# why is an open faithfully-flat morphism fpqc?

Why is an open faithfully-flat morphism fpqc?

In other words, why must an open faithfully flat morphism $X\rightarrow Y$ have the property that around every $x\in X$, there is an open nbhd $U$ of $x\in X$ such that $f(U)$ is open, and the restriction of $f$ to $U\rightarrow f(U)$ is quasi-compact?

thanks,

• will
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See mathoverflow.net/questions/39211, there Qing Liu has answered this question. –  Martin Brandenburg Sep 15 '13 at 8:15