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I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$)

I am studying a little about distinguished triangle and I want to know that what I write below is correct or not!

We have natural morphism $f^*\Omega_Y \to\Omega_C(p_1+...+p_n)$ if we consider these two term in two complex (put these two term in $0$-place and other elements of complex are $0$) we will have distinguished triangle

$[...\to 0 \to f^*\Omega_Y \to 0 \to ...] \to [... \to 0 \to (0,f^*\Omega_Y,0),0) \to (f^*\Omega_Y,0,\Omega_C(p_1+...+p_n) \to0 \to...] \to [...\to 0\to(f^*\Omega_Y,0)\to (0,\Omega_C(p_1+...+p_n))\to 0 \to...]$

Here I used the fact that if we have $f:K^* \to L^*$(map between complexes) then we have distinguished triangle $K^* \to Cyl(f) \to C(f) \to K[1]^*$

Now the first term of distinguished triangle is clear.the third term is same as $ f^*\Omega_Y \to \Omega_C(p_1+...+p_n)$ (via definition of map in $C(f)$ which i didnt imply) and the first is I think will be isomorph to complex [$\Omega_C(p_1+...+p_n)]$. Now if we apply $Ext(-,O_C) $ we will get the long exact sequence in my question.

I want to know is it correct or not?

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    $\begingroup$ Those Ext groups are hyperderived functors defined for every bounded above complex of $\mathcal{O}_C$-modules, i.e., for every object of the derived category. One keyword is "Cartan-Eilenberg resolution". The short exact sequence you write down (missing a dual somewhere) is actually part of a distinguished triangle in the derived category, not actually a short exact sequence. There is a long exact sequence of hyperderived functors for every distinguished triangle in the derived category. $\endgroup$
    – Jason Starr
    Commented Sep 25, 2018 at 16:30
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    $\begingroup$ Can you introduce some references? I am not familiar with these things.thanks $\endgroup$
    – Tom
    Commented Sep 25, 2018 at 16:56
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    $\begingroup$ Crossposted from MSE: math.stackexchange.com/questions/2929846/… $\endgroup$
    – KReiser
    Commented Sep 25, 2018 at 18:30
  • $\begingroup$ Some good references for hyperderived functors, and homological algebra in general, are the book by Cartan and Eilenberg, the book by Charles Weibel, and the book by Yuri Manin. $\endgroup$
    – Jason Starr
    Commented Sep 25, 2018 at 20:16
  • $\begingroup$ Yes, that is correct. The complex that we use to compute deformations is a mapping cone as you write. This is part of what is called the "distinguished triangle of transitivity" in references about the cotangent complex. $\endgroup$
    – Jason Starr
    Commented Sep 27, 2018 at 16:10

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