Consider over $\mathbb C$. Let $Artin/\mathbb C$ denote the category whose objects are Artinian local $\mathbb C$-algebras with residue field $\mathbb C$, and morphisms are local ring maps preserving residue fields.
Let $X$ be a projective scheme. Let $\mathcal E$ be a coherent $\mathcal O_X$-module. Let $(\mathcal I^\bullet,\operatorname{d}^\bullet)$ be a complex of injective $\mathcal O_X$-modules such that $\mathcal E\to\mathcal I^\bullet$ is a resolution. For $A\in Artin/\mathbb C$, denote $X_A:=X\times_{\mathbb C}\operatorname{Spec}(A)$. Let $\mathcal F$ be a coherent $\mathcal O_{X_A}$-module, flat over $A$, and extending $\mathcal E$, i.e. $\mathcal F\otimes_A(A/\mathfrak m_A)\cong\mathcal E$.
My question is about proving the following.
For any $A\in Artin/\mathbb C$ and $\mathcal F$ over $A$ extending $\mathcal E$, there exist differentials $\operatorname{d}_A^i:\mathcal I^i\otimes_{\mathbb C}A\to \mathcal I^{i+1}\otimes_{\mathbb C}A$ extending $\operatorname{d}^i$ such that $\mathcal F\cong H^0(\mathcal I^\bullet\otimes_{\mathbb C}A,\operatorname{d}_A^i)$.
This is a statement in 2.A.6 in Daniel Huybrechts and Manfred Lehn.
The following is my attempt. Consider the following situation.
Let $X$ be as above. Let $r:\mathcal E\to \mathcal I$ be a morphism of $\mathcal O_X$-modules such that $\mathcal I$ is injective. Let $B\to A$ be a small extension in $Artin/\mathbb C$. Assume $\mathcal F$ is an $\mathcal O_{X_B}$-module, flat over $B$, extending $\mathcal E$, whose pullback to $A$ is denoted by $\mathcal F_A$. Assume we have an $\mathcal O_{X_A}$-morphisms $r_A:\mathcal F_A\to \mathcal I\otimes_\Bbbk A$ extending $r$, i.e. the right square commutes
$$\require{AMScd}\begin{CD} \mathcal F@>\textrm{surjective}>>\mathcal F_A@>\textrm{surjective}>>\mathcal E\\ @.@Vr_AVV@VrVV\\ \mathcal I\otimes B@>>>\mathcal I\otimes A@>>>\mathcal I \end{CD}$$
Can we find $\mathcal O_{X_B}$-morphism $r_B:\mathcal F\to \mathcal I\otimes_{\mathbb C}B$ completing the left square?
We say a surjective ring map $B\to A$ in $Artin/\mathbb C$ is a small extension if there exists $t\in B\setminus \{0\}$ such that the following is exact $$0\to\mathfrak m_B\to B\xrightarrow{t}B\to A\to 0.$$
By homological algebra, the existence of $r_B:\mathcal F\to \mathcal I\otimes_{\mathbb C}B$ is equivalent to that $$r_A\in \ker\big(\operatorname{Hom}_B(\mathcal F_A,\mathcal I\otimes A\big)\to \operatorname{Ext}_B^1(\mathcal F,\mathcal I\otimes tB)\big)$$ where $tB:=\ker(B\to A)$. Note that as a $B$-module, we have $tB\cong B/\mathfrak m_B\cong \mathbb C$. Then we are asking $$r_A\in\ker\big(\operatorname{Hom}_B(\mathcal F_A,\mathcal I\otimes A\big)\to \operatorname{Ext}_B^1(\mathcal F,\mathcal I)\big). $$
I would guess $\operatorname{Ext}_B^1(\mathcal F,\mathcal I)=0$. However, $\mathcal I$ is injective as an $\mathcal O_X$-module, not as an $\mathcal O_{X_B}$-module. But $X$ and $X_B$ have the same underlying topological space, which may help in some aspect.
