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If we have a surjective morphism $f：X\to Y$, where $X$ is $n$ dimensional projective variety and $Y$ is $m$ dimensional projective variety.

If $m<n$, Can we choose a general hyperplane section $H$ of $X$, such that $\text{dim} f(H)=m$?（or can we take $H$ such that $f(H)=Y$?)

If $m=n$, Can we choose a general hyperplane section $H$ of $X$, such that $\text{dim} f(H)=n-1$?

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$m<n$. Assuming $X$ irreducible (otherwise, there is a trivial counter-example), the answer is kind of obvious. There exist an open $\varnothing\ne U\subset Y$ such that $\dim f^{-1}(u)=n-m$ for all $u\in U$. Pick a smooth (i.e., generic) point $p\in f^{-1}(u)$ for a generic $u\in U$ and a generic hyperplane $H\ni p$ transversal to $f^{-1}(u)$. So, $\dim_p(H\cap f^{-1}(u))=n-m-1$. Since a fibre of $f|_{X\cap H}$ has dimension $n-m-1$ and $\dim(X\cap H)=n-1$, the image of $f|_{X\cap H}$ has dimension $\ge(n-1)-(n-m-1)=m$.

$n=m$. Any $H$ intersecting $f^{-1}(U)$ will provide $\dim f(X\cap H)=n-1$ because the morphism $f^{-1}(U)\to U$ is finite.

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  • $\begingroup$ Thanks，but there is still a point unclear to me: suppose $X\subset \mathbb{P}(V)$, for each $u$, we can choose a general $H$ such that $\text{dim}{H\cap f^{-1}(u)}=n-m-1$, that means: $\text{dim}{H\cap f^{-1}(u)}=n-m-1$ holds for $H$ in a Zariski dense subset $U_u\subset \mathbb{P}（V^*)$.But now we hope $f|_{X\cap_H}$ have fiber of dimension $n-m-1$ on a open set of $T=f(X\cap H)$, so we ask $H \in \cap_{u\in T}{U_u}$, why does this intersection contain a open subset set?（As we hope to prove for a general $H$, \text{dim}(H\cap X)=m） $\endgroup$
    – user39380
    Commented Jun 2, 2014 at 13:24

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