$\DeclareMathOperator\Coh{Coh}\DeclareMathOperator\ev{ev}\DeclareMathOperator\cone{cone}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Ext{Ext}$This is a specific question concerning a statement in [Symplectic structures on moduli spaces of sheaves via the Atiyah class](https://www.sciencedirect.com/science/article/pii/S0393044009000503).


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](https://www.sciencedirect.com/science/article/pii/S0393044009000503), 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](https://doi.org/10.1016/j.geomphys.2009.03.008).



Homomorphism between Ext induced by the left mutation functor
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.


