For a finite type morphism $f:X\to S$, $X$ is a regular scheme, should there always exist an open (dense) subscheme $U\subset S$ such that the fibre of $f$ at each Zariski point of $U$ is regular? All schemes are excellent.

If the anwser is 'yes', then: could one choose such an $U$ such that the preimage of any regular subscheme of $U$ is regular? Are these conditions on $U$ equivalent?

For a finite type morphism $f:X\to S$, $X$ is a regular scheme, should there always exist an open (dense) subscheme $U\subset S$ such that the fibre of $f$ at each Zariski point of $U$ is regular? All schemes are excellent.

If the answer is 'yes', then: could one choose such an $U$ such that the preimage of any regular subscheme of $U$ is regular? Are these conditions on $U$ equivalent?

For a finite type morphism $f:X\to S$, $X$ is a regular scheme, should there always exist an open (dense) subscheme $U\subset S$ such that the fibre of $f$ at each Zariski point of $U$ is regular? All schemes are excellent.

For a finite type morphism $f:X\to S$, $X$ is a regular scheme, should there always exist an open (dense) subscheme $U\subset S$ such that the fibre of $f$ at each Zariski point of $U$ is regular? All schemes are excellent.

If the anwser is 'yes', then: could one choose such an $U$ such that the preimage of any regular subscheme of $U$ is regular? Are these conditions on $U$ equivalent?