Proof. We first treat the special case where $X$ and $Y$ are affine schemes and where the morphism $f$ is flat. Say $X = \mathop{\mathrm{Spec}}(R)$ and $Y = \mathop{\mathrm{Spec}}(S)$. Then $f$ corresponds to a flat ring map $R \to S$. Moreover, $Z \subset X$ is cut out by a finitely generated ideal $I \subset R$. Choose generators $f_1, \ldots , f_ t \in I$. By the description of quasi-coherent modules in terms of modules (Schemes, Section 26.7), we see that the category $\textit{QCoh}(Y \to X, Z)$ is canonically equivalent to the category $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ of More on Algebra, Remark 15.89.10 such that the functors (80.10.6.1) and (80.10.6.2) correspond to the functors $\text{Can}$ and $H^0$. Hence the result follows from More on Algebra, Proposition 15.89.15 in this case.

We return to the general case. Let $\mathcal{F}$ be a quasi-coherent module on $X$. We will show that

$\alpha : \mathcal{F} \longrightarrow \mathop{\mathrm{Ker}}\left(j_*\mathcal{F}|_ U \oplus f_*f^*\mathcal{F} \to (f \circ j')_*f^*\mathcal{F}|_ V\right)$

is an isomorphism. Let $(\mathcal{H}, \mathcal{G}, \varphi )$ be an object of $\mathit{QCoh}(Y \to X, Z)$. We will show that

$\beta : f^*\mathop{\mathrm{Ker}}\left( j_*\mathcal{H} \oplus f_*\mathcal{G} \to (f \circ j')_*\mathcal{G}|_ V \right) \longrightarrow \mathcal{G}$

and

$\gamma : j^*\mathop{\mathrm{Ker}}\left( j_*\mathcal{H} \oplus f_*\mathcal{G} \to (f \circ j')_*\mathcal{G}|_ V \right) \longrightarrow \mathcal{H}$

are isomorphisms. To see these statements are true it suffices to look at stalks. Let $\overline{y}$ be a geometric point of $Y$ mapping to the geometric point $\overline{x}$ of $X$.

Fix an object $(\mathcal{H}, \mathcal{G}, \varphi )$ of $\mathit{QCoh}(Y \to X, Z)$. By Lemma 80.10.2 and a diagram chase (omitted) the canonical map

$\mathop{\mathrm{Ker}}(j_*\mathcal{H} \oplus f_*\mathcal{G} \to (f \circ j')_*\mathcal{G}|_ V)_{\overline{x}} \longrightarrow \mathop{\mathrm{Ker}}( j_*\mathcal{H}_{\overline{x}} \oplus \mathcal{G}_{\overline{y}} \to j'_*\mathcal{G}_{\overline{y}} )$

is an isomorphism.

In particular, if $\overline{y}$ is a geometric point of $V$, then we see that $j'_*\mathcal{G}_{\overline{y}} = \mathcal{G}_{\overline{y}}$ and hence that this kernel is equal to $\mathcal{H}_{\overline{x}}$. This easily implies that $\alpha _{\overline{x}}$, $\beta _{\overline{x}}$, and $\beta _{\overline{y}}$ are isomorphisms in this case.

Next, assume that $\overline{y}$ is a point of $f^{-1}Z$. Let $I_{\overline{x}} \subset \mathcal{O}_{X, \overline{x}}$, resp. $I_{\overline{y}} \subset \mathcal{O}_{Y, \overline{y}}$ be the stalk of the ideal cutting out $Z$, resp. $f^{-1}Z$. Then $I_{\overline{x}}$ is a finitely generated ideal, $I_{\overline{y}} = I_{\overline{x}}\mathcal{O}_{Y, \overline{y}}$, and $\mathcal{O}_{X, \overline{x}} \to \mathcal{O}_{Y, \overline{y}}$ is a flat local homomorphism inducing an isomorphism $\mathcal{O}_{X, \overline{x}}/I_{\overline{x}} = \mathcal{O}_{Y, \overline{y}}/I_{\overline{y}}$. At this point we can bootstrap using the diagram of categories

$\xymatrix{ \mathit{QCoh}(\mathcal{O}_ X) \ar[r]_-{(0AEW)} \ar[d] & \mathit{QCoh}(Y \to X, Z) \ar[d] \ar@/_2pc/[l]^{(0AEX)} \\ \text{Mod}_{\mathcal{O}_{X, \overline{x}}} \ar[r]^-{\text{Can}} & \text{Glue}(\mathcal{O}_{X, \overline{x}} \to \mathcal{O}_{Y, \overline{y}}, f_1, \ldots , f_ t) \ar@/^2pc/[l]_{H^0} }$

Namely, as in the first paragraph of the proof we identify

$\text{Glue}(\mathcal{O}_{X, \overline{x}} \to \mathcal{O}_{Y, \overline{y}}, f_1, \ldots , f_ t) = \mathit{QCoh}(\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, \overline{y}}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \overline{x}}), V(I_{\overline{x}}))$

The right vertical functor is given by pullback, and it is clear that the inner square is commutative. Our computation of the stalk of the kernel in the third paragraph of the proof combined with Lemma 80.10.3 implies that the outer square (using the curved arrows) commutes. Thus we conclude using the case of a flat morphism of affine schemes which we handled in the first paragraph of the proof. $\square$

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