Lemma 51.15.5. Let $X$ be a Noetherian scheme which locally has a dualizing complex. Let $T' \subset T \subset X$ be subsets stable under specialization such that if $x \leadsto x'$ is an immediate specialization of points in $X$ and $x' \in T'$, then $x \in T$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Then there exists a unique map $\mathcal{F} \to \mathcal{F}''$ of coherent $\mathcal{O}_ X$-modules such that

1. $\mathcal{F}_ x \to \mathcal{F}''_ x$ is an isomorphism for $x \not\in T$,

2. $\mathcal{F}_ x \to \mathcal{F}''_ x$ is surjective and $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}''_ x) \geq 1$ for $x \in T$, $x \not\in T'$, and

3. $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}''_ x) \geq 2$ for $x \in T'$.

If $f : Y \to X$ is a Cohen-Macaulay morphism with $Y$ Noetherian, then $f^*\mathcal{F} \to f^*\mathcal{F}''$ satisfies the same properties with respect to $f^{-1}(T') \subset f^{-1}(T) \subset Y$.

Proof. First, let $\mathcal{F} \to \mathcal{F}'$ be the quotient of $\mathcal{F}$ constructed in Lemma 51.15.1 using $T$. Second, let $\mathcal{F}' \to \mathcal{F}''$ be the unique map of coherent modules construction in Lemma 51.15.4 using $T'$. Then $\mathcal{F} \to \mathcal{F}''$ is as desired. $\square$

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