Lemma 13.18.3 (013K)—The Stacks project

Lemma 13.18.3. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives.

Any object of $\mathcal{A}$ has an injective resolution.
If $H^ n(K^\bullet ) = 0$ for all $n \ll 0$ then $K^\bullet $ has an injective resolution.
If $K^\bullet $ is a complex with $K^ n = 0$ for $n < a$, then there exists an injective resolution $\alpha : K^\bullet \to I^\bullet $ with $I^ n = 0$ for $n < a$ such that each $\alpha ^ n : K^ n \to I^ n$ is injective.

Proof. Proof of (1). First choose an injection $A \to I^0$ of $A$ into an injective object of $\mathcal{A}$. Next, choose an injection $I_0/A \to I^1$ into an injective object of $\mathcal{A}$. Denote $d^0$ the induced map $I^0 \to I^1$. Next, choose an injection $I^1/\mathop{\mathrm{Im}}(d^0) \to I^2$ into an injective object of $\mathcal{A}$. Denote $d^1$ the induced map $I^1 \to I^2$. And so on. By Lemma 13.18.2 part (2) follows from part (3). Part (3) is a special case of Lemma 13.15.5. $\square$

Comments (2)

Comment #8413 by Elías Guisado on May 19, 2023 at 02:54

I think one can refer the whole proof to 13.15.5. For part (1), one uses the fact that an injective resolution of $A$ is the same as a quasi-isomorphism into a complex with injective terms $A[0]\to I^\bullet$ such that $I^n=0$ for $n<0$ . Part (2) is 13.15.5, (2).

Comment #9038 by Stacks project on May 03, 2024 at 11:28

Yes, but I think it is fine as is too.

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