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. $\square$

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