= Solution
Here the <degree-zero Picard group of a curve> uses tensor product as its group operation. The map is the <Abel map of a pointed smooth projective curve>. If $\alpha(p)=\alpha(q)$, then $\mathcal O(p-q)$ is trivial. By the divisor description of <line bundles>, there is a nonzero <rational function> $f$ with <principal divisor> $(f)=p-q$.
If $p\ne q$, this function has exactly one pole, of order one. It defines a nonconstant <morphism of algebraic varieties> $X\to\mathbb P^1$: locally at a pole one uses $1/f$ as the target coordinate. The <degree of a morphism of curves> is the total pole degree, so it is one. The map is finite, since a nonconstant map of projective integral curves has finite fibers and is proper, and a degree-one finite map to the normal <projective line> is an isomorphism. Algebraically, on a target affine chart the finite coordinate algebra lies inside the common function field and is integral over the integrally closed coordinate ring, so it equals that ring. Thus $X\cong\mathbb P^1$ and its genus is zero, contradicting $g\ge1$. Consequently
$$
\boxed{g\ge1\Longrightarrow\alpha\text{ is injective}.}
$$
This spells out why a <principal divisor with one simple zero and one simple pole> cannot occur on a curve of positive genus.
If $g=0$, the chosen point $p_0$ is enough to identify $X$ with $\mathbb P^1$, even over a nonclosed field. By the <Riemann-Roch theorem>, $h^0(\mathcal O(p_0))=2$, since the complementary canonical twist has degree $-3$ and no nonzero section. Besides constants there is a <rational function> with its sole pole a simple pole at $p_0$, giving the same degree-one isomorphism. On $\mathbb P^1$, every degree-zero <divisor> is principal: over an algebraically closed field write a product of powers of linear forms for its finite point coefficients, with the degree-zero condition giving the coefficient at infinity. Over a general field the same construction uses irreducible polynomial factors for closed points, weighted by their degrees. Therefore
$$
\boxed{\operatorname{Pic}^0(X)=0\qquad(g=0).}
$$
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