Homomorphism between Ext induced by the left mutation functor

Question

$\DeclareMathOperator\Coh{Coh}\DeclareMathOperator\ev{ev}\DeclareMathOperator\cone{cone}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Ext{Ext}$This is a specific question concerning a statement in Symplectic structures on moduli spaces of sheaves via the Atiyah class.

Let $Y\subset\mathbb{P}^5$ be a smooth cubic 4-fold, and $\mathcal{I}_\ell$ be the ideal sheaf of a line $\ell$. Consider the functor $$\mathbb{L}\colon D^b(\Coh(Y))\rightarrow D^b(\Coh(Y)),\quad E\mapsto \cone(\ev:\mathcal{O}_Y\otimes\Hom^{\bullet}(\mathcal{O}_Y,E)\rightarrow E)$$ then one has $\mathbb{L}(\mathcal{I}_\ell (1))\cong\mathcal{F}_\ell[1]$ in $D^b(\Coh(Y))$ where $\mathcal{F}_\ell$ is defined by $$0\rightarrow\mathcal{F}_\ell\rightarrow\mathcal{O}_Y^{\oplus 4}\rightarrow\mathcal{I}_\ell(1)\rightarrow0$$ as $\mathcal{I}_\ell(1)$ is generated by global sections. Applying $\Hom(\mathcal{I}_\ell(1),-)$ one has a long exact sequence $$\cdots\rightarrow0\rightarrow\Ext^1(\mathcal{I}_\ell(1),\mathcal{I}_\ell(1))\stackrel{\alpha}{\rightarrow}\Ext^2(\mathcal{I}_\ell(1),\mathcal{F}_\ell)\rightarrow\cdots$$ It states in [1, Proposition 5.4] that the map $$\mathbb{L}\colon\Ext^1(\mathcal{I}_\ell(1),\mathcal{I}_\ell(1))\rightarrow\Ext^1(\mathcal{F}_\ell,\mathcal{F}_\ell)$$ coincides with the composition $$\Ext^1(\mathcal{I}_\ell(1),\mathcal{I}_\ell(1))\stackrel{\alpha}{\rightarrow}\Ext^2(\mathcal{I}_\ell(1),\mathcal{F}_\ell)\cong\Ext^1(\mathcal{F}_\ell,\mathcal{F}_\ell).$$

However, I can not see why it is true. Why they should coincide?

[1] Kuznetsov and Markushevich - Symplectic structures on moduli spaces of sheaves via the Atiyah class.

Answer 1

To say that mutation is a functor, we must define its action on morphisms. It is defined as follows: any morphism $f \colon F_1 \to F_2$ in the derived category induces $H^\bullet(f) \colon H^\bullet(F_1) \to H^\bullet(F_2)$ and there is a unique morphism $\mathbb{L}(f)$ that fits into a commutative diagram $$ \require{AMScd} \begin{CD} H^\bullet(F_1) \otimes \mathcal{O} @>>> F_1 @>>> \mathbb{L}(F_1) \\ @VH^\bullet(f)VV @VfVV @V\mathbb{L}(f)VV \\ H^\bullet(F_2) \otimes \mathcal{O} @>>> F_2 @>>> \mathbb{L}(F_2) \end{CD} $$ Applying this to $F_1 = I_\ell(1)$ and $F_2 = I_\ell(1)[1]$ gives a commutative square $$ \require{AMScd} \begin{CD} I_\ell(1) @>>> F_\ell[1] \\ @VfVV @V\mathbb{L}(f)VV \\ I_\ell(1)[1] @>>> F_\ell[2]. \end{CD} $$ It remains to note that the map $\alpha$ is given by composition with the bottom arrow, and the isomorphism in the question is given by composition with the top arrow.