Lemma 19.12.5 (079N)—The Stacks project

Proof. Choose a functorial injective embeddings $i_ M : M \to I(M)$, see Theorem 19.11.7. For every complex $M^\bullet $ denote $J^\bullet (M^\bullet )$ the complex with terms $J^ n(M^\bullet ) = I(M^ n) \oplus I(M^{n + 1})$ and differential

\[ d_{J^\bullet (M^\bullet )} = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) \]

There exists a canonical injective map of complexes $u_{M^\bullet } : M^\bullet \to J^\bullet (M^\bullet )$ by mapping $M^ n$ to $I(M^ n) \oplus I(M^{n + 1})$ via the maps $i_{M^ n} : M^ n \to I(M^ n)$ and $i_{M^{n + 1}} \circ d : M^ n \to M^{n + 1} \to I(M^{n + 1})$. Hence a short exact sequence of complexes

\[ 0 \to M^\bullet \xrightarrow {u_{M^\bullet }} J^\bullet (M^\bullet ) \xrightarrow {v_{M^\bullet }} Q^\bullet (M^\bullet ) \to 0 \]

functorial in $M^\bullet $. Set

\[ \mathbf{N}^\bullet (M^\bullet ) = C(v_{M^\bullet })^\bullet [-1]. \]

Note that

\[ \mathbf{N}^ n(M^\bullet ) = Q^{n - 1}(M^\bullet ) \oplus J^ n(M^\bullet ) \]

with differential

\[ \left( \begin{matrix} - d^{n - 1}_{Q^\bullet (M^\bullet )} & - v^ n_{M^\bullet } \\ 0 & d^ n_{J^\bullet (M)} \end{matrix} \right) \]

Hence we see that there is a map of complexes $j_{M^\bullet } : M^\bullet \to \mathbf{N}^\bullet (M^\bullet )$ induced by $u$. It is injective and factors through an injective subobject by construction. The map $j_{M^\bullet }$ is a quasi-isomorphism as one can prove by looking at the long exact sequence of cohomology associated to the short exact sequences of complexes above. $\square$

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Comment #9521 by Elías Guisado on July 23, 2024 at 07:09

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For those who care, here's the proof that $j_{M^\bullet}$ is a qis (as it is hinted the last sentence of the proof).

Definition. Extending the terminology introduced in Derived Categories, Remark 13.4.4, if $\mathcal{A}$ is an abelian category, a triangle $A\to B\to C\to A[1]$ in $K(\mathcal{A})$ or in $D(\mathcal{A})$ is said to be $H^*$ -special if its induced long sequence in cohomology is exact.

We know that all distinguished triangles in $K(\mathcal{A})$ and in $D(\mathcal{A})$ are $H^*$ -special. The converse isn't true in general: Let $\varepsilon_i\in\{1,-1\}$ be any signs satisfying $\varepsilon_1\varepsilon_2\varepsilon_3=-1$ . If $A\xrightarrow{a}B\xrightarrow{b}C\xrightarrow{c} A[1]$ is a d.t. in $K(\mathcal{A})$ or in $D(\mathcal{A})$ , then the anti-distinguished triangle $A\xrightarrow{\varepsilon_1 a}B\xrightarrow{\varepsilon_2 b}C\xrightarrow{\varepsilon_3 c} A[1]$ is again $H^*$ -special but not necessarily distinguished. Note that if $(f,g,h)$ is a morphism of $H^*$ -special triangles and at least two among $f,g,h$ are quasi-isomorphisms, then so is the third.

Now, consider the following diagram in $D(\mathcal{A})$ : $\require{AMScd} \begin{CD} J^\bullet(M^\bullet)@>>> C(u_{M^\bullet})@>-p>>M^\bullet[1]@>u_{M^\bullet}>>J^\bullet(M^\bullet)[1]\\ @|@VVqV@|@|\\ J^\bullet(M^\bullet)@>\smash{v_{M^\bullet}}>>Q^\bullet(M^\bullet)@>\smash{-pq^{-1}}>>M^\bullet[1]@>\smash{u_{M^\bullet}}>>J^\bullet(M^\bullet)[1]\\ @|@|@VVj_{M^\bullet}[1]V@|\\ J^\bullet(M^\bullet)@>\smash{v_{M^\bullet}}>>Q^\bullet(M^\bullet)@>\smash i>>C(v_{M^\bullet})@>\smash{(0,1)}>>J^\bullet(M^\bullet)[1] \end{CD}$ where $j_{M^\bullet}=<script type="math/tex; mode=display">\begin{pmatrix}0\\u_{M^\bullet}\end{pmatrix}$ , $p=(0,1)$ and $q=(v_{M^\bullet},0)$ is a quasi-isomorphism as it is argued in Derived Categories, Section 13.12, before Lemma 13.12.1. In the diagram, the top row is a distinguished triangle. Since the three upper squares commute and $q$ is an iso in $D(\mathcal{A})$ , the middle row is a distinguished triangle too (in $D(\mathcal{A})$ ). On the other hand, the bottom row is anti-distinguished, so it is $H^*$ -special. Thus, to see that $j_{M^\bullet}[1]$ is a qis, it suffices to see that the three lower squares commute. Clearly the left and right lower squares commute. Commutativity of the middle square amounts to $i\circ q=-j_{M^\bullet}[1]\circ p$ . It suffices to verify this equality in $K(\mathcal{A})$ . This is the content of the following

Lemma. Let $A\xrightarrow{a} B\xrightarrow{b} C$ be morphisms in some additive category. Then the morphisms $<script type="math/tex; mode=display">\begin{pmatrix}b&0\\0&0\end{pmatrix}$ , $\begin{pmatrix}0&0\\0&-a[1]\end{pmatrix}$ :C(a)\rightrightarrows C(b) are chain homotopic.

Proof. A homotopy is given by $h^n=\begin{pmatrix}0&0\\1_{B^n}&0\end{pmatrix}:C(a)^n=B^n\oplus A^{n+1}\to C^{n-1}\oplus B^n=C(b)^{n-1}.\quad\square$

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