I am trying to prove the following Bertini type theorem:


Given a non-constant morphism $f:X \rightarrow C$, where $X \subset \mathbb{P}^n$ is a smooth irreducible variety and $C$ is a smooth curve, then the set of hyperplanes $H \subset \mathbb{P}^n$ such that $f|\_{X \cap H}$ is still non-constant, is zariski open in $(\mathbb{P}^n)^\*=$ the set of hyperplanes in $\mathbb{P}^n$.


The statement seems natural and obvious but I'm stuck. Any suggestion?



Bertini type theorem
First observe that we may assume that $X$ is projective: Let $\bar X$ be the closure of $X$ in $\mathbb P^n$. Since $f$ maps to a curve, it extends to $\bar X$, call that $\bar f$ (otherwise one could resolve the indeterminacies) and $X\cap H$ is dense in $\bar X\cap H$ for a general $H$, so if $\bar f$ is non-constant, so is $f$. Se we'll assume that $X$ is projective.


Next, notice that $f|\_{X\cap H}$ is constant if and only if $X\cap H$ consists of only a fiber of $f$. However, $X\cap H$ is very ample. If $\dim X>1$, this means immediately that it has to intersect all fibers, so it cannot consist of a single fiber (in other words, in this case a fiber is not even ample). If $\dim X=1$, then the divisors corresponding to the fibers form a $1$-dimensional subset of the corresponding linear system, but unless $X=\mathbb P^1=\mathbb P^n$, that linear system has to be larger. Consequently a general $H$ will intersect $X$ in something different than a fiber, in other words $f|\_{X\cap H}$ is not constant. (Of course, this also means that there is a trivial counter-example to your statement: If $X=\mathbb P^1=\mathbb P^n$, then $X\cap H$ is a single point for any $H$, so $f|\_{X\cap H}$ is constant. However, the above shows that that's the only way it fails and obviously you would want to exclude that case anyway.)


