The theorem of the square and the Mumford homomorphism into the Picard group

Pull the identity of [F1] back along $A_{k^{\mathrm{\prime }\mathrm{\prime }}}\to (A_{k^{\mathrm{\prime }\mathrm{\prime }}}{)}^3$, $x\mapsto (x,a,b)$. Its factors involving $x$ are $t_{a+b}^{\ast }\mathcal{L}$, $\mathcal{L}$, $t_a^{\ast }\mathcal{L}$ and $t_b^{\ast }\mathcal{L}$. The remaining factors are the constant line bundles with fibres ${\mathcal{L}}_a$, ${\mathcal{L}}_b$ and ${\mathcal{L}}_{a+b}$, each a one-dimensional $k^{\mathrm{\prime }\mathrm{\prime }}$-vector space and therefore isomorphic to the trivial line bundle. Removing these constant factors gives $t_{a+b}^{\ast }\mathcal{L}\otimes \mathcal{L}\cong t_a^{\ast }\mathcal{L}\otimes t_b^{\ast }\mathcal{L}$. Although trivializations of the constant factors need not be canonical, the resulting equality of Picard classes is canonical and is preserved by field extension.

The square identity shows that ${\phi }_{\mathcal{L}}(a+b)=[t_{a+b}^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{-1}]=[t_a^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{-1}][t_b^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{-1}]={\phi }_{\mathcal{L}}(a){\phi }_{\mathcal{L}}(b)$ in $\mathrm{Pic}(A_{k^{\mathrm{\prime }\mathrm{\prime }}})$: expand the first factor using the square identity and cancel $\mathcal{L}\otimes {\mathcal{L}}^{-1}$. Hence ${\phi }_{\mathcal{L}}$ is a group homomorphism, and it is natural in $k^{\mathrm{\prime }\mathrm{\prime }}$ because the constructions and $\mathcal{L}$ are defined over $k$.

For degree zero: ${\phi }_{\mathcal{L}}(a)$ is represented by the difference of the two line bundles $t_a^{\ast }\mathcal{L}$ and $\mathcal{L}$, which occur as fibres of the connected family $\mathcal{L}$ over $A$ under the translation family; hence its geometric-fibre restrictions are algebraically equivalent to zero, and ${\phi }_{\mathcal{L}}$ lands in the algebraically trivial subfunctor of [F2], i.e. in the degree-zero part of the rigidified Picard functor.