Map of flat sheaves is onto

Question

Suppose we have a map of coherent sheaves $f:F\rightarrow G$ on $X\times S$ (where $X$ and $S$ are schemes over some base field). Assume that both $F$ and $G$ are flat over $S$. If the induced map $f|_{X\times\{s\}}$ is surjective for all $s$, can we deduce that $f$ itself is surjective? This would be true if the cokernel of $f$ is also flat over $S$. It looks as though this comes down to relating (in general) the stalk $F_{(x,s)}$ to $(F|_{X\times\{s\}})_x$.

Answer 1

Let $K$ be the cokernel of $f$, we want to show that $K=0$. The sequence $F \rightarrow G \rightarrow K \rightarrow 0$ of sheaves is exact, thus so is the sequence of the restrictions to any $X \times \{s\}$. By the assumption, we know that for any $s \in S$, $K _{|X \times \{s\}}=0$.

Let $U \subset X,V \subset S$ be affine open subsets $A=\mathcal{O}_X(U),B=\mathcal{O}_S(V)$. $K_1=K(U\times V)$ which is an $A \otimes B$-module, so that $K_{|U \times V}=\tilde{K_1}$. We know that for any $s \in V$, $(\tilde{K_1})_{|U \times \{s\}}=K_{|U \times \{s\}}=(K_{|X \times\{s\}})_{|U \times \{s\}}=0$.

In particular, this means that for any prime ideal $\mathfrak{p}$ of $B$, $0=K_1 \otimes_{A \otimes B}(A \otimes \kappa(\mathfrak{p}))$. In particular, for any $x \in U \times V$, $K_1 \otimes_{A \otimes B}\kappa(x)=0$.

So $K_1$ is a coherent $C$-module of finite type with $C=A\otimes B$ an algebra over a field, and such that, for each $\mathfrak{p}$ prime ideal of $C$, $K_1 \otimes \kappa(\mathfrak{p})=0$.

In particular, for any prime ideal $\mathfrak{p}$ and any $x \in K_1$, there exists $s \in C \backslash \mathfrak{p}$ such that $sx\in \mathfrak{p}K_1$. In particular, $(K_1)_{\mathfrak{p}}=\mathfrak{p}(K_1)_{\mathfrak{p}}$, and thus by Nakayama $(K_1)_{\mathfrak{p}}=0$ (for any $\mathfrak{p}$ prime ideal of $C$). In other words, all the stalks of $K_{|U \times V}$ are zero so $K_{|U \times V}=0$.

So $K=0$ and $f$ is surjective.