The Stacks project

Proof. Let $\mathcal{N}$ be a left graded $\mathcal{A}$-module. By Lemma 24.10.2 we have

as graded modules. Since both restriction to $U$ and $j_!$ are exact this proves condition (1). The same argument works for (2) using Lemma 24.19.2.

Consider a morphism $(f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ of ringed topoi, a differential graded $\mathcal{O}'$-algebra $\mathcal{A}'$, and a map $\varphi : f^{-1}\mathcal{A} \to \mathcal{A}'$ of differential graded $f^{-1}\mathcal{O}$-algebras. We have to show that

satisfies (1) and (2) for the ringed topos $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ endowed with the sheaf of differential graded $\mathcal{O}'$-algebras $\mathcal{A}'$. To prove this we may replace $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ and $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ by equivalent ringed topoi. Thus by Modules on Sites, Lemma 18.7.2 we may assume that $f$ comes from a morphism of sites $f : \mathcal{C} \to \mathcal{C}'$ given by the continuous functor $u : \mathcal{C} \to \mathcal{C}'$. In this case, set $U' = u(U)$ and denote $j' : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}'/U') \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}')$ the corresponding localization morphism. We obtain a commutative square of morphisms of ringed topoi

and we have $f'_*(j')^{-1} = j^{-1}f_*$. See Modules on Sites, Lemma 18.20.1. By uniqueness of adjoints we obtain $f^{-1}j_! = j'_!(f')^{-1}$. Thus we obtain

The first equation is the definition of the pullback of $j_!\mathcal{A}_ U$ to a differential graded module over $\mathcal{A}'$. The second equation because $f^{-1}j_! = j'_!(f')^{-1}$. The third equation by Lemma 24.19.2 applied to the ringed site $(\mathcal{C}', f^{-1}\mathcal{O})$ with sheaf of differential graded algebras $f^{-1}\mathcal{A}$ and with differential graded modules $(f')^{-1}\mathcal{A}_ U$ on $\mathcal{C}'/U'$ and $\mathcal{A}'$ on $\mathcal{C}'$. The fourth equation holds because of course we have $(f')^{-1}\mathcal{A}_ U = f^{-1}\mathcal{A}|_{U'}$. Hence we see that the pullback is another module of the same kind and we've proven conditions (1) and (2) for it above. $\square$

Proof. For condition (1) this is immediate as the sequence is a direct sum at the graded level. For condition (2) note that for any left differential graded $\mathcal{A}$-module, the sequence

is an admissible short exact sequence of differential graded $\mathcal{O}$-modules (since forgetting the differential the tensor product is just taken in the category of graded modules). Hence if two out of three are exact as complexes of $\mathcal{O}$-modules, so is the third. Finally, the same argument shows that given a morphism $(f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ of ringed topoi, a differential graded $\mathcal{O}'$-algebra $\mathcal{A}'$, and a map $\varphi : f^{-1}\mathcal{A} \to \mathcal{A}'$ of differential graded $f^{-1}\mathcal{O}$-algebras we have that

is an admissible short exact sequence of differential graded $\mathcal{A}'$-modules and the same argument as above applies here. $\square$

Proof. Omitted. Hint: direct sums and filtered colimits commute with tensor products and with pullbacks. $\square$

Proof. Consider triples $(U, k, x)$ where $U$ is an object of $\mathcal{C}$, $k \in \mathbf{Z}$, and $x$ is a section of $\mathcal{M}^ k$ over $U$ with $\text{d}_\mathcal {M}(x) = 0$. Then we obtain a unique morphism of differential graded $\mathcal{A}_ U$-modules $\varphi _ x : \mathcal{A}_ U[-k] \to \mathcal{M}|_ U$ mapping $1$ to $x$. This is adjoint to a morphism $\psi _ x : j_{U!}\mathcal{A}_ U[-k] \to \mathcal{M}$. Observe that $1 \in \mathcal{A}_ U(U)$ corresponds to a section $1 \in j_{U!}\mathcal{A}_ U[-k](U)$ of degree $k$ whose differential is zero and which is mapped to $x$ by $\psi _ x$. Thus if we consider the map

then we will have conditions (2) and (3). Namely, the objects $j_{U!}\mathcal{A}_ U[-k]$ are good (Lemma 24.23.1) and any direct sum of good objects is good (Lemma 24.23.3).

Next, consider triples $(U, k, x)$ where $U$ is an object of $\mathcal{C}$, $k \in \mathbf{Z}$, and $x$ is a section of $\mathcal{M}^ k$ (not necessarily annihilated by the differential). Then we can consider the cone $C_ U$ on the identity map $\mathcal{A}_ U \to \mathcal{A}_ U$ as in Remark 24.22.5. The element $x$ will determine a map $\varphi _ x : C_ U[-k - 1] \to \mathcal{A}_ U$, see Remark 24.22.5. Now, since we have an admissible short exact sequence

we conclude that $j_{U!}C_ U$ is a good module by Lemma 24.23.2 and the already used Lemma 24.23.1. As above we conclude that the direct sum of the maps $\psi _ x : j_{U!}C_ U \to \mathcal{M}$ adjoint to the $\varphi _ x$

is surjective. Taking the direct sum with the map produced in the first paragraph we conclude. $\square$

Proof. Let $\mathcal{N}$ be a left graded $\mathcal{A}$-module. Then we have

where $\mathcal{N}[\mathcal{S}$ is the graded $\mathcal{O}$-module whose degree $n$ part is the sheaf associated to the presheaf

It is clear that $\mathcal{N} \to \mathcal{N}[\mathcal{S}]$ is an exact functor, hence $\mathcal{A}[\mathcal{S}$ is flat as a graded $\mathcal{A}$-module. Next, suppose that $\mathcal{N}$ is a differential graded left $\mathcal{A}$-module. Then we have

as graded sheaves of $\mathcal{O}$-modules, which by the flatness (over $\mathcal{O})$ is equal to

as a graded $\mathcal{O}$-module. Hence if $\mathcal{N}$ is acyclic, then $\mathcal{A}[\mathcal{S}] \otimes _\mathcal {A} \mathcal{N}$ is acyclic.

Finally, consider a morphism $(f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ of ringed topoi, a differential graded $\mathcal{O}'$-algebra $\mathcal{A}'$, and a map $\varphi : f^{-1}\mathcal{A} \to \mathcal{A}'$ of differential graded $f^{-1}\mathcal{O}$-algebras. Then it is straightforward to see that

which finishes the proof that our module is good. $\square$

First proof. Let $\mathcal{S}_0$ be the sheaf of graded sets (Remark 24.23.5) whose degree $n$ part is $\mathop{\mathrm{Ker}}(\text{d}_\mathcal {M}^ n)$. Consider the homomorphism of differential graded modules

where the left hand side is as in Remark 24.23.5 and the map sends a local section $s$ of $\mathcal{S}_0$ to the corresponding local section of $\mathcal{M}^{\deg (s)}$ (which is in the kernel of the differential, so our map is a map of differential graded modules indeed). By construction the induced maps on cohomology sheaves $H^ n(\mathcal{P}_0) \to H^ n(\mathcal{M})$ are surjective. We are going to inductively construct maps

Observe that of course $H^*(\mathcal{P}_ i) \to H^*(\mathcal{M})$ will be surjective for all $i$. Given $\mathcal{P}_ i \to \mathcal{M}$ denote $\mathcal{S}_{i + 1}$ the sheaf of graded sets whose degree $n$ part is

Then we set

as graded $\mathcal{A}$-module with differential and map to $\mathcal{M}$ defined as follows

1. for local sections of $\mathcal{P}_ i$ use the differential on $\mathcal{P}_ i$ and the given map to $\mathcal{M}$,

2. for a local section $s = (p, m)$ of $\mathcal{S}_{i + 1}$ we set $\text{d}(s)$ equal to $p$ viewed as a section of $\mathcal{P}_ i$ of degree $\deg (s) + 1$ and we map $s$ to $m$ in $\mathcal{M}$, and

3. extend the differential uniquely so that the Leibniz rule holds.

This makes sense because $\text{d}(m)$ is the image of $p$ and $\text{d}(p) = 0$. Finally, we set $\mathcal{P} = \mathop{\mathrm{colim}}\nolimits \mathcal{P}_ i$ with the induced map to $\mathcal{M}$.

The map $\mathcal{P} \to \mathcal{M}$ is a quasi-isomorphism: we have $H^ n(\mathcal{P}) = \mathop{\mathrm{colim}}\nolimits H^ n(\mathcal{P}_ i)$ and for each $i$ the map $H^ n(\mathcal{P}_ i) \to H^ n(\mathcal{M})$ is surjective with kernel annihilated by the map $H^ n(\mathcal{P}_ i) \to H^ n(\mathcal{P}_{i + 1})$ by construction. Each $\mathcal{P}_ i$ is good because $\mathcal{P}_0$ is good by Lemma 24.23.6 and each $\mathcal{P}_{i + 1}$ is in the middle of the admissible short exact sequence $0 \to \mathcal{P}_ i \to \mathcal{P}_{i + 1} \to \mathcal{A}[\mathcal{S}_{i + 1}] \to 0$ whose outer terms are good by induction. Hence $\mathcal{P}_{i + 1}$ is good by Lemma 24.23.2. Finally, we conclude that $\mathcal{P}$ is good by Lemma 24.23.3. $\square$

Second proof. We urge the reader to read the proof of Differential Graded Algebra, Lemma 22.20.4 before reading this proof. Set $\mathcal{M} = \mathcal{M}_0$. We inductively choose short exact sequences

where the maps $\mathcal{P}_ i \to \mathcal{M}_ i$ are chosen as in Lemma 24.23.4. This gives a “resolution”

Then we let $\mathcal{P}$ be the differential graded $\mathcal{A}$-module defined as follows

1. as a graded $\mathcal{A}$-module we set $\mathcal{P} = \bigoplus _{a \leq 0} \mathcal{P}_{-a}[-a]$, i.e., the degree $n$ part is given by $\mathcal{P}^ n = \bigoplus \nolimits _{a + b = n} \mathcal{P}_{-a}^ b$,

2. the differential on $\mathcal{P}$ is as in the construction of the total complex associated to a double complex given by

   \[ \text{d}_\mathcal {P}(x) = f_{-a}(x) + (-1)^ a \text{d}_{\mathcal{P}_{-a}}(x) \]

   for $x$ a local section of $\mathcal{P}_{-a}^ b$.

With these conventions $\mathcal{P}$ is indeed a differential graded $\mathcal{A}$-module; we omit the details. There is a map $\mathcal{P} \to \mathcal{M}$ of differential graded $\mathcal{A}$-modules which is zero on the summands $\mathcal{P}_{-a}[-a]$ for $a < 0$ and the given map $\mathcal{P}_0 \to \mathcal{M}$ for $a = 0$. Observe that we have

where $F_ i\mathcal{P} \subset \mathcal{P}$ is the differential graded $\mathcal{A}$-submodule whose underlying graded $\mathcal{A}$-module is

It is immediate that the maps

are all admissible monomorphisms and we have admissible short exact sequences

By induction and Lemma 24.23.2 we find that $F_ i\mathcal{P}$ is a good differential graded $\mathcal{A}$-module. Since $\mathcal{P} = \mathop{\mathrm{colim}}\nolimits F_ i\mathcal{P}$ we find that $\mathcal{P}$ is good by Lemma 24.23.3.

Finally, we have to show that $\mathcal{P} \to \mathcal{M}$ is a quasi-isomorphism. If $\mathcal{C}$ has enough points, then this follows from the elementary Homology, Lemma 12.26.2 by checking on stalks. In general, we can argue as follows (this proof is far too long — there is an alternative argument by working with local sections as in the elementary proof but it is also rather long). Since filtered colimits are exact on the category of abelian sheaves, we have

We claim that for each $i \geq 0$ and $d \in \mathbf{Z}$ we have (a) a short exact sequence

where the second arrow comes from $F_ i\mathcal{P} \to \mathcal{P} \to \mathcal{M}$ and (b) the composition

is zero. It is clear that the claim suffices to finish the proof.

Proof of the claim. For any $i \geq 0$ there is a map $\mathcal{M}_{i + 1}[i] \to F_ i\mathcal{P}$ coming from the inclusion of $\mathcal{M}_{i + 1}$ into $\mathcal{P}_ i$ as the kernel of $f_ i$. Consider the short exact sequence

of complexes of $\mathcal{O}$-modules defining $C_ i$. Observe that $C_0 = \mathcal{M}_0 = \mathcal{M}$. Also, observe that $C_ i$ is the total complex associated to the double complex $C_ i^{\bullet , \bullet }$ with columns

in degree $-i, -i + 1, \ldots , 0$. There is a map of double complexes $C_ i^{\bullet , \bullet } \to C_{i - 1}^{\bullet , \bullet }$ which is $0$ on the column in degree $-i$, is the surjection $\mathcal{P}_{i - 1} \to \mathcal{M}_{i - 1}$ in degree $-i + 1$, and is the identity on the other columns. Hence there are maps of complexes

These maps are surjective quasi-isomorphisms because the kernel is the total complex on the double complex with columns $\mathcal{M}_ i, \mathcal{M}_ i$ in degrees $-i, -i + 1$ and the identity map between these two columns. Using the resulting identifications $H^ d(C_ i) = H^ d(C_{i - 1} = \ldots = H^ d(\mathcal{M})$ this already shows we get a long exact sequence

from the short exact sequence of complexes above. However, we also have the commutative diagram

where $T_{i + 1}$ is the total complex on the double complex with columns $\mathcal{P}_{i + 1}, \mathcal{M}_{i + 1}$ placed in degrees $-i - 1$ and $-i$. In other words, $T_{i + 1}$ is a shift of the cone on the map $\mathcal{P}_{i + 1} \to \mathcal{M}_{i + 1}$ and we find that $a$ is a quasi-isomorphism and the map $a^{-1} \circ b$ is a shift of the third map of the distinguished triangle in $D(\mathcal{O})$ associated to the short exact sequence

The map $H^ d(\mathcal{P}_{i + 1}) \to H^ d(\mathcal{M}_{i + 1})$ is surjective because we chose our maps such that $\mathop{\mathrm{Ker}}(\text{d}_{\mathcal{P}_{i + 1}}) \to \mathop{\mathrm{Ker}}(\text{d}_{\mathcal{M}_{i + 1}})$ is surjective. Thus we see that $a^{-1} \circ b$ is zero on cohomology sheaves. This proves part (b) of the claim. Since $T_{i + 1}$ is the kernel of the surjective map of complexes $F_{i + 1}\mathcal{P} \to C_ i$ we find a map of long exact cohomology sequences

Here we know, by the discussion above, that the vertical maps on the outside are zero. Hence the maps $H^ d(F_{i + 1}\mathcal{P}) \to H^ d(\mathcal{M})$ are surjective and part (a) of the claim follows. More precisely, the claim follows for $i > 0$ and we leave the claim for $i = 0$ to the reader (actually it suffices to prove the claim for all $i \gg 0$ in order to get the lemma). $\square$

Proof. Proof of (1). By Lemma 24.23.7 we can choose a good left differential graded $\mathcal{Q}$ and a quasi-isomorphism $\mathcal{Q} \to \mathcal{N}$. Then $\mathcal{P} \otimes _\mathcal {A} \mathcal{Q}$ is acyclic because $\mathcal{Q}$ is good. Let $\mathcal{N}'$ be the cone on the map $\mathcal{Q} \to \mathcal{N}$. Then $\mathcal{P} \otimes _\mathcal {A} \mathcal{N}'$ is acyclic because $\mathcal{P}$ is good and because $\mathcal{N}'$ is acyclic (as the cone on a quasi-isomorphism). We have a distinguished triangle

in $K(\textit{Mod}(\mathcal{A}, \text{d}))$ by our construction of the triangulated structure. Since $\mathcal{P} \otimes _\mathcal {A} -$ sends distinguished triangles to distinguished triangles, we obtain a distinguished triangle

in $K(\textit{Mod}(\mathcal{O}))$. Thus we conclude.

Proof of (2). Observe that $f^*\mathcal{P}$ is good by our definition of good modules. Recall that $f^*\mathcal{P} = f^{-1}\mathcal{P} \otimes _{f^{-1}\mathcal{A}} \mathcal{A}'$. Then $f^{-1}\mathcal{P}$ is a good acyclic (because $f^{-1}$ is exact) differential graded $f^{-1}\mathcal{A}$-module. Hence we see that $f^*\mathcal{P}$ is acyclic by part (1). $\square$