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Consider a lower term of local-to-global spectral sequence

$0 \to H^1(X,\mathcal{Hom}(\mathcal{F},\mathcal{G})) \to Ext^1(\mathcal{F},\mathcal{G}) \to H^0(X,\mathcal{Ext}^1(\mathcal{F},\mathcal{G})) \to H^2(X,\mathcal{Hom}(\mathcal{F},\mathcal{G}))$

Where $X$ is a smooth projective variety, $\mathcal{F},\mathcal{G}$ are coherent sheaves on $X$.

Now consider a morphism $\mathcal{G} \to \mathcal{G'}$. Then we have an induced diagram. $ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$

$$ \begin{array}{c} 0 & \ra{} & H^1(X,\mathcal{Hom}(\mathcal{F},\mathcal{G})) & \ra{} & Ext^1(\mathcal{F},\mathcal{G}) & \ra{} & H^0(X,\mathcal{Ext}^1(\mathcal{F},\mathcal{G})) & \ra{} & H^2(X,\mathcal{Hom}(\mathcal{F},\mathcal{G})) \\ & & \da{} & & \da{} & & \da{} & & \da{} \\ 0 & \ra{} & H^1(X,\mathcal{Hom}(\mathcal{F},\mathcal{G'})) & \ra{} & Ext^1(\mathcal{F},\mathcal{G'}) & \ra{} & H^0(X,\mathcal{Ext}^1(\mathcal{F},\mathcal{G'})) & \ra{} & H^2(X,\mathcal{Hom}(\mathcal{F},\mathcal{G'})) \end{array} $$

So, is this diagram commutes?

Improve this question
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    $\begingroup$ Yes, the diagram commutes. The Grothendieck spectral sequence is a spectral sequence in the "Abelian category of additive functors". One way to see this is to check compatibility of Cartan-Eilenberg resolutions, cf. Exercise 5.7.2, p.146, "An Introduction to Homological Algebra", Charles A. Weibel, Camb. Studies in Adv. Math. 38. $\endgroup$
    – Jason Starr
    Commented Sep 22, 2015 at 13:52
  • $\begingroup$ Thanks, then I think I made a stupid mistake in the following mistake. Let Y be a smooth projective space and Consider a deformation long exact sequence of the stable map space $M_0,0(Y,\beta)$. Consider $C$ is a nodal curve which is a union of two $\mathbb{P}^1$. And consider a stable map $f : C \to Y$. Then we have a long exact sequence $Ext^i_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C) \to Ext^i_C(\Omega_C,\mathcal{O}_C) \to Ext^i_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{i+1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C) \to ...$ for $i \geq 0$. Then for $i=1$, we can show that $\endgroup$
    – free-object
    Commented Sep 22, 2015 at 14:56
  • $\begingroup$ $Ext^1_C(\Omega_C,\mathcal{O}_C) \to Ext^1_C(f^*\Omega_Y,\mathcal{O}_C)$ is zero map. Then It means $Ext^1_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C) \to Ext^1_C(\Omega_C,\mathcal{O}_C)$ is surjective. It means that stable map $f$ can be smoothable. I think it is very absurd because it means every stable map from $C$ is smoothable and it is not true. I cannot find where is wrong but I'm sure that I made a stupid mistake somewhere. $\endgroup$
    – free-object
    Commented Sep 22, 2015 at 15:03
  • $\begingroup$ You write, "Let $Y$ be a smooth projective space ...". For projective space, it is indeed the case that $\text{Ext}_C^1(f^*\Omega_Y,\mathcal{O}_C)$ is zero for every genus $0$ stable map. Every genus $0$ stable map to projective space does deform to stable maps with smooth, irreducible domain. For $Y$ some other projective variety, $\text{Ext}^1_C(f^*\Omega_Y,\mathcal{O}_C)$ may be nonzero. $\endgroup$
    – Jason Starr
    Commented Sep 22, 2015 at 16:07
  • $\begingroup$ Sorry, 'smooth projective space' is a typo of 'smooth projective variety'. Yes $Ext^1_C(f^*\Omega_Y,\mathcal{O}_C)$ may be nonzero, but using the above spectral sequence, then $Ext^1_C(\Omega_C,\mathcal{O}_C) \to Ext^1_C(f^*\Omega_Y,\mathcal{O}_C)$ is zero. which means I think that the map $f$ is smoothable since $Ext^1_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C) \to Ext^1_C(\Omega_C,\mathcal{O}_C)$ is surjective and $Ext^1_C(\Omega_C,\mathcal{O}_C)$ corresponds to deformation smoothing nodes and $Ext^1_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C)$ correspond s to deformation of the map $f$ $\endgroup$
    – free-object
    Commented Sep 22, 2015 at 17:35

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