The Stacks project

Proof. If there exists a $t \in \mathbf{Z}$ such that $A_0^ q = 0$ for $q < t$, then this follows immediately from Lemma 12.25.4 (with $p$ and $q$ swapped as in the final statement of that lemma). OK, but for every $t \in \mathbf{Z}$ we have a complex

of stupid truncations. Denote $A(t)^{\bullet , \bullet }$ the corresponding double complex. Every element $\xi $ of $H^ n(\text{Tot}(A^{\bullet , \bullet }))$ is the image of an element of $H^ n(\text{Tot}(A(t)^{\bullet , \bullet }))$ for some $t$ (look at explicit representatives of cohomology classes). Hence $\xi $ is in the image of $H^ n(\sigma _{\geq t}M^\bullet )$. Thus the map $H^ n(M^\bullet ) \to H^ n(\text{Tot}(A^{\bullet , \bullet }))$ is surjective. It is injective because for all $t$ the map $H^ n(\sigma _{\geq t}M^\bullet ) \to H^ n(\text{Tot}(A(t)^{\bullet , \bullet }))$ is injective and similar arguments. $\square$

Proof. Using the short exact sequences $0 \to \mathop{\mathrm{Ker}}(d^ p_{A_ n^\bullet }) \to A_ n^ p \to \mathop{\mathrm{Im}}(d^ p_{A_ n^\bullet }) \to 0$ and the assumptions we see that

is exact for all $p \in \mathbf{Z}$. Repeating with the exact sequences $0 \to \mathop{\mathrm{Im}}(d^{p - 1}_{A_ n^\bullet }) \to \mathop{\mathrm{Ker}}(d^ p_{A_ n^\bullet }) \to H^ p(A_ n^\bullet ) \to 0$ we find that

is exact for all $p \in \mathbf{Z}$.

Write $T^\bullet = \text{Tot}(A^{\bullet , \bullet })$. We will show that $H^0(T^\bullet ) \to H^0(M^\bullet )$ is an isomorphism. The same argument works for other degrees. Let $x \in \mathop{\mathrm{Ker}}(\text{d}_{T^\bullet }^0)$ represent an element $\xi \in H^0(T^\bullet )$. Write $x = \sum _{i = n, \ldots , 0} x_ i$ with $x_ i \in A_ i^ i$. Assume $n > 0$. Then $x_ n$ is in the kernel of $d_{A_ n^\bullet }^ n$ and maps to zero in $H^ n(A_{n - 1}^\bullet )$ because it maps to an element which is the boundary of $x_{n - 1}$ up to sign. By the first paragraph of the proof, we find that $x_ n \bmod \mathop{\mathrm{Im}}(d^{n - 1}_{A_ n^\bullet })$ is in the image of $H^ n(A_{n + 1}^\bullet ) \to H^ n(A_ n^\bullet )$. Thus we can modify $x$ by a boundary and reach the situation where $x_ n$ is a boundary. Modifying $x$ once more we see that we may assume $x_ n = 0$. By induction we see that every cohomology class $\xi $ is represented by a cocycle $x = x_0$. Finally, the condition on exactness of kernels tells us two such cocycles $x_0$ and $x_0'$ are cohomologous if and only if their image in $H^0(M^\bullet )$ are the same. $\square$

Proof. Abbreviating $T^\bullet = \text{Tot}_\pi (A^{\bullet , \bullet })$ we define

where $f_ p^\bullet : A_ p^\bullet \to A_{p + 1}^\bullet $ are the maps of complexes in the lemma.

We will show that $H^0(M^\bullet ) \to H^0(T^\bullet )$ is an isomorphism. The same argument works for other degrees. Let $x \in \mathop{\mathrm{Ker}}(\text{d}_{T^\bullet }^0)$ represent $\xi \in H^0(T^\bullet )$. Write $x = (x_ i)$ with $x_ i \in A_ i^{-i}$. Note that $x_0$ maps to zero in $\mathop{\mathrm{Coker}}(A_1^{-1} \to A_1^0)$. Hence we see that $x_0 = m_0 + d_{A_0^\bullet }^{-1}(y)$ for some $m_0 \in M^0$ and $y \in A_0^{-1}$. Then $d_{M^\bullet }(m_0) = 0$ because $\text{d}_{A_0^\bullet }(x_0) = 0$ as $\text{d}_{T^\bullet }(x) = 0$. Thus, replacing $\xi $ by something in the image of $H^0(M^\bullet ) \to H^0(T^\bullet )$ we may assume that $x_0$ is in $\mathop{\mathrm{Im}}(d^{-1}_{A_0^\bullet })$.

Assume $x_0 \in \mathop{\mathrm{Im}}(d^{-1}_{A_0^\bullet })$. We claim that in this case $\xi = 0$. To prove this we find, by induction on $n$ elements $y_0, y_1, \ldots , y_ n$ with $y_ i \in A_ i^{-i - 1}$ such that $x_0 = \text{d}_{A_0}^{-1}(y_0)$ and $x_ j = f_{j - 1}^{-j}(y_{j - 1}) + (-1)^ j d^{-j - 1}_{A_ j^\bullet }(y_ j)$ for $j = 1, \ldots , n$. This is clear for $n = 0$. Proof of induction step: suppose we have found $y_0, \ldots , y_{n - 1}$. Then $w_ n = x_ n - f_{n - 1}^{-n}(y_{n - 1})$ is in the kernel of $d^{-n}_{A_ n^\bullet }$ and maps to zero in $H^{-n}(A_{n + 1}^\bullet )$ (because it maps to an element which is the boundary of $x_{n + 1}$ up to sign). Exactly as in the proof of Lemma 12.26.2 the assumptions of the lemma imply that

is exact for all $p \in \mathbf{Z}$. Thus after changing $y_{n - 1}$ by an element in $\mathop{\mathrm{Ker}}(d^{n - 1}_{A_{n - 1}^\bullet })$ we may assume that $w_ n$ maps to zero in $H^{-n}(A_ n^\bullet )$. This means we can find $y_ n$ as desired. Observe that this procedure does not change $y_0, \ldots , y_{n - 2}$. Hence continuing ad infinitum we find an element $y = (y_ i)$ in $T^{n - 1}$ with $d_{T^\bullet }(y) = \xi $. This shows that $H^0(M^\bullet ) \to H^0(T^\bullet )$ is surjective.

Suppose that $m_0 \in \mathop{\mathrm{Ker}}(d^0_{M^\bullet })$ maps to zero in $H^0(T^\bullet )$. Say it maps to the differential applied to $y = (y_ i) \in T^{-1}$ . Then $y_0 \in A_0^{-1}$ maps to zero in $\mathop{\mathrm{Coker}}(d^{-2}_{A_1^\bullet })$. By assumption this means that $y_0 \bmod \mathop{\mathrm{Im}}(d^{-2}_{A_0^\bullet })$ is the image of some $z \in M^{-1}$. It follows that $m_0 = d^{-1}_{M^\bullet }(z)$. This proves injectivity and the proof is complete. $\square$

Proof. Abbreviating $T^\bullet = \text{Tot}_\pi (A^{\bullet , \bullet })$ we define

where $f_ p^\bullet : A_ p^\bullet \to A_{p - 1}^\bullet $ are the maps of complexes in the lemma. We will show that $T^\bullet $ is acyclic when $M^\bullet $ is the zero complex. This will suffice by the following trick. Set $B_ n^\bullet = A_{n + 1}^\bullet $ and $B_0^\bullet = M^\bullet $. Then we have an exact sequence

as in the lemma. Let $S^\bullet = \text{Tot}_\pi (B^{\bullet , \bullet })$. Then there is an obvious short exact sequence of complexes

and we conclude by the long exact cohomology sequence. Some details omitted.

Assume $M^\bullet = 0$. We will show $H^0(T^\bullet ) = 0$. The same argument works for other degrees. Let $x =(x_ n) \in \mathop{\mathrm{Ker}}(d_{T^\bullet })$ map to $\xi \in H^0(T^\bullet )$ with $x_ n \in A^{-n, n} = A_ n^ n$. Since $M^0 = 0$ we find that $x_0 = f_1^0(y_0)$ for some $y_0 \in A_1^0$. Then $x_1 - d^0_{A_1^\bullet }(y_0) = f_2^1(y_1)$ because it is mapped to zero by $f_1^1$ as $x$ is a cocycle. for some $y_1 \in A_2^1$. Continuing, using induction, we find $y = (y_ i) \in T^{-1}$ with $d_{T^\bullet }(y) = x$ as desired. $\square$