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 $0$.