= Solution
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 varieties>
$$
\Phi:J(C)\longrightarrow X
$$
whose 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 \b[$\Phi$ is surjective], as expressed by the principle that an <ample curve on an abelian surface generates the surface>.
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