Lemma 13.32.1. Let $\mathcal{A}$ be an abelian category. Let $d : \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}) \to \{ 0, 1, 2, \ldots , \infty \} $ be a function. Assume that

every object of $\mathcal{A}$ is a subobject of an object $A$ with $d(A) = 0$,

$d(A \oplus B) \leq \max \{ d(A), d(B)\} $ for $A, B \in \mathcal{A}$, and

if $0 \to A \to B \to C \to 0$ is short exact, then $d(C) \leq \max \{ d(A) - 1, d(B)\} $.

Let $K^\bullet $ be a complex such that $n + d(K^ n)$ tends to $-\infty $ as $n \to -\infty $. Then there exists a quasi-isomorphism $K^\bullet \to L^\bullet $ with $d(L^ n) = 0$ for all $n \in \mathbf{Z}$.

**Proof.**
By Lemma 13.15.5 we can find a quasi-isomorphism $\sigma _{\geq 0}K^\bullet \to M^\bullet $ with $M^ n = 0$ for $n < 0$ and $d(M^ n) = 0$ for $n \geq 0$. Then $K^\bullet $ is quasi-isomorphic to the complex

\[ \ldots \to K^{-2} \to K^{-1} \to M^0 \to M^1 \to \ldots \]

Hence we may assume that $d(K^ n) = 0$ for $n \gg 0$. Note that the condition $n + d(K^ n) \to -\infty $ as $n \to -\infty $ is not violated by this replacement.

We are going to improve $K^\bullet $ by an (infinite) sequence of elementary replacements. An *elementary replacement* is the following. Choose an index $n$ such that $d(K^ n) > 0$. Choose an injection $K^ n \to M$ where $d(M) = 0$. Set $M' = \mathop{\mathrm{Coker}}(K^ n \to M \oplus K^{n + 1})$. Consider the map of complexes

\[ \xymatrix{ K^\bullet : \ar[d] & K^{n - 1} \ar[d] \ar[r] & K^ n \ar[d] \ar[r] & K^{n + 1} \ar[d] \ar[r] & K^{n + 2} \ar[d] \\ (K')^\bullet : & K^{n - 1} \ar[r] & M \ar[r] & M' \ar[r] & K^{n + 2} } \]

It is clear that $K^\bullet \to (K')^\bullet $ is a quasi-isomorphism. Moreover, it is clear that $d((K')^ n) = 0$ and

\[ d((K')^{n + 1}) \leq \max \{ d(K^ n) - 1, d(M \oplus K^{n + 1})\} \leq \max \{ d(K^ n) - 1, d(K^{n + 1})\} \]

and the other values are unchanged.

To finish the proof we carefuly choose the order in which to do the elementary replacements so that for every integer $m$ the complex $\sigma _{\geq m}K^\bullet $ is changed only a finite number of times. To do this set

\[ \xi (K^\bullet ) = \max \{ n + d(K^ n) \mid d(K^ n) > 0\} \]

and

\[ I = \{ n \in \mathbf{Z} \mid \xi (K^\bullet ) = n + d(K^ n) \text{ and } d(K^ n) > 0\} \]

Our assumption that $n + d(K^ n)$ tends to $-\infty $ as $n \to -\infty $ and the fact that $d(K^ n) = 0$ for $n >> 0$ implies $\xi (K^\bullet ) < +\infty $ and that $I$ is a finite set. It is clear that $\xi ((K')^\bullet ) \leq \xi (K^\bullet )$ for an elementary transformation as above. An elementary transformation changes the complex in degrees $\leq \xi (K^\bullet ) + 1$. Hence if we can find finite sequence of elementary transformations which decrease $\xi (K^\bullet )$, then we win. However, note that if we do an elementary transformation starting with the smallest element $n \in I$, then we either decrease the size of $I$, or we increase $\min I$. Since every element of $I$ is $\leq \xi (K^\bullet )$ we see that we win after a finite number of steps.
$\square$

## Comments (2)

Comment #1292 by JuanPablo on

Comment #1304 by Johan on