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Given a flat and projective morphism $f:X\rightarrow Y$ of noetherian schemes over some algebraically closed $k$ and $F$, $G$ coherent $O_X$-modules, flat over $Y$.

Then the base change theroem for the relative $Ext$ sheaves reads: Let $y\in Y$ and assume $\tau^i(y): \mathcal{E}xt_f^i(F,G)\otimes k(y)\rightarrow Ext_{X_y}^i(F_y,G_y)$ is surjective. Then:

i) There is a neighbourhood $U$ of $y$ s.t. $\tau^i(y')$ is an isomorphism for all $y' \in U$

ii) $\tau^{i-1}(y)$ is surjective if and only if $\mathcal{E}xt_f^i(F,G)$ is locally free in a neighbourhood of $y$

So now assume, we are in the following situation: $Y$ is a smooth projective surface and $X$ is the product of $Y$ with another projective smooth surface and $f$ is the projection. We have $Ext_{X_y}^i(F_y,G_y)=0$ for all $i\geq3$ and all $y\in Y$. Furthermore we have: $Ext_{X_y}^i(F_y,G_y)=0$ for $i=0,1,2$ and all $y\in Y\setminus \{y_0\}$ for a fixed $y_0 \in Y$. Finally $Ext_{X_{y_0}}^0(F_{y_0},G_{y_0})=Ext_{X_{y_0}}^2(F_{y_0},G_{y_0})=k^s$ and $Ext_{X_{y_0}}^1(F_{y_0},G_{y_0})=k^{2s}$ for some $s\geq1$.

The claim is that we get (with an application of the base change theorem):

(a)$\mathcal{E}xt_f^i(F,G)=0$ for $i=0,1$ and (b)$\mathcal{E}xt_f^2(F,G)\otimes k(y_0)=k^s$

So, since $Ext_{X_y}^3(F_y,G_y)=0$ for all y, we have by i) that $\tau^3(y)$ is surjective for all y, so it is an isomorphism for all $y\in Y$ and $\mathcal{E}xt_f^3(F,G)=0$, which this is locally free on $Y$. So $\tau^2(y)$ is an isomorphism for all $y\in Y$ by ii). So we get (b).

But how about (a). We have that $\tau^1(y)$ is surjective for all $y\in Y\setminus y_0$. But what about $\tau^1(y_0)$. What can be said about $\mathcal{E}xt_f^2(F,G)$? Is it localy free somewhere? How do we get the vanishing of $\mathcal{E}xt^1_f(F,G)$ with the base change theorem?

Background: I'm still working on the Example $\href{http://books.google.de/books?id=_mYV1q0RVzIC&printsec=frontcover&dq=%22geometry+of+moduli+spaces+of+sheaves%22&source=bl&ots=ZGL9LDSVjV&sig=p9l1BZ-UOXZdlxakuC3k15sTtBk&hl=de&ei=flHpTOHyPI3oOaXj7IcK&sa=X&oi=book_result&ct=result&resnum=2&sqi=2&ved=0CCMQ6AEwAQ#v=onepage&q=base%20change&f=false}{here}$ at the bottom of page 169.

Edit: Sasha gave an answer that uses derived categories. But i'm also interested if one can prove this, just by using i) and ii) in the base change theorem.

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It is always useful to look from the derived category point of view. What you want is the object of $D(Y)$ which is $Rf_*R{\mathcal H}om(F,G)$ --- the sheaves ${\mathcal E}xt^i_f$ are its sheaf cohomology. The base change tells you that $Li^*Rf_*R{\mathcal H}om(F,G) \cong RHom(F_{y_0},G_{y_0})$, where $i$ is the embedding of the point $y_0$. Since you know $Ext^2(F_{y_0},G_{y_0}) = k^s$, you have a map $RHom(F_{y_0},G_{y_0}) \to k^s[-2]$. It gives by adjunction a map $Rf_*R{\mathcal H}om(F,G) \to O_{y_0}^s[-2]$. You want to prove that it is an isomorphism. Include this morphism into exact triangle $$ Rf_*R{\mathcal H}om(F,G) \to O_{y_0}^s[-2] \to C. $$ You want to check that $C = 0$. You already know that $C$ is zero on the complement of $y_0$. Then it suffices to check that $Li^* C = 0$. Apply $Li^*$ to the triangle, you will get $$ RHom(F_{y_0},G_{y_0}) \to Li^* O_{y_0}^s[-2] \to Li^* C. $$ Note that $Li^* O_{y_0} = Li^* i_* k$ has cohomology $k$, $k^2$ and $k$ in degrees $0$, $-1$ and $-2$. Hence you see that the first two terms of the sequence have the same cohomology. So, it remains to check that the map induces their isomorphism. This is the tricky part.

To do this, you need more information about the situation. In your case, remember that $L_1i^* O_{y_0} = T_{y_0}^* Y$, and since $Y$ is the moduli space of sheaves $G$, it is isomorphic to $Ext^1(G_{y_0},G_{y_0})^*$. So, you have to check that $Ext^\bullet(F_{y_0},G_{y_0})$ as a module over $Ext^\bullet(G_{y_0},G_{y_0})$ is free, which is evident in your example.

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Okay. I'm not so good in derived categorys, but i will try to get this argument. Now i'm having two questions, just to avoid confusions: $O_{y_0}$ denotes the structure sheaf of the point $y_0$? And why do we have $L_1 i^{\*} O_{y_0} =T^{\*}_{y_0} Y$, the cotangent space, i guess? Do you know some literature concerning the use of derived categories in algebraic geometry? –  TonyS Nov 21 '10 at 21:17
    
A really nice friendly introduction to the derived category in AG is the article "Derived Categories of Sheaves: A Skimming" by Caldararu. –  Matt Nov 22 '10 at 0:21
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Yes, $O_{y_0}$ is the structure sheaf of the point. And if $Z \subset Y$ is a locally complete intersection then $L_k i^* O_Z = \Lambda^k N^*_{Z/Y}$ (this is called "local fundamental isomorphism" or something of this sort), can be checked e.g. by a local computation. –  Sasha Nov 22 '10 at 3:38
    
Concerning the literature, I would suggest another book of Daniel Huybrechts --- "Fourier-Mukai transforms in algebraic geometry". –  Sasha Nov 22 '10 at 3:40
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You have to check that the map $Ext^i(F_{y_0},G_{y_0}) \to Ext^2(F_{y_0},G_{y_0}) \otimes Ext^{2-i}(G_{y_0},G_{y_0})^*$ is an isomorphism, the map is induced by the Yoneda multiplication $Ext^i(F_{y_0},G_{y_0}) \otimes Ext^{2-i}(G_{y_0},G_{y_0}) \to Ext^2(F_{y_0},G_{y_0})$). Since $Ext^\bullet(G_{y_0},G_{y_0})$ is a Frobenius algebra (the exterior algebra on two variables), it is equivalent to $Ext^\bullet(F_{y_0},G_{y_0})$ being a free module (with respect to the action given by the Yoneda multiplication). The latter is clear since $F_{y_0}$ is a multiple of $G_{y_0}$ (ro vice versa). –  Sasha Nov 22 '10 at 12:57
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