Application of the base change theorem

Question

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 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 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.

Answer 1

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.