 Solution

If $L\in {\mathrm{Pic}}^0(X)$, part (iii) makes $\Lambda (L)$ trivial. Pulling it back along $(f,g):Y\to X\times X$ givesTaking $Y=X$, $f={\mathrm{id}}_X$, and $g=i$, where $i$ is inversion, givesInduction with $f=[n]$ and $g={\mathrm{id}}_X$ proves $[n{]}^{\ast }L\simeq L^{\otimes n}$ for $n\ge 0$; combining this with inversion proves

Conversely, suppose $i^{\ast }L\simeq L^{\vee }$. For $M={\varphi }_L(x)$, part (ii) gives $M\in {\mathrm{Pic}}^0(X)$, so the result just proved yields $i^{\ast }M\simeq M^{\vee }$. On the other hand,whereas $i^{\ast }L\simeq L^{\vee }$ gives ${\varphi }_{i^{\ast }L}(x)={\varphi }_{L^{\vee }}(x)=M^{\vee }$. Hence $M^{\otimes 2}$ is trivial for every $x$, so ${\varphi }_{L^{\otimes 2}}$ is trivial and $L^{\otimes 2}\in {\mathrm{Pic}}^0(X)$. The torsion-freeness proved in part (ii) now implies