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A smooth projective curve $C\subset A$ generates $A$ by its point differences if and only if it meets every prime Weil divisor. If a prime Weil divisor avoids $C$, its restriction to $C$ has degree zero; constancy of line bundle degree in a family shows that every translate either contains $C$ or misses it. This forces invariance under $C-C$ and puts the generated subgroup inside the line bundle translation stabilizer. If this subgroup were all of $A$, the divisor would be algebraically trivial, contradicting its positive intersection with an ample line bundle. Conversely a proper closed generated subgroup admits a disjoint divisor by the pole divisor avoiding a fiber construction, after translating the fiber to contain $C$.

For the curve generation criterion for an abelian variety, first suppose $Y=X$ and an irreducible Weil divisor $D$ avoids $C$. The preceding part gives $K({\mathcal{O}}_X(D))=X$, so ${\varphi }_{{\mathcal{O}}_X(D)}=0$ and ${\mathcal{O}}_X(D)\in {\mathrm{Pic}}^0(X)$. A nonzero effective divisor cannot have this property: if $H$ is an ample line bundle and $g=\dim X$, then the intersection product of Cartier divisors $D\cdot H^{g-1}$ is positive, whereas an algebraically trivial line bundle has zero intersection with every curve. The latter follows from constancy of degree along a connected family defining algebraic equivalence; the former is the positive projective degree of $D$ after replacing $H$ by a very ample power. This contradiction proves that $C$ meets every irreducible Weil divisor.

Conversely suppose $Y\ne X$. Fix $c_0\in C$, so $C\subseteq c_0+Y$. Use the permitted fiber theorem to choose a morphism of varieties $f:X\to Z$ with $f^{-1}(z_0)=Y$. This morphism is nonconstant because $Y$ is proper. Choose an affine neighborhood $U$ of $z_0$. The irreducible image $f(X)$ has positive dimension, so its intersection with $U$ does too; some regular function $h$ on $U$ is nonconstant on this intersection. Then $F=h\circ f$ is a nonconstant rational function on $X$, regular on $f^{-1}(U)$.

The pole divisor avoiding a fiber construction applies: its pole Weil divisor is nonzero. Indeed on the smooth, hence normal, projective variety $X$, a rational function with no codimension-one poles extends to a global regular function; every global regular function on a connected projective variety is constant. Every pole component $D_0$ lies outside $f^{-1}(U)$ and hence avoids $Y$. Therefore $D_0+c_0$ is an irreducible Weil divisor disjoint from $c_0+Y$, and in particular from $C$. This proves the converse using precisely the fiber fact allowed in the question, without requiring a projective target $Z$. The criterion is