= Solution
For a <smooth projective curve> $C$ of <geometric genus> one, the <Riemann-Roch theorem> says
$$
\ell(D)-\ell(K_C-D)=\deg D.
$$
The <canonical divisor> has degree zero and is principal because a nonzero regular differential has no zeros. Hence $K_C\sim0$, so equivalently
$$
\ell(D)-\ell(-D)=\deg D.
$$
In particular, $\ell(D)=\deg D$ when $\deg D>0$.
The group $\operatorname{Pic}^0(E)$ is the group of degree-zero <divisor class>[divisor classes] on $E$, with addition induced by addition of divisors. Consider
$$
\iota:E\longrightarrow\operatorname{Pic}^0(E),
\qquad P\longmapsto[(P)-(O_E)].
$$
For surjectivity, let $D$ have degree zero. Since $\deg(D+(O_E))=1$, Riemann--Roch gives $\ell(D+(O_E))=1$. A nonzero element of this space makes $D+(O_E)$ linearly equivalent to an effective divisor of degree one, necessarily $(P)$ for some point $P$. Thus $[D]=[(P)-(O_E)]$.
For injectivity, suppose $(P)-(O_E)$ is a <principal divisor>. If $P\ne O_E$, its defining function would be nonconstant and would have at most one simple pole, whereas Riemann--Roch gives $\ell((O_E))=1$, so every such function is constant. Therefore $P=O_E$, and $\iota$ is a bijection.
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