expensive, found a demo for this question.

Set $m: A \times A \rightarrow A$ be the group law over $A$, $p_{12}: \textrm{Pic}^0(A) \times A \times A \rightarrow \textrm{Pic}^0(A) \times A$ be the projection over the first and second factors, $p_{13}:\textrm{Pic}^0(A) \times A \times A \rightarrow \textrm{Pic}^0(A) \times A$ be the projection over the first and third factors and $L$ be the Poincaré bundle over $A \times \textrm{Pic}^0(A)$ trivialized in neutral element $e$ of $A$, this is, $L \mid _{ \{e\} \times \textrm{Pic}^0(A)}$ is trivial.

Now, define $ (\textrm{Id},m):\textrm{Pic}^0(A) \times A \times A \rightarrow \textrm{Pic}^0(A) \times A$, where Id is the identity map, and consider $$M:= (\textrm{Id},m)^*L \otimes P_{12}^*L^{-1} \otimes p_{13}^*L^{-1}$$ line bundle over $\textrm{Pic}^0(A) \times A \times A $. Then, since $M \mid_{\{\mathscr{O}_{\textrm{Pic}^0(A)}\} \times A \times A}, \ M \mid _{ \textrm{Pic}^0(A) \times \{e\} \times A} \ \textrm{and} \ M \mid _{ \textrm{Pic}^0(A) \times A \times \{e\}}$ are trivial, follow of the Cube theorem that $M$ is trivial. Hence, $M \mid _{\{\mathscr{N}\} \times \{x\} \times A} \cong t^*_x\mathscr{N} \otimes \mathscr{N}^{-1}$ is trivial for any $\mathscr{N} \in \textrm{Pic}^0(A)$ and $x \in A$, this is, $\textrm{Pic}^0(A) \subseteq \Sigma :=\{ \mathscr{L} \in \textrm{Pic}(A) \mid t_x^* \mathscr{L} \cong \mathscr{L} \forall \ x \in X\}.$ So, since $\Sigma$ is an Abelian variety, see [Mumford section 8], and $\textrm{Pic}^0(A)$ is open and closed in $\textrm{Pic}(A)$, see [FAG proposition 0.5.3], follow that $ \textrm{Pic}^0(A)=\Sigma$.

defines$\textrm{Pic}^0$ in this way: see Chapter II, Section 8 $\endgroup$ – Francesco Polizzi May 22 '11 at 8:46