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

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

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

3. 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. $\square$

Comment #1292 by JuanPablo on

In the second paragraph, when defining elementary replacement, it says that $d(K'^{n+1})\leq \text{max}\{d(K^n)-1,d(K^{n+1})\}$. I do not see why that is.

From the short exact sequence $0\rightarrow K^n\rightarrow K^{n+1}\oplus M\rightarrow K'^{n+1}\rightarrow 0$, I see that $d(K'^{n+1})\leq\text{max}\{d(K^n\oplus M)-1,d(K^{n+1})\}$.

If $d$ satisfies additionally that for $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ short exact $d(B)\leq \text{max}\{d(A),d(C)\}$, then $d(K^n\oplus M)\leq d(K^n)$.

If $d$ is as in the proof of Lemma 13.30.2 (tag 07K7), it comes from a $\delta$-functor $F^i$, $d(M)=\text{Sup}\{0\}\cup\{i \| F^i(M)\neq 0\}$, then $d$ satisfies (2) of this lemma, the property in this comment and additionally $d(A)\leq\text{max}\{d(B),d(C)+1\}$.

Finally, assuming the weaker property: "For $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ short exact, $d(A)=d(C)=0$ implies $d(B)=0$". Then I can replace $d$ by a minimal function $d'$ such that $d'(A)=0$ iff $d(A)=0$, and then $d'$ satisfies the three inequalities mentioned so far. So the result follows. The details of the construction of $d'$ are a bit long for a comment and maybe just assuming $d(B)\leq \text{max}\{d(A),d(C)\}$ is good enough.

Comment #1304 by on

Wonderful! Thanks so much for finding this. I decided to fix it by adding another assumption to the lemma concerning the value of $d$ on direct sums. See here.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).