Lemma 52.25.3. We have $\chi (\mathcal{G}, \mathcal{G}_0, \beta ) = \chi (\mathcal{F}, \mathcal{F}_0, \alpha ) + \chi (\mathcal{H}, \mathcal{H}_0, \gamma )$ if

$0 \to (\mathcal{F}, \mathcal{F}_0, \alpha ) \to (\mathcal{G}, \mathcal{G}_0, \beta ) \to (\mathcal{H}, \mathcal{H}_0, \gamma ) \to 0$

is a short exact sequence of coherent triples.

Proof. Choose $\mathcal{G}', \beta ', \beta '_0$ as in Lemma 52.25.1 for the triple $(\mathcal{G}, \mathcal{G}_0, \beta )$. Denote $j : U \to X$ the inclusion morphism. Let $\mathcal{F}' \subset \mathcal{G}'$ be the kernel of the composition

$\mathcal{G}' \xrightarrow {\beta '} j_*\mathcal{G} \to j_*\mathcal{H}$

Observe that $\mathcal{H}' = \mathcal{G}'/\mathcal{F}'$ is a coherent subsheaf of $j_*\mathcal{H}$ and hence $f : \mathcal{H}' \to \mathcal{H}'$ is injective. Hence by the snake lemma we obtain a short exact sequence

$0 \to \mathcal{F}'/f\mathcal{F}' \to \mathcal{G}'/f\mathcal{G}' \to \mathcal{H}'/f\mathcal{H}' \to 0$

We have isomorphisms $\alpha ' : \mathcal{F}'|_ U \to \mathcal{F}$, $\beta ' : \mathcal{G}'|_ U \to \mathcal{G}$, and $\gamma ' : \mathcal{H}'|_ U \to \mathcal{H}$ by construction. To finish the proof we'll need to construct maps $\alpha '_0 : \mathcal{F}'/f\mathcal{F}' \to \mathcal{F}_0$ and $\gamma '_0 : \mathcal{H}'/f\mathcal{H}' \to \mathcal{H}_0$ as in Lemma 52.25.1 and fitting into a commutative diagram

$\xymatrix{ 0 \ar[r] & \mathcal{F}'/f\mathcal{F}' \ar[r] \ar@{..>}[d]^{\alpha '_0} & \mathcal{G}'/f\mathcal{G}' \ar[r] \ar[d]^{\beta '_0} & \mathcal{H}'/f\mathcal{H}' \ar[r] \ar@{..>}[d]^{\gamma '_0} & 0 \\ 0 \ar[r] & \mathcal{F}_0 \ar[r] & \mathcal{G}_0 \ar[r] & \mathcal{H}_0 \ar[r] & 0 }$

However, this may not be possible with our initial choice of $\mathcal{G}'$. From the displayed diagram we see the obstruction is exactly the composition

$\delta : \mathcal{F}'/f\mathcal{F}' \to \mathcal{G}'/f\mathcal{G}' \xrightarrow {\beta '_0} \mathcal{G}_0 \to \mathcal{H}_0$

Note that the restriction of $\delta$ to $U_0$ is zero by our choice of $\mathcal{F}'$ and $\mathcal{H}'$. Hence by Cohomology of Schemes, Lemma 30.10.5 there exists an $k > 0$ such that $\delta$ vanishes on $\mathfrak m^ k \cdot (\mathcal{F}'/f\mathcal{F}')$. For $n > k$ set $\mathcal{G}'_ n = \mathfrak m^ n \mathcal{G}'$, $\mathcal{F}'_ n = \mathcal{G}'_ n \cap \mathcal{F}'$, and $\mathcal{H}'_ n = \mathcal{G}'_ n/\mathcal{F}'_ n$. Observe that $\beta '_0$ can be composed with $\mathcal{G}'_ n/f\mathcal{G}'_ n \to \mathcal{G}'/f\mathcal{G}'$ to give a map $\beta '_{n, 0} : \mathcal{G}'_ n/f\mathcal{G}'_ n \to \mathcal{G}_0$ as in Lemma 52.25.1. By Artin-Rees (Algebra, Lemma 10.51.2) we may choose $n$ such that $\mathcal{F}'_ n \subset \mathfrak m^ k \mathcal{F}'$. As above the maps $f : \mathcal{F}'_ n \to \mathcal{F}'_ n$, $f : \mathcal{G}'_ n \to \mathcal{G}'_ n$, and $f : \mathcal{H}'_ n \to \mathcal{H}'_ n$ are injective and as above using the snake lemma we obtain a short exact sequence

$0 \to \mathcal{F}'_ n/f\mathcal{F}'_ n \to \mathcal{G}'_ n/f\mathcal{G}'_ n \to \mathcal{H}'_ n/f\mathcal{H}'_ n \to 0$

As above we have isomorphisms $\alpha '_ n : \mathcal{F}'_ n|_ U \to \mathcal{F}$, $\beta '_ n : \mathcal{G}'_ n|_ U \to \mathcal{G}$, and $\gamma '_ n : \mathcal{H}'_ n|_ U \to \mathcal{H}$. We consider the obstruction

$\delta _ n : \mathcal{F}'_ n/f\mathcal{F}'_ n \to \mathcal{G}'_ n/f\mathcal{G}'_ n \xrightarrow {\beta '_{n, 0}} \mathcal{G}_0 \to \mathcal{H}_0$

as before. However, the commutative diagram

$\xymatrix{ \mathcal{F}'_ n/f\mathcal{F}'_ n \ar[r] \ar[d] & \mathcal{G}'_ n/f\mathcal{G}'_ n \ar[r]_{\beta '_{n, 0}} \ar[d] & \mathcal{G}_0 \ar[r] \ar[d] & \mathcal{H}_0 \ar[d] \\ \mathcal{F}'/f\mathcal{F}' \ar[r] & \mathcal{G}'/f\mathcal{G}' \ar[r]^{\beta '_0} & \mathcal{G}_0 \ar[r] & \mathcal{H}_0 }$

our choice of $n$ and our observation about $\delta$ show that $\delta _ n = 0$. This produces the desired maps $\alpha '_{n, 0} : \mathcal{F}'_ n/f\mathcal{F}'_ n \to \mathcal{F}_0$, and $\gamma '_{n, 0} : \mathcal{H}'_ n/f\mathcal{H}'_ n \to \mathcal{H}_0$. OK, so we may use $\mathcal{F}'_ n, \alpha '_ n, \alpha '_{n, 0}$, $\mathcal{G}'_ n, \beta '_ n, \beta '_{n, 0}$, and $\mathcal{H}'_ n, \gamma '_ n, \gamma '_{n, 0}$ to compute $\chi (\mathcal{F}, \mathcal{F}_0, \alpha )$, $\chi (\mathcal{G}, \mathcal{G}_0, \beta )$, and $\chi (\mathcal{H}, \mathcal{H}_0, \gamma )$. Now finally the lemma follows from an application of the snake lemma to

$\xymatrix{ 0 \ar[r] & \mathcal{F}'_ n/f\mathcal{F}'_ n \ar[r] \ar[d] & \mathcal{G}'_ n/f\mathcal{G}'_ n \ar[r] \ar[d] & \mathcal{H}'_ n/f\mathcal{H}'_ n \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}_0 \ar[r] & \mathcal{G}_0 \ar[r] & \mathcal{H}_0 \ar[r] & 0 }$

and additivity of lengths (Algebra, Lemma 10.52.3). $\square$

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