What is the co-kernel of the morphism of vector bundles?

Let $X$ be a surface, and $i:C\subset X$ be a smooth curve. Let $A$ be a line bundle on $C$, and $E$ be a vector bundle of rank $r$ on $X$.

Suppose there is a surjection: $E\longrightarrow i_*A\longrightarrow 0\\$, then we get an exact sequence $\qquad\qquad\qquad\qquad\qquad0\longrightarrow F\longrightarrow E\longrightarrow i_*A\longrightarrow 0$.

Now $i_*A$ is torsion, we have $(i_*A)^*=0$. Therefore taking the dual sequence, we get

$\qquad\qquad\qquad\qquad\qquad 0\longrightarrow E^*\longrightarrow F^*\longrightarrow Q\longrightarrow 0$.

$Q$ is $\mathcal{Ext}^1(i_*A,\mathcal{O}_X)$. But what is it? I saw somewhere that it is $\mathcal{O}_C(C)\otimes A^\vee$. Is that right? How do we get it? Thanks in advance!

• Grothendieck duality gives isomorphisms $R\mathcal{H}om_X(i_*A,\mathcal{O}_X)\cong i_*R\mathcal{H}om_C(A,i^!\mathcal{O}_X)$ and $i^!\mathcal{O}_X\cong \mathcal{O}_C(C))[-1]$, hence your claim. See for instance Hartshorne's "Residues and duality". – abx Apr 6 '15 at 19:54
• Calculate the ext group using the exact sequence, $$0\to \mathcal{O}_X\to\mathcal{O}_X(C)\to\mathcal{O}_C(C)\to 0,$$ and taking $\mathcal{H}om (A,*)$. – Mohan Apr 6 '15 at 21:00
• @poorna, Which part do you have trouble with? Are you ok with the above exact sequence? If so, when you apply the hom functor, you end up with $$0\to \mathcal{H}om(A,O_C(C))\to \mathcal{E}xt^1(A,O_X)\to\mathcal{E}xt^1(A,O_X(C))$$ exact. Easy to check that the last map is zero and thus you get the required isomorphism. – Mohan Apr 8 '15 at 18:26
• The last map is zero because both the sheaves are supported completely on $C$ and the map is just (locally) multiplication by the equation defining $C$ in $X$ and hence zero. – Mohan Apr 9 '15 at 2:43
• It is induced by the map $O_X\to O_X(C)$ which is just multiplication by the equation defining $C$ in $X$. But this equation acts as zero on any sheaf supported (scheme theoretically) on $C$. – Mohan Apr 9 '15 at 17:27