**Proof.**
Observe that the statement makes sense as finitely presented $\mathcal{O}_ X$-modules are quasi-coherent. Since $\mathcal{A}$ is a Serre subcategory closed under extensions and direct sums and since $\mathcal{F}$ is an object of $\mathcal{A}$ we see that $\mathcal{B} \subset \mathcal{A}$. Thus it remains to show that $\mathcal{A}$ is contained in $\mathcal{B}$.

Let $\mathcal{E}$ be an object of $\mathcal{A}$. There exists a maximal submodule $\mathcal{E}' \subset \mathcal{E}$ which is in $\mathcal{B}$. Namely, suppose $\mathcal{E}_ i \subset \mathcal{E}$, $i \in I$ is the set of subobjects which are objects of $\mathcal{B}$. Then $\bigoplus \mathcal{E}_ i$ is in $\mathcal{B}$ and so is

\[ \mathcal{E}' = \mathop{\mathrm{Im}}(\bigoplus \mathcal{E}_ i \longrightarrow \mathcal{E}) \]

This is clearly the maximal submodule we were looking for.

Now suppose that we have a nonzero map $\mathcal{G} \to \mathcal{E}/\mathcal{E}'$ with $\mathcal{G}$ in $\mathcal{B}$. Then $\mathcal{G}' = \mathcal{E} \times _{\mathcal{E}/\mathcal{E}'} \mathcal{G}$ is in $\mathcal{B}$ as an extension of $\mathcal{E}'$ and $\mathcal{G}$. Then the image $\mathcal{G}' \to \mathcal{E}$ would be strictly bigger than $\mathcal{E}'$, contradicting the maximality of $\mathcal{E}'$. Thus it suffices to show the claim in the following paragraph.

Let $\mathcal{E}$ be an nonzero object of $\mathcal{A}$. We claim that there is a nonzero map $\mathcal{G} \to \mathcal{E}$ with $\mathcal{G}$ in $\mathcal{B}$. We will prove this by induction on the minimal number $n$ of affine opens $U_ i$ of $X$ such that $\text{Supp}(\mathcal{E}) \subset U_1 \cup \ldots \cup U_ n$. Set $U = U_ n$ and denote $j : U \to X$ the inclusion morphism. Denote $\mathcal{E}' = \mathop{\mathrm{Im}}(\mathcal{E} \to j_*\mathcal{E}|_ U)$. Then the kernel $\mathcal{E}''$ of the surjection $\mathcal{E} \to \mathcal{E}'$ has support contained in $U_1 \cup \ldots \cup U_{n - 1}$. Thus if $\mathcal{E}''$ is nonzero, then we win. In other words, we may assume that $\mathcal{E} \subset j_*\mathcal{E}|_ U$. In particular, we see that $\mathcal{E}|_ U$ is nonzero. By Lemma 56.6.2 there exists a nonzero map $\mathcal{F}|_ U \to \mathcal{E}|_ U$. This corresponds to a map

\[ \varphi : \mathcal{F} \longrightarrow j_*(\mathcal{E}|_ U) \]

whose restriction to $U$ is nonzero. Setting $\mathcal{G} = \varphi ^{-1}(\mathcal{E})$ we conclude.
$\square$

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