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Let $C$ be a smooth connected ample curve in an abelian surface $A$. Translate its chosen basepoint to zero and apply the universal property of the Jacobian variety. The image of $J(C)\to A$ is a closed connected algebraic subgroup containing the translated curve. If it had dimension one, that curve would be the subgroup itself; a distinct coset is disjoint and numerically equivalent, forcing $C^2=0$. This contradicts ampleness, so the homomorphism is surjective.

Choose a projective embedding of the abelian surface $X$. By Bertini's theorem, a general hyperplane section $C$ is a smooth projective curve that is connected. It is ample, so $C^2>0$. Choose $p\in C$ and translate the inclusion so it sends $p$ to the identity of $X$. The universal property of the Jacobian variety then gives a homomorphism of abelian varietieswhose composition with the based Abel map of an algebraic curve is this translated inclusion.

The image is a closed connected algebraic subgroup, since $J(C)$ is projective, and it contains a translate of $C$. If the image had dimension one, that irreducible curve would be the image subgroup itself. A distinct coset is disjoint from it and has the same numerical divisor class: translation acts trivially on cohomology, being homotopic to the identity on the complex torus. Its self-intersection would therefore be zero, contradicting $C^2>0$. The image has dimension two and is all of $X$. Thus $\Phi$ is surjective, as expressed by the principle that an ample curve on an abelian surface generates the surface.

A minimal algebraic surface is a smooth projective surface containing no exceptional curve of the first kind, namely no smooth rational curve of self-intersection $-1$. An abelian surface contains no rational curve because every morphism from ${\mathbb{P}}^1$ to an abelian variety is constant. It therefore has no $(-1)$-curve and is minimal.

For every $n\ge 2$, the Minimal Hirzebruch surface ${\mathbb{F}}_n$ is a rational minimal surface. These surfaces are pairwise nonisomorphic: the negative section is the unique irreducible curve of negative self-intersection and has square $-n$, so an isomorphism would recover $n$. Thus there are infinitely many nonisomorphic minimal rational surfaces.

A K3 surface is a smooth projective surface $X$ with $K_X\cong {\mathcal{O}}_X$ and $H^1(X,{\mathcal{O}}_X)=0$. For a smooth curve $C\subset X$ of geometric genus $g$, the adjunction formula givesand hence

An elliptic surface is a smooth projective surface with a morphism to a smooth curve whose generic fiber is a smooth genus-one curve. If $E$ is an elliptic curve, projectionis an elliptic fibration. Its canonical bundle is pulled back from $K_{{\mathbb{P}}^1}$, so every positive pluricanonical space vanishes and the Kodaira dimension is $-\infty$. This supplies the requested negative-Kodaira-dimension example.

For an elliptically fibered K3 surface, choose a smooth quartic $X\subseteq {\mathbb{P}}^3$ containing a line $L$. The canonical bundle of a smooth projective hypersurface formula makes $K_X$ trivial, and the standard cohomology sequence gives $H^1(X,{\mathcal{O}}_X)=0$, so $X$ is K3. The pencil of planes through $L$ cuts $X$ into $L$ plus a residual plane cubic. The residual linear system $|H-L|$ is basepoint-free, has square zero and defines a morphism $X\to {\mathbb{P}}^1$ whose generic fiber is a smooth plane cubic. This is the Elliptic K3 surface from a quartic containing a line.

A surface of general type is a smooth projective surface of Kodaira dimension two. Let $B\subseteq {\mathbb{P}}^2$ be a smooth plane curve of degree eight and letbe the degree-two cover branched along $B$. The branch-cover canonical-bundle formula givesThis divisor is ample, so $X$ is a double plane of general type and $\pi$ is the required finite morphism of degree two.