## 12.26 Double complexes of abelian groups

In this section we put some results on double complexes of abelian groups for which do not (yet) have the analogues results for general abelian categories. Please be careful not to use these lemmas except when the underlying abelian category is the category of abelian groups or some such (e.g., the cateogry of modules over a ring). Some of the arguments will be difficult to follow without drawing “zig-zags” on a napkin – compare with the proof of Algebra, Lemma 10.74.3.

Lemma 12.26.1. Let $M^\bullet$ be a complex of abelian groups. Let

$0 \to M^\bullet \to A_0^\bullet \to A_1^\bullet \to A_2^\bullet \to \ldots$

be an exact complex of complexes of abelian groups. Set $A^{p, q} = A_ p^ q$ to obtain a double complex. Then the map $M^\bullet \to \text{Tot}(A^{\bullet , \bullet })$ induced by $M^\bullet \to A_0^\bullet$ is a quasi-isomorphism.

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

$0 \to \sigma _{\geq t}M^\bullet \to \sigma _{\geq t}A_0^\bullet \to \sigma _{\geq t}A_1^\bullet \to \sigma _{\geq t}A_2^\bullet \to \ldots$

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. $\square$

Lemma 12.26.2. Let $M^\bullet$ be a complex of abelian groups. Let

$\ldots \to A_2^\bullet \to A_1^\bullet \to A_0^\bullet \to M^\bullet \to 0$

be an exact complex of complexes of abelian groups such that for all $p \in \mathbf{Z}$ the complexes

$\ldots \to \mathop{\mathrm{Ker}}(d_{A_2^\bullet }^ p) \to \mathop{\mathrm{Ker}}(d_{A_1^\bullet }^ p) \to \mathop{\mathrm{Ker}}(d_{A_0^\bullet }^ p) \to \mathop{\mathrm{Ker}}(d_{M^\bullet }^ p) \to 0$

are exact as well. Set $A^{p, q} = A_{-p}^ q$ to obtain a double complex. Then $\text{Tot}(A^{\bullet , \bullet }) \to M^\bullet$ induced by $A_0^\bullet \to M^\bullet$ is a quasi-isomorphism.

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

$\ldots \to \mathop{\mathrm{Im}}(d_{A_2^\bullet }^ p) \to \mathop{\mathrm{Im}}(d_{A_1^\bullet }^ p) \to \mathop{\mathrm{Im}}(d_{A_0^\bullet }^ p) \to \mathop{\mathrm{Im}}(d_{M^\bullet }^ p) \to 0$

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

$\ldots \to H^ p(A_2^\bullet ) \to H^ p(A_1^\bullet ) \to H^ p(A_0^\bullet ) \to H^ p(M^\bullet ) \to 0$

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. $\square$

Lemma 12.26.3. Let $M^\bullet$ be a complex of abelian groups. Let

$0 \to M^\bullet \to A_0^\bullet \to A_1^\bullet \to A_2^\bullet \to \ldots$

be an exact complex of complexes of abelian groups such that for all $p \in \mathbf{Z}$ the complexes

$0 \to \mathop{\mathrm{Coker}}(d_{M^\bullet }^ p) \to \mathop{\mathrm{Coker}}(d_{A_0^\bullet }^ p) \to \mathop{\mathrm{Coker}}(d_{A_1^\bullet }^ p) \to \mathop{\mathrm{Coker}}(d_{A_2^\bullet }^ p) \to \ldots$

are exact as well. Set $A^{p, q} = A_ p^ q$ to obtain a double complex. Let $\text{Tot}_\pi (A^{\bullet , \bullet })$ be the product total complex associated to the double complex (see proof). Then the map $M^\bullet \to \text{Tot}_\pi (A^{\bullet , \bullet })$ induced by $M^\bullet \to A_0^\bullet$ is a quasi-isomorphism.

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

$T^ n = \prod \nolimits _{p + q = n} A^{p, q} = \prod \nolimits _{p + q = n} A_ p^ q \quad \text{with}\quad \text{d}_{T^\bullet }^ n = \prod \nolimits _{n = p + q} (f_ p^ q + (-1)^ pd_{A_ p^\bullet }^ q)$

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 a boundary 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

$0 \to H^ p(M^\bullet ) \to H^ p(A_0^\bullet ) \to H^ p(A_1^\bullet ) \to H^ p(A_2^\bullet ) \to \ldots$

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. $\square$

Lemma 12.26.4. Let $M^\bullet$ be a complex of abelian groups. Let

$\ldots \to A_2^\bullet \to A_1^\bullet \to A_0^\bullet \to M^\bullet \to 0$

be an exact complex of complexes of abelian groups. Set $A^{p, q} = A_{-p}^ q$ to obtain a double complex. Let $\text{Tot}_\pi (A^{\bullet , \bullet })$ be the product total complex associated to the double complex (see proof). Then the map $\text{Tot}_\pi (A^{\bullet , \bullet }) \to M^\bullet$ induced by $A_0^\bullet \to M^\bullet$ is a quasi-isomorphism.

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

$T^ n = \prod \nolimits _{p + q = n} A^{p, q} = \prod \nolimits _{p + q = n} A_{-p}^ q \quad \text{with}\quad \text{d}_{T^\bullet }^ n = \prod \nolimits _{n = p + q} (f_{-p}^ q + (-1)^ pd_{A_{-p}^\bullet }^ q)$

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

$\ldots \to B_2^\bullet \to B_1^\bullet \to B_0^\bullet \to 0 \to 0$

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

$0 \to M^\bullet \to S^\bullet \to T^\bullet  \to 0$

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. $\square$

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