For a surjective morphism $f \colon X \to Z$ of smooth projective varieties, a line bundle $L$ on $X$ is $f$-nef if is nef on the fibers of $f$. More precisely, for every curve $C$ that is mapped to a point, the degree of $L$ restricted to $C$ should be larger than zero. 


Now it is not true that an $f$-nef bundle becomes nef if I tensor it with a sufficiently large power of an ample line bundle on $Z$. There is a counterexample in the Positivity book of Lazarsfeld, Example 1.7.12.


Are there any conditions I can put on $L$ such that this statement becomes true? I'm interested in the case where $L = \mathcal{O}\_X (-D)$ for an effective divisor D.



When does f-nef imply nef (after twisting?)
Let $f=\alpha\circ g$ the Stein factorization of $f$ with $g:X\to W$ having connected fibers and $\alpha:W\to Z$ finite.


Let $x\in {\rm supp}\, D\subset X$ and $w=f(x)$ and let $C\subseteq X\_w=g^{-1}(w)$ such that $x\in C$. Since $-D\cdot C\geq 0$ this means that $C\subseteq {\rm supp}\, D$. Therefore, ${\rm supp}\, D$ is a union of $g$-fibers. In particular $g({\rm supp}\, D)$ is *not* dense in $W$ and hence $f({\rm supp}\, D)$ is not dense in $Z$. Now let $B$ be an effective reduced Weil divisor on $Z$ containing $f({\rm supp}\, D)$. Further let $A$ be a sufficiently very ample divisor on $Z$ such that $A-B$ is effective.


Now $f^\*(mA)-D$ for $m\gg 0$ can be written as the pull-back of a very ample divisor from $Z$ plus an effective divisor supported entirely on fibers of $f$. This implies that it will have non-negative intersection with any curve that is not contained in a fiber. On the other hand $f^\*({\rm anything})$ is trivial on fibers, so the intersection of $f^\*(mA)-D$ with any curve that is contained in a fiber is the same as the intersection of $-D$ with that curve and hence non-negative. This shows that $f^\*(mA)-D$ is indeed nef.


**EDIT:**
1) added applying Stein factorization first to account for the possibility that the fibers of $f$ are not necessarily connected.
2) Corrected the statement that $f({\rm supp}\, D)$ is a divisor to that it is *contained* in a divisor. 3) Thanks for the constructive comments, especially Artie!


