The Stacks project

Proof. Let $M^\bullet $ be an acyclic complex. We are going to construct by induction on the ordinal $\alpha $ an acyclic subcomplex $K_\alpha ^\bullet \subset M^\bullet $ as follows. For $\alpha = 0$ we set $K_0^\bullet = 0$. For $\alpha > 0$ we proceed as follows:

1. If $\alpha = \beta + 1$ and $K_\beta ^\bullet = M^\bullet $ then we choose $K_\alpha ^\bullet = K_\beta ^\bullet $.

2. If $\alpha = \beta + 1$ and $K_\beta ^\bullet \not= M^\bullet $ then $M^\bullet /K_\beta ^\bullet $ is a nonzero acyclic complex. We choose a subcomplex $N_\alpha ^\bullet \subset M^\bullet /K_\beta ^\bullet $ as in Lemma 19.12.2. Finally, we let $K_\alpha ^\bullet \subset M^\bullet $ be the inverse image of $N_\alpha ^\bullet $.

3. If $\alpha $ is a limit ordinal we set $K_\beta ^\bullet = \mathop{\mathrm{colim}}\nolimits K_\alpha ^\bullet $.

It is clear that $M^\bullet = K_\alpha ^\bullet $ for a suitably large ordinal $\alpha $. We will prove that

is zero by transfinite induction on $\alpha $. It holds for $\alpha = 0$ since $K_0^\bullet $ is zero. Suppose it holds for $\beta $ and $\alpha = \beta + 1$. In case (1) of the list above the result is clear. In case (2) there is a short exact sequence of complexes

Since each component of $I^\bullet $ is injective we see that we obtain an exact sequence

By induction the term on the left is zero and by assumption on $I^\bullet $ the term on the right is zero. Thus the middle group is zero too. Finally, suppose that $\alpha $ is a limit ordinal. Then we see that

with notation as in More on Algebra, Section 15.71. These complexes compute morphisms in $K(\mathcal{A})$ by More on Algebra, Equation (15.71.0.1). Note that the transition maps in the system are surjective because $I^ j$ is surjective for each $j$. Moreover, for a limit ordinal $\alpha $ we have equality of limit and value (see displayed formula above). Thus we may apply Homology, Lemma 12.31.8 to conclude. $\square$