$\operatorname{Hom}_{D_{\mathfrak{U},\mathbb{Q}}^{\dagger}}(\mathcal{E}|_{\mathfrak{U}}, \mathcal{F}|_{\mathfrak{U}})$. Let $D_0$ be a divisor of $P_0$ such that $X_0 \setminus D_0$ is smooth and dense in a connected component of $X_0$ of dimension $\dim(X_0)$ and such that both $\mathcal{E}(\dagger D_0)$ and $\mathcal{F}(\dagger D_0)$ belong to $\mathrm{MIC}_F^{\dagger\dagger}(Y_0, \mathfrak{P}, Z_0 \cup D_0)$. We have the exact triangle of $H_F(Y_0, \mathfrak{P}, Z_0)$ of the form

$$ (8.3) \qquad 0 \to \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{E}) \xrightarrow{a_{\mathcal{E}}} \mathcal{E} \xrightarrow{b_{\mathcal{E}}} \mathcal{E}(\dagger D_0) \xrightarrow{c_{\mathcal{E}}} \mathcal{H}_{D_0}^{\dagger,1}(\mathcal{E}) \to 0 $$

and similarly for $\mathcal{E}$ replaced by $\mathcal{F}$.

(ii) Arguing as in [DE22, Cor. 5.7], the functor

$$ |\mathfrak{U}: \mathrm{MIC}_F^{\dagger\dagger}(Y_0, \mathfrak{P}, Z_0 \cup D_0) \to \mathrm{MIC}_F^{\dagger\dagger}(Y_0, \mathfrak{U}, D_0 \cap U_0) $$

is fully faithful. Hence, there exists a (unique) morphism $\theta_1: \mathcal{E}(\dagger D_0) \to \mathcal{F}(\dagger D_0)$ such that $\theta_1|_{\mathfrak{U}} = (\dagger D_0)(\psi)$. By the induction hypothesis, there exists a (unique) morphism $\theta_2: \mathcal{H}_{D_0}^{\dagger,1}(\mathcal{E}) \to \mathcal{H}_{D_0}^{\dagger,1}(\mathcal{F})$ of $H_F(Y_0 \cap D_0, \mathfrak{P}, Z_0)$ such that $\theta_2|_{\mathfrak{U}} = \mathcal{H}_{D_0 \cap U_0}^{\dagger,1}(\psi)$.

Consider the diagram of $H_F(Y_0, \mathfrak{P}, Z_0)$

$$ (8.4) \qquad \begin{tikzcd} \mathcal{E}(\dagger D_0) \arrow[r, "c_{\mathcal{E}}"] & \mathcal{H}_{D_0}^{\dagger,1}(\mathcal{E}) \arrow[d, "\theta_1"] & \mathcal{H}_{D_0}^{\dagger,1}(\mathcal{F}) \arrow[d, "\theta_2"] \\ \mathcal{F}(\dagger D_0) \arrow[r, "c_{\mathcal{F}}"] & \mathcal{H}_{D_0}^{\dagger,1}(\mathcal{F}) \arrow[d, "\theta_2"] & \mathcal{F}(\dagger D_0) \arrow[r, "c_{\mathcal{F}}"] & \mathcal{H}_{D_0}^{\dagger,1}(\mathcal{F}) \arrow[d, "\theta_1"'] \end{tikzcd} $$

Using [Car25, 15.3.7.7], the diagram (8.4) is commutative because so is on $\mathfrak{U}$. Since $\mathrm{Ker}c_{\mathcal{E}} = \mathrm{Im}b_{\mathcal{E}}$ and $\mathrm{Ker}c_{\mathcal{F}} = \mathrm{Im}b_{\mathcal{F}}$, we get therefore a morphism $\theta: \mathrm{Im}b_{\mathcal{E}} \to \mathrm{Im}b_{\mathcal{F}}$ such that $\theta|_{\mathfrak{U}}$ is the morphism induced by $\psi$.

(iii) Let $\mathcal{G}$ be an object of $H_F(\mathfrak{P}, Z_0)$. Consider the commutative diagram:

$$ \begin{tikzcd} \operatorname{Hom}_{D_{\mathfrak{P},\mathbb{Q}}^{\dagger}}(\mathcal{G}, \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F})) \arrow[r, "\sim"']_{(8.2)} & \operatorname{Hom}_{D_{\mathfrak{P},\mathbb{Q}}^{\dagger}}(\mathcal{H}_{D_0}^{\dagger,0}(\mathcal{G}^*), \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F})) \arrow[d] \\ \operatorname{Hom}_{D_{\mathfrak{U},\mathbb{Q}}^{\dagger}}(\mathcal{G}|_{\mathfrak{U}}, \mathcal{H}_{D_0 \cap U_0}^{\dagger,0}(\mathcal{F}|_{\mathfrak{U}})) \arrow[r, "\sim"']_{(8.2)} & \operatorname{Hom}_{D_{\mathfrak{U},\mathbb{Q}}^{\dagger}}(\mathcal{H}_{D_0 \cap U_0}^{\dagger,0}((\mathcal{G}|_{\mathfrak{U}})^*), \mathcal{H}_{D_0 \cap U_0}^{\dagger,0}(\mathcal{F}|_{\mathfrak{U}})). \arrow[d] \end{tikzcd} $$

By induction hypothesis, the right arrow is an isomorphism. Then so is the left one.

(iv) The map

$$ \mathrm{Ext}_{D_{\mathfrak{P},\mathbb{Q}}^{\dagger}}^{1}(\mathcal{E}, \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F})) \to \mathrm{Ext}_{D_{\mathfrak{U},\mathbb{Q}}^{\dagger}}^{1}(\mathcal{E}|_{\mathfrak{U}}, \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F})|_{\mathfrak{U}}) $$

is injective. Indeed, let

$$ 0 \to \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F}) \to \mathcal{G} \to \mathcal{E} \to 0 $$

be an exact sequence of $H_F(\mathfrak{P}, Z_0)$. If $\mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F})|_{\mathcal{U}} \to \mathcal{G}|_{\mathcal{U}}$ has a retract, then it follows from (iii) and the faithfulness of $\mathcal{U}$ that $\mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F}) \to \mathcal{G}$ has also a retract and we are done.
(v) Consider the commutative diagram of K-vector spaces:

$$ (8.5) \qquad \begin{tikzcd} \operatorname{Hom}_{D_{\mathfrak{P},Q}^{\dagger}} (\mathcal{E}, \mathcal{F}) \arrow[r] \arrow[d] & \operatorname{Hom}_{D_{\mathfrak{P},Q}^{\dagger}} (\mathcal{E}, \operatorname{Im} b_{\mathcal{F}}) \arrow[r] \arrow[d] & \operatorname{Ext}_{D_{\mathfrak{P},Q}^{\dagger}}^{1} (\mathcal{E}, \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F})) \arrow[d] \\ \operatorname{Hom}_{D_{\mathfrak{P},Q}^{\dagger}} (\mathcal{E}|_{\mathfrak{U}}, \mathcal{F}|_{\mathfrak{U}}) \arrow[r] & \operatorname{Hom}_{D_{\mathfrak{P},Q}^{\dagger}} (\mathcal{E}|_{\mathfrak{U}}, \operatorname{Im} b_{\mathcal{F}}|_{\mathfrak{U}}) \arrow[r] & \operatorname{Ext}_{D_{\mathfrak{P},Q}^{\dagger}}^{1} (\mathcal{E}|_{\mathfrak{U}}, \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F})|_{\mathfrak{U}}) \end{tikzcd} $$

whose rows are exact. According to (iv), the right arrow of (8.5) is injective.

Let $\eta: \mathcal{E} \to \operatorname{Im} b_{\mathcal{F}}$ be the composition of $\mathcal{E} \to \operatorname{Im} b_{\mathcal{E}}$ with the morphism $\theta$ constructed in (ii). Since the bottom row of (8.5) is exact, since $\eta|_{\mathfrak{U}}$ is the composition of $\psi$ with $\mathcal{F}|_{\mathfrak{U}} \to \operatorname{Im} b_{\mathcal{F}}|_{\mathfrak{U}}$, then the image of $\eta|_{\mathfrak{U}}$ in $\operatorname{Ext}_{D_{\mathfrak{P},Q}^{\dagger}}^{1}(\mathcal{E}|_{\mathfrak{U}}, \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F})|_{\mathfrak{U}})$ vanishes. Since the right arrow of (8.5) is injective, this yields that the image of $\eta$ in $\operatorname{Ext}_{D_{\mathfrak{P},Q}^{\dagger}}^{1}(\mathcal{E}, \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F}))$ vanishes. By the exactness of the top row, we get that $\eta$ comes from an element $\phi$ of $\operatorname{Hom}_{D_{\mathfrak{P},Q}^{\dagger}}(\mathcal{E}, \mathcal{F})$. Since the image of $\psi - \phi|_{\mathfrak{U}}$ in $\operatorname{Hom}_{D_{\mathfrak{P},Q}^{\dagger}}(\mathcal{E}|_{\mathfrak{U}}, \operatorname{Im} b_{\mathcal{F}}|_{\mathfrak{U}})$ is null, then it comes from an element $\beta$ of $\operatorname{Hom}_{D_{\mathfrak{P},Q}^{\dagger}}(\mathcal{E}|_{\mathfrak{U}}, \mathcal{H}_{D_0 \cap U_0}^{\dagger,0}(\mathcal{F})|_{\mathfrak{U}})$. It follows from (iii) that there exists $\alpha \in \operatorname{Hom}_{D_{\mathfrak{P},Q}^{\dagger}}(\mathcal{E}, \mathcal{H}_{D_0}^{\dagger,0}(\mathcal{F}))$ such that $\alpha|_{\mathfrak{U}} = \beta$. Hence, $\psi - \phi|_{\mathfrak{U}} = (a_{\mathcal{E}} \circ \alpha)|_{\mathfrak{U}}$, i.e., $\psi = (\phi + a_{\mathcal{E}} \circ \alpha)|_{\mathfrak{U}}$. Hence, we are done. $\square$

**Corollary 8.2.5.** Let $\mathcal{E}, \mathcal{F}$ be two objects of $\mathbf{H}_F(Y_0, \mathfrak{P}, Z_0)$. The map

$$ \operatorname{Ext}_{D_{\mathfrak{P},Q}^{\dagger}}^{1}(\mathcal{E}, \mathcal{F}) \to \operatorname{Ext}_{D_{\mathfrak{P},Q}^{\dagger}}^{1}(\mathcal{E}|_{\mathfrak{U}}, \mathcal{F}|_{\mathfrak{U}}) $$

is injective.

*Proof.* This follows from Theorem 8.2.4 as in Theorem 3.2.1. $\square$

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UNIVERSITY OF CAEN NORMANDIE, LMNO, 14000 CAEN, FRANCE

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