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.

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.

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.

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.