Here $\kappa(X)$ denotes the Kodaira dimension of a smooth projective variety $X$.

**Question 1:**

I would like to solve Exercise VII.7 (3) from the Beauville book "Complex Algebraic Surfaces":

Let $\pi \colon X\to Y$ be a surjective morphism, then $\kappa(X)\geq \kappa(Y)$. Moreover, if $\pi$ is étale, then the equality holds.

I have already looked around and I have found these two questions that could be related to my problem

Iitaka dimension is invariant under surjective morphism between smooth projective varieties

Rational maps and Kodaira dimension

However, I don't understand so much what they mean, so I decided to ask once more here.

My idea is simply the following:

By the Hurwitz formula, we have $dK_X=\pi^*(dK_Y)+D$ for any $d\geq 1$, where $D$ is an effective divisor (in particular, $D$ is $d$-times the ramification divisor of $\pi$).

Therefore, the pullback $\pi^*\colon H^0(Y, dK_Y)\hookrightarrow H^0(X, \pi_1^*(dK_Y))\subseteq H^0(X, dK_X)$ is injective. In other words, $h^0(dK_X)\geq h^0(dK_Y)$. Define $k:=\kappa(S)$; if $k\leq 0$, then the thesis follows immediately by the definition of the Kodaira dimension. Otherwise, we would get $$ \limsup_{d\to \infty}\frac{h^0(dK_X)}{d^k}\geq \limsup_{d\to \infty}\frac{h^0(dK_Y)}{d^k}, $$ hence $k(Y)\leq k=k(X)$. Suppose now that $\pi$ is étale. This means for me just that $f$ is unramified, and so basically $D=0$. How can I conclude from here that $\kappa(X)=\kappa(Y)$?

Is maybe true that $h^0(dK_X)=h^0(dK_Y)$ for each $d\geq 1$?

What I have thought is that if $D=0$, then there exists at each point an open neighborhood $U$ (not necessarily Zariski open) such that $f\colon U\to f(U)=:V $ is an isomorphism, and therefore $\pi^*\colon H^0(V, dK_V)\to H^0(U, dK_U)$ is an isomorphism too. How could I conclude from here that then $\pi^*\colon H^0(Y, dK_Y)\to H^0(X, dK_X)$ is an isomorphism?

**Question 2:**

The second question is to generalize the previous result as follows:

Let $\pi\colon X\to Y$ be a surjective morphism of algebraic varieties, with $Y$ and $X$ not necessarily smooth. Then $\kappa(X)\geq \kappa(Y)$.

I think I have found a solution in the case of surfaces, but I don't know which should be the strategy in the higher dimension:

Let $\rho_X\colon \widehat{X}\to X$ and $\rho_Y\colon \widehat{Y}\to Y$ be two resolutions of the singularities of $X$ and $Y$. Consider the natural rational map $\widehat{X}\dashrightarrow \widehat{Y}$ and resolve its indeterminacy by a finite number of blow-ups $b\colon \smash{\widehat{X}}'\to \widehat{X}$. We have therefore a morphism $\pi\colon \smash{\widehat{X}}'\to \widehat{Y}$, which is surjective since $f$ is surjective too. Apply now the previous result to the map $\pi$ to get $\kappa(\smash{\widehat{X}}')\geq \kappa(\widehat{Y})$. However, the Kodaira dimension is a birational invariant, so

$$\kappa(X)=\kappa(\widehat{X})=\kappa(\smash{\widehat{X}}')\geq \kappa(\widehat{Y})=\kappa(Y).$$

Is it possible to generalize this? If not, which is the idea to prove it in a higher dimension?