The Stacks project

Lemma 19.12.5. Let $\mathcal{A}$ be a Grothendieck abelian category. There exists a functor $M^\bullet \mapsto \mathbf{N}^\bullet (M^\bullet )$ and a natural transformation $j_{M^\bullet } : M^\bullet \to \mathbf{N}^\bullet (M^\bullet )$ such

  1. $j_{M^\bullet }$ is a (termwise) injective quasi-isomorphism, and

  2. for every $n \in \mathbf{Z}$ the map $M^ n \to \mathbf{N}^ n(M^\bullet )$ factors through a subobject $I^ n \subset \mathbf{N}^ n(M^\bullet )$ where $I^ n$ is an injective object of $\mathcal{A}$.

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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