functor, we see that the properties (1) and (2) for the counit $\mathcal{A}_{U, gr} \to F_ U(G_ U(\mathcal{A}_{U, gr}))$ which we've seen in the previous part of the proof imply properties (1) and (2) for the counit $j_{U!}\mathcal{A}_{U, gr} \to F(G(j_{U!}\mathcal{A}_{U, gr})) = j_{U!}F_ U(G_ U(\mathcal{A}_{U, gr}))$.

In the proof of Lemma 24.11.1 we have seen that any object of $\textit{Mod}(\mathcal{A})$ is a quotient of a direct sum of copies of $j_{U!}\mathcal{A}_{U, gr}$. Since $G$ is a left adjoint, we see that $G$ commutes with direct sums. Thus properties (1) and (2) hold for direct sums of objects for which they hold. Thus we see that every object $\mathcal{N}$ of $\textit{Mod}(\mathcal{A})$ fits into an exact sequence

\[ \mathcal{N}_1 \to \mathcal{N}_0 \to \mathcal{N} \to 0 \]

such that (1) and (2) hold for $\mathcal{N}_1$ and $\mathcal{N}_0$. We leave it to the reader to deduce (1) and (2) for $\mathcal{N}$ using that $G$ is right exact. $\square$

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