Lemma 13.15.6 (05T8)—The Stacks project

Lemma 13.15.6. In Situation 13.15.1. Let $\mathcal{I} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset with the following properties:

every object of $\mathcal{A}$ is a subobject of an element of $\mathcal{I}$,
for any short exact sequence $0 \to P \to Q \to R \to 0$ of $\mathcal{A}$ with $P, Q \in \mathcal{I}$, then $R \in \mathcal{I}$, and $0 \to F(P) \to F(Q) \to F(R) \to 0$ is exact.

Then every object of $\mathcal{I}$ is acyclic for $RF$.

Proof. Pick $A \in \mathcal{I}$. Let $A[0] \to K^\bullet $ be a quasi-isomorphism with $K^\bullet $ bounded below. Then we can find a quasi-isomorphism $K^\bullet \to I^\bullet $ with $I^\bullet $ bounded below and each $I^ n \in \mathcal{I}$, see Lemma 13.15.5 ¹. Hence we see that these resolutions are cofinal in the category $A[0]/\text{Qis}^{+}(\mathcal{A})$. To finish the proof it therefore suffices to show that for any quasi-isomorphism $A[0] \to I^\bullet $ with $I^\bullet $ bounded below and $I^ n \in \mathcal{I}$ we have $F(A)[0] \to F(I^\bullet )$ is a quasi-isomorphism. To see this suppose that $I^ n = 0$ for $n < n_0$. Of course we may assume that $n_0 < 0$. Starting with $n = n_0$ we prove inductively that $\mathop{\mathrm{Im}}(d^{n - 1}) = \mathop{\mathrm{Ker}}(d^ n)$ and $\mathop{\mathrm{Im}}(d^{-1})$ are elements of $\mathcal{I}$ using property (2) and the exact sequences

\[ 0 \to \mathop{\mathrm{Ker}}(d^ n) \to I^ n \to \mathop{\mathrm{Im}}(d^ n) \to 0. \]

Moreover, property (2) also guarantees that the complex

\[ 0 \to F(I^{n_0}) \to F(I^{n_0 + 1}) \to \ldots \to F(I^{-1}) \to F(\mathop{\mathrm{Im}}(d^{-1})) \to 0 \]

is exact. The exact sequence $0 \to \mathop{\mathrm{Im}}(d^{-1}) \to I^0 \to I^0/\mathop{\mathrm{Im}}(d^{-1}) \to 0$ implies that $I^0/\mathop{\mathrm{Im}}(d^{-1})$ is an element of $\mathcal{I}$. The exact sequence $0 \to A \to I^0/\mathop{\mathrm{Im}}(d^{-1}) \to \mathop{\mathrm{Im}}(d^0) \to 0$ then implies that $\mathop{\mathrm{Im}}(d^0) = \mathop{\mathrm{Ker}}(d^1)$ is an elements of $\mathcal{I}$ and from then on one continues as before to show that $\mathop{\mathrm{Im}}(d^{n - 1}) = \mathop{\mathrm{Ker}}(d^ n)$ is an element of $\mathcal{I}$ for all $n > 0$. Applying $F$ to each of the short exact sequences mentioned above and using (2) we observe that $F(A)[0] \to F(I^\bullet )$ is an isomorphism as desired. $\square$

[1] By (1) we see that $\mathcal{I}$ is nonempty. Pick $P$ in $\mathcal{I}$. Then the short exact sequence $0 \to P \to P \to 0 \to 0$ and assumption (2) shows that $0$ is in $\mathcal{I}$. Thus the lemma applies.

Comments (2)

Comment #8397 by Elías Guisado on May 14, 2023 at 13:42

A little (maybe pedantic) remark: At the beginning of the proof I think we should say "we may add $0$ to $\mathcal{I}$ if necessary and suppose that $\mathcal{I}$ is closed under isomorphisms" (i.e., if $P\to R$ is an isomorphism in $\mathcal{A}$ with $P$ in $\mathcal{I}$ , then also $R\in\mathcal{I}$ ), equivalently, we could just say "we may replace $\mathcal{I}$ by the essential image of $\{0\}\cup\mathcal{I}\to\mathcal{A}$ if necessary". I say this because after only adding $0$ to $\mathcal{I}$ , then we may consider a short exact sequence $0\to 0\to P\to R\to 0$ , hence the condition "closure under isomorphisms" for (2) to still hold.

Comment #9007 by Stacks project on April 17, 2024 at 14:56

Thanks. I moved the discussion of the existence of $0$ to a footnote. See here.

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7 comment(s) on Section 13.15: Derived functors on derived categories

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