Let $D\subset X$ be an effective smooth divisor in a smooth projective variety $X$. Assume that $h^0(X,D)=1$. In particular $D$ spans an extremal ray of the effective cone of $X$. 


Now, let $f:X\rightarrow Y$ be a morphism. Assume that $D\_Y = f(D)\subset Y$ is a divisor. Could we say that $h^0(Y,D\_Y) = 1$ as well?



Rigid effective divisors
No. Let $X$ be the blow-up of $\mathbb P^2$ at two points, and let $f : X \to Y$ be the map down to $\mathbb P^2$. Let $D$ be the strict transform on $X$ of the line between the two points you blew up. This is a $(-1)$-curve, hence $h^0(X,D) = 1$. But $D\_Y$ is a line in $\mathbb P^2$, which has larger $h^0$.


In fact, the answer is maximally negative, in the sense that given any divisor $D\_Y \subset Y$, you can find a birational morphism $f : X \to Y$ such that the strict transform of $D\_Y$ on $X$ has $h^0(X,D) = 1$. (Just blow up lots of points on $D\_Y$.)


I might add $h^0(X,D)=1$ does not imply $D$ is extremal; you need to assume that $h^0(X,nD) = 1$ for all $n$. But this is a somewhat orthogonal question; it's not the issue in the example. (Added later: even $h^0(X,nD)=1$ doesn't mean it's extremal, but it at least tells you it's on the boundary.)


