## Return to Answer

3 cleaned up leftover line of html

The following is maybe obvious, and isn't exactly what you asked, but perhaps is still worth spelling out.

As Sándor's answer shows, if $L=O_X(-D)$ is $f$-nef (where $D$ is effective) then in fact $D$ must be pulled back from $Z$, so $L$ belongs to the subspace $f^\ast(N^1(Z))$ of $N^1(X)$. In general, for any class $f^\ast L$ in $f^\ast(N^1(Z))$ we can twist with a line bundle $f^\ast(A)$ (where $A$ is a sufficiently ample bundle on $Z$) to get something nef, because $A+L$ is ample on $Z$ for sufficiently ample, and then $f^\ast(A+L)$ is nef --- indeed, it's even semi-ample.

So we only have a problem when there are f-nef classes which are not pulled back from $Z$. In particular, we can have a problem when the subspace $K$ of f-numerically trivial classes is strictly bigger than $f^\ast(N^1(Z))$.

This is what happens in the example you cite from Lazarsfeld: there $K$ has dimension 2, whereas $f^\ast(N^1(Z))$ has dimension 1 (since $Z$ is a curve). Drawing a picture of what happens there is instructive, and explains why the f-ample and f-nef cases behave differently. The nef cone of $X$ is a 3-dimensional round cone, and $K$ is a 2-dimensional subspace of $N^1(X)$ which lies tangent to the cone. They meet in a single ray, which is exactly $f^\ast(Nef(Z))$. It's geometrically clear that starting from a point on $K$ which is not on the line $f^\ast(N^1(Z))$ and adding elements of $f^\ast(Nef(Z))$, we can never get into the nef cone; on the other hand, starting with an f-ample class, which means exactly a point in the open half-space on the same side of $K$ as the nef cone, and moving in the direction of $f^\ast(Nef(Z))$, we eventually end up in the interior of $Nef(X)$, i.e. in the ample cone.

Here is my amateurish attempt to illustrate that:

2 tried to fix the link

The following is maybe obvious, and isn't exactly what you asked, but perhaps is still worth spelling out.

As Sándor's answer shows, if $L=O_X(-D)$ is $f$-nef (where $D$ is effective) then in fact $D$ must be pulled back from $Z$, so $L$ belongs to the subspace $f^\ast(N^1(Z))$ of $N^1(X)$. In general, for any class $f^\ast L$ in $f^\ast(N^1(Z))$ we can twist with a line bundle $f^\ast(A)$ (where $A$ is a sufficiently ample bundle on $Z$ Z$) to get something nef, because$A+L$is ample on$Z$for sufficiently ample, and then$f^\ast(A+L)$is nef --- indeed, it's even semi-ample. So we only have a problem when there are f-nef classes which are not pulled back from$Z$. In particular, we can have a problem when the subspace$K$of f-numerically trivial classes is strictly bigger than$f^\ast(N^1(Z))$. This is what happens in the example you cite from Lazarsfeld: there$K$has dimension 2, whereas$f^\ast(N^1(Z))$has dimension 1 (since$Z$is a curve). Drawing a picture of what happens there is instructive, and explains why the f-ample and f-nef cases behave differently. The nef cone of$X$is a 3-dimensional round cone, and$K$is a 2-dimensional subspace of$N^1(X)$which lies tangent to the cone. They meet in a single ray, which is exactly$f^\ast(Nef(Z))$. It's geometrically clear that starting from a point on$K$which is not on the line$f^\ast(N^1(Z))$and adding elements of$f^\ast(Nef(Z))$, we can never get into the nef cone; on the other hand, starting with an f-ample class, which means exactly a point in the open half-space on the same side of$K$as the nef cone, and moving in the direction of$f^\ast(Nef(Z))$, we eventually end up in the interior of$Nef(X)$, i.e. in the ample cone. Here is my amateurish attempt to illustrate that: 1 The following is maybe obvious, and isn't exactly what you asked, but perhaps is still worth spelling out. As Sándor's answer shows, if$L=O_X(-D)$is$f$-nef (where$D$is effective) then in fact$D$must be pulled back from$Z$, so$L$belongs to the subspace$f^\ast(N^1(Z))$of$N^1(X)$. In general, for any class$f^\ast L$in$f^\ast(N^1(Z))$we can twist with a line bundle$f^\ast(A)$(where$A$is a sufficiently ample bundle on$Z$to get something nef, because$A+L$is ample on$Z$for sufficiently ample, and then$f^\ast(A+L)$is nef --- indeed, it's even semi-ample. So we only have a problem when there are f-nef classes which are not pulled back from$Z$. In particular, we can have a problem when the subspace$K$of f-numerically trivial classes is strictly bigger than$f^\ast(N^1(Z))$. This is what happens in the example you cite from Lazarsfeld: there$K$has dimension 2, whereas$f^\ast(N^1(Z))$has dimension 1 (since$Z$is a curve). Drawing a picture of what happens there is instructive, and explains why the f-ample and f-nef cases behave differently. The nef cone of$X$is a 3-dimensional round cone, and$K$is a 2-dimensional subspace of$N^1(X)$which lies tangent to the cone. They meet in a single ray, which is exactly$f^\ast(Nef(Z))$. It's geometrically clear that starting from a point on$K$which is not on the line$f^\ast(N^1(Z))$and adding elements of$f^\ast(Nef(Z))$, we can never get into the nef cone; on the other hand, starting with an f-ample class, which means exactly a point in the open half-space on the same side of$K$as the nef cone, and moving in the direction of$f^\ast(Nef(Z))$, we eventually end up in the interior of$Nef(X)\$, i.e. in the ample cone.

Here is my amateurish attempt to illustrate that: