# Stronger version of Bertini's theorem

In char 0, is there a generalised version of Bertini's theorem that will ensure that for a proper map $f: Y\rightarrow X$ between smooth projective varieties and for every point $x\in X$ we can find an open set $U \ni x$ (maybe analytic) around $x$, such that a general divisor in a base point free linear system $|L|$ on $Y$ will intersect every fibre over a point in $U$ transversally. Assume in addition that $|L|$ contains horizontal divisors.

Here, I would like to think of transversal intersection with a singular fibre as that the tangent spaces at each point in the intersection add up to give tangent space of $Y$ at that point.

However, I would like it even if a general divisor $D\in |L|$ intersect only the smooth fibres close to the fibre over $x$ transversally. The fibre over $x$ may itself be singluar.

• I am a little puzzled. Consider the map f as defined by the linear system |L| which is say a pencil, so that X is P^1 and the fibers are the divisors in |L|. Then it looks as if the divisors do not intersect the fibers, i.e. the divisors, transversely in your sense. That is they are either disjoint or coincide. Am I missing something obvious? Maybe the quantifiers needed in your question are not quite present. – roy smith May 20 '16 at 5:29
• Yes you're absolutely right. I should add the additional assumption that the linear system does contain horizontal divisors. Thank you for pointing it out. I have edited the question. – yagna May 20 '16 at 5:33