Let $f:X\rightarrow Y$ be a morphism of smooth projective varieties, and $NE(X),NE(Y)$ the Mori cones of curves of $X$ and $Y$. Assume that $NE(X)$ is finitely generated. Then is $NE(Y)$ finitely generated as well?


If $f$ is birational I think the answer is positive. Let $C\subset Y$ be an irreducible curve and $\Gamma$ its strict transform in $X$. Then $\Gamma\sim a\_1\Gamma\_1+\dots + a\_r\Gamma\_r$ with $a\_1,...,a\_r\geq 0$ and where $\Gamma\_1,\dots,\Gamma\_r$ are the generators of $NE(X)$. So $C\sim a\_1f\_{\*}\Gamma\_1 + \dots + a\_rf\_{\*}\Gamma\_r$ and hence $NE(Y)$ is generated by $f\_{\*}\Gamma\_1,\dots,f\_{\*}\Gamma\_r$. Is this argument correct?


If $f$ is not birational could it happen that $NE(Y)$ is not finitely generated even if $NE(X)$ is?


Thank you very much.



Mori cones and projective morphisms
It suffices to assume that $f$ is surjective (equivalently, dominant). Then for $C \subset Y$ any irreducible curve there exists an irreducible curve $D \subset X$ such that $f(D) = C$. (A schemy proof: let $D$ be the closure in $X$ of any closed point in $f^{-1}(c)$, where $c$ is the generic point of $C$.) It follows that $f$ induces a surjection from $NE(X)$ to $NE(Y)$, so if $NE(X)$ is finitely generated then so is $NE(Y)$.


The same argument also implies that if $\overline{NE(X)}$ is finitely generated then so is $\overline{NE(Y)}$.


