The Stacks project

Lemma 19.12.2. Let $\mathcal{A}$ be a Grothendieck abelian category. There exists a cardinal $\kappa $ such that given any acyclic complex $M^\bullet $ we have

  1. if $M^\bullet $ is nonzero, there is a nonzero subcomplex $N^\bullet $ which is bounded above, acyclic, and $|N^ n| \leq \kappa $,

  2. there exists a surjection of complexes

    \[ \bigoplus \nolimits _{i \in I} M_ i^\bullet \longrightarrow M^\bullet \]

    where $M_ i^\bullet $ is bounded above, acyclic, and $|M_ i^ n| \leq \kappa $.

Proof. Choose a generator $U$ of $\mathcal{A}$. Denote $c$ the function of Lemma 19.12.1. Set $\kappa = \sup \{ c^ n(|U|), n = 1, 2, 3, \ldots \} $. Let $n \in \mathbf{Z}$ and let $\psi : U \to M^ n$ be a morphism. In order to prove (1) and (2) it suffices to prove there exists a subcomplex $N^\bullet \subset M^\bullet $ which is bounded above, acyclic, and $|N^ m| \leq \kappa $, such that $\psi $ factors through $N^ n$. To do this set $N^ n = \mathop{\mathrm{Im}}(\psi )$, $N^{n + 1} = \mathop{\mathrm{Im}}(U \to M^ n \to M^{n + 1})$, and $N^ m = 0$ for $m \geq n + 2$. Suppose we have constructed $N^ m \subset M^ m$ for all $m \geq k$ such that

  1. $\text{d}(N^ m) \subset N^{m + 1}$, $m \geq k$,

  2. $\mathop{\mathrm{Im}}(N^{m - 1} \to N^ m) = \mathop{\mathrm{Ker}}(N^ m \to N^{m + 1})$ for all $m \geq k + 1$, and

  3. $|N^ m| \leq c^{\max \{ n - m, 0\} }(|U|)$

for some $k \leq n$. Because $M^\bullet $ is acyclic, we see that the subobject $\text{d}^{-1}(\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})) \subset M^{k - 1}$ surjects onto $\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})$. Thus we can choose $N^{k - 1} \subset M^{k - 1}$ surjecting onto $\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})$ with $|N^{k - 1}| \leq c^{n - k + 1}(|U|)$ by Lemma 19.12.1. The proof is finished by induction on $k$. $\square$


Comments (3)

Comment #8590 by nkym on

Unnecessary period at the end of (3).

Comment #9164 by on

I am always in favor or changes to decrease the total number of characters while preserving the content of the text! Change here.

Comment #9507 by on

In the proof, I got a little bit confused because means different things in “let and be a morphism” and “ for some .” Maybe in the former case one could write “let and be a morphism” instead? If I got it right, the base case of the downward induction is and one fixes some .


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