Lemma 52.25.2. The quantity $\chi (\mathcal{F}, \mathcal{F}_0, \alpha )$ in (52.25.1.1) does not depend on the choice of $\mathcal{F}', \alpha ', \alpha '_0$ as in Lemma 52.25.1.

Proof. Let $\mathcal{F}', \alpha ', \alpha '_0$ and $\mathcal{F}'', \alpha '', \alpha ''_0$ be two such choices. For $n > 0$ set $\mathcal{F}'_ n = \mathfrak m^ n \mathcal{F}'$. By Cohomology of Schemes, Lemma 30.10.5 for some $n$ there exists an $\mathcal{O}_ X$-module map $\mathcal{F}'_ n \to \mathcal{F}''$ agreeing with the identification $\mathcal{F}''|_ U = \mathcal{F}'|_ U$ determined by $\alpha '$ and $\alpha ''$. Then the diagram

$\xymatrix{ \mathcal{F}'_ n/f\mathcal{F}'_ n \ar[r] \ar[d] & \mathcal{F}'/f\mathcal{F}' \ar[d]^{\alpha _0'} \\ \mathcal{F}''/f\mathcal{F}'' \ar[r]^{\alpha _0''} & \mathcal{F}_0 }$

is commutative after restricting to $U_0$. Hence by Cohomology of Schemes, Lemma 30.10.5 it is commutative after restricting to $\mathfrak m^ l(\mathcal{F}'_ n/f\mathcal{F}'_ n)$ for some $l > 0$. Since $\mathcal{F}'_{n + l}/f\mathcal{F}'_{n + l} \to \mathcal{F}'_ n/f\mathcal{F}'_ n$ factors through $\mathfrak m^ l(\mathcal{F}'_ n/f\mathcal{F}'_ n)$ we see that after replacing $n$ by $n + l$ the diagram is commutative. In other words, we have found a third choice $\mathcal{F}''', \alpha ''', \alpha '''_0$ such that there are maps $\mathcal{F}''' \to \mathcal{F}''$ and $\mathcal{F}''' \to \mathcal{F}'$ over $X$ compatible with the maps over $U$ and $X_0$. This reduces us to the case discussed in the next paragraph.

Assume we have a map $\mathcal{F}'' \to \mathcal{F}'$ over $X$ compatible with $\alpha ', \alpha ''$ over $U$ and with $\alpha '_0, \alpha ''_0$ over $X_0$. Observe that $\mathcal{F}'' \to \mathcal{F}'$ is injective as it is an isomorphism over $U$ and since $f : \mathcal{F}'' \to \mathcal{F}''$ is injective. Clearly $\mathcal{F}'/\mathcal{F}''$ is supported on $\{ \mathfrak m\}$ hence has finite length. We have the maps of coherent $\mathcal{O}_{X_0}$-modules

$\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}' \xrightarrow {\alpha '_0} \mathcal{F}_0$

whose composition is $\alpha ''_0$ and which are isomorphisms over $U_0$. Elementary homological algebra gives a $6$-term exact sequence

$\begin{matrix} 0 \to \mathop{\mathrm{Ker}}(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}') \to \mathop{\mathrm{Ker}}(\alpha ''_0) \to \mathop{\mathrm{Ker}}(\alpha '_0) \to \\ \mathop{\mathrm{Coker}}(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}') \to \mathop{\mathrm{Coker}}(\alpha ''_0) \to \mathop{\mathrm{Coker}}(\alpha '_0) \to 0 \end{matrix}$

By additivity of lengths (Algebra, Lemma 10.52.3) we find that it suffices to show that

$\text{length}_ A( \mathop{\mathrm{Coker}}(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}')) - \text{length}_ A( \mathop{\mathrm{Ker}}(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}')) = 0$

This follows from applying the snake lemma to the diagram

$\xymatrix{ 0 \ar[r] & \mathcal{F}'' \ar[r]_ f \ar[d] & \mathcal{F}'' \ar[r] \ar[d] & \mathcal{F}''/f\mathcal{F}'' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}' \ar[r]^ f & \mathcal{F}' \ar[r] & \mathcal{F}'/f\mathcal{F}' \ar[r] & 0 }$

and the fact that $\mathcal{F}'/\mathcal{F}''$ has finite length. $\square$

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