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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.