Is surjectivity preserved in open neighborhoods? Let $X,S$ be schemes of finite type over a field and let $f:X\times S\to S$ be the projection. Suppose we have a morphism of coherent sheaves $\phi:\mathscr E\to \mathscr F$ on $X\times S$.

Is it true that if the restriction $\phi|_{X\times\{s\}}$ to the fiber
  over $s$ is surjective (or injective), then there is an open
  neighborhood $U\subset S$ of $s$ such that $\phi|_{X\times U}$ is
  surjective (or injective)?

A similar statement is true for stalks: for a locally noetherian scheme $S$, if a morphism of coherent $\mathscr O_S$-modules is such that $\phi_s:\mathscr E_s\to \mathscr F_s$ is surjective at $s$, then it is surjective in a neighborhood of $s$. 
Thanks for any help!
 A: It's simpler to consider the support of the kernel or cokernel sheaf, call it $\mathscr G$ -- we wish to show that it vanishes appropriately.
Edit: the original argument I posted was incorrect (I'll leave it for posterity; see below and in the comments...). Here is a simpler argument:
The statement is false in general without some kind of properness assumption. For example, if $X = S = \mathbb{A}^1$, then the structure sheaf of the hyperbola $\{xs = 1\}$ is zero on the fiber $s = 0$ but is nonzero on every other fiber.
However, if $X$ is complete, then $f : X \times S \to S$ is proper (more generally, if $f : \mathcal{X} \to S$ is any proper map), then we can make the following argument: let $Z$ be the support locus of $\mathscr G$. Then $Z$ is closed because $\mathscr G$ is coherent. By properness, $f(Z)$ is closed in $S$, and by hypothesis $s \notin f(Z)$. Consequently, $U = S - f(Z)$ is an open neighborhood of $s$ and $\mathscr G|_{f^{-1}(U)} = 0$.
(Original argument:)
We can assume $S$ is affine; and we can actually just have $\mathcal{X} \to S$ be any finite type $S$-scheme.
Now the statement is: if a sheaf vanishes on the fiber over $s \in S$, we wish to show that it vanishes on the preimage of a neighborhood of $s$. At this point we can also assume $X$ is affine (reason: cover with finitely-many affine opens, solve the problem there, and intersect the corresponding opens of $S$...)
Now we're reduced to an easy commutative algebra statement! For $B$ a finitely-generated $A$-algebra and $M$ a finite $B$-module, if $M \otimes A_P = 0$, then there exists $f\notin P$ ($f \in A$) such that $M \otimes A[1/f] = 0$. I leave this part to you.
