Part iii already proves injectivity. Pullback along $i_j$ sends translation-invariant line bundles to translation-invariant line bundles, so it defines
$$
\rho:\operatorname{Pic}^0(X_1\times X_2)\to
\operatorname{Pic}^0X_1\times\operatorname{Pic}^0X_2.
$$
Clearly $\rho\tau=1$.

For $L\in\operatorname{Pic}^0(X_1\times X_2)$, put $L_j=i_j^*L$ and
$$
M=L\otimes\operatorname{pr}_1^*L_1^{-1}\otimes\operatorname{pr}_2^*L_2^{-1}.
$$
Then $M$ is trivial on both coordinate axes. Since $L$ and the two correcting factors lie in <Identity component of the Picard group>[$\operatorname{Pic}^0$], translation by $(x_1,e_2)$ leaves $M$ invariant. Hence every restriction $M|_{\{x_1\}\times X_2}$ is trivial. The <Seesaw theorem> and triviality on $X_1\times\{e_2\}$ imply $M\cong\mathcal O_X$. Thus $L=\tau(L_1,L_2)$, so
$$
\operatorname{Pic}^0X_1\times\operatorname{Pic}^0X_2
\xrightarrow{\sim}\operatorname{Pic}^0(X_1\times X_2).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 126 / 4 / iv / Solution

{c}
{wiki=Theorem_of_the_cube#The_see-saw_principle}

The seesaw theorem determines a line bundle on a product from its restrictions to fibers: if it is trivial on all fibers over one factor and on one transverse section, then it is trivial.


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 Incoming links: Seesaw theorem