Let $\pi: X \to Y$ be a morphism between projective varieties and $H$ be an ample divisor on $Y$. I want to "show" $\pi^*H$ is ample on $X$ by the following argument:
Let $N_1(X)_\mathbb{R},N_1(Y)_\mathbb{R}$ be vector spaces of 1-cycles module numerical equivalence, and let $NE(X),NE(Y)$ be the cones of effective 1-cycles in $N_1(X)_\mathbb{R},N_1(Y)_\mathbb{R}$ with $\overline{NE(X)},\overline{NE(Y)}$ being their closures.
Then we have a continuous map
$\pi_*: N_1(X)_\mathbb{R} \to N_1(Y)_\mathbb{R}$ defined by $[C] \mapsto [\pi_* C]$.
Where if $C$ is an irreducible curve, then $\pi_* C$ is defined to be $\deg(\pi)\pi(C)$ when $\pi$ does not contract $C$, and $0$ otherwise.
$\pi_*$ induces a continuous map
$\pi_*: \overline{NE(X)} \to \overline{NE(Y)}$.
Kleiman's criterion shows: $H $ is ample $\iff$ $\forall ~C' \in \overline{NE(Y)}, H\cdot C'>0$ .
Let $C \in \overline{NE(X)}$, and $C_i \to C$. Then by projection formula, we have
$\pi^* H \cdot C_i = H \cdot \pi_* C_i > 0$, and hence $\pi^* H \cdot C>0$.
By Kleiman's criterion again, $\pi^*H$ is an ample divisor.
I know ampleness does not behave quite well under pull back, one need some extra condition (e.g. finite morphism) to make sure $\pi^*H$ preserve ampleness. There must be something wrong in the above argument, and I will be very appreciated if someone can point it to me!!
