Kodaira dimensions of push-forward via finite map

Let $$f:X \to Y$$ be a finite map from a normal projective variety to a smooth projective variety, $$D$$ be a Cartier divisor on $$X$$. Do we have any relation between $$\kappa(X,D)$$ and $$\kappa(Y,f_*D)$$?

There is an obvious relation: the pushforward map $$f_*:|mD|\rightarrow |f_*(mD)|=|mf_*D|$$ is injective, hence $$\kappa (X,D)\leq \kappa (X,f_*D)$$. It is easy to see that you cannot get more: for instance, take for $$f$$ a general projection from a cubic surface $$X\subset \mathbb{P}^3$$ to $$\mathbb{P}^2$$, and for $$D$$ a line in $$S$$. Then $$\kappa (S,D)=0$$, but $$f_*D$$ is a line in $$\mathbb{P}^2$$, so $$\kappa (\mathbb{P}^2,f_*D)= 2$$.
• why $f_*$ is injective? Is there a reference? – Hu Zhengyu Feb 26 at 17:05
• I can imagine that $|mD| \to |mf^*f_*D|$ is injective but I am not sure if $f_*$ is injective. – Hu Zhengyu Feb 26 at 17:31
• @Hu Zhengyu: It is finite because $f$ is finite, on the other hand it is linear. – abx Feb 26 at 17:32
• I think $f_*$ is not injective. If you take $D=0$, then $f_*$ would map some nonzero rational function of $X$ to zero since $K(X)$ is obviously bigger than $K(Y)$. You just said $f$ is finite? What is the definition of finite for a linear map? – Hu Zhengyu Feb 27 at 3:01