Neighborhood of fiber contains the full pre-image of an open set Let  $ f :  X  \to   Y    $  be a morphism of schemes $ X $ and $ Y $. Suppose that $ f $ is quasi-compact. Is it true that  for any $ y \in Y $ and any open subset $ U $ of $ X $ which contains $ f^{-1} ( y ) $,  there is an open subset $ V $ of $ Y $ containing $y$ such that $  f^{-1} ( V )   \subset  U   $?  
Remark. I have often run into situations where I have to deal with properties of schemes which are local on target and such a result would be helpful in those situations. I don't think it's true in the stated form, but it is true if $ U  $  also contains the  pre-image of all generizations of $ y $ e.g. when $ y $ is a generic point of $ Y $.        

No.  For instance, working over your favorite field, let $X=\mathbb{A}^2\setminus\{(0, 0)\}$ and let $f:X\to\mathbb{A}^1$ be the first projection.  Then the open set $U=\mathbb{A}^1\times(\mathbb{A}^1\setminus\{0\})\subset X$ contains the entire fiber of $f$ over $0$, but does not contain the preimage of any neighborhood of $0$.
