The Stacks project

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). $\square$