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Let $m:B\times B\to B$ be addition, and let $p_1,p_2$ be the projections. Consider the line bundleIts restriction to $B\times \{y\}$ has Picard group class $[T_y^{\ast }L\otimes L^{-1}]={\varphi }_L(y)$; the factor $L_y^{-1}$ is a constant line and has trivial class. Its restriction to $\{0\}\times B$ is also trivial. Choose a trivialization of the fiber $L_0$ and use it to normalize along the zero sections. The defining universal property of the normalized Poincaré line bundle then identifiesMore explicitly, both sides have the same restrictions to all fibers of the second projection, and both are normalized on $\{0\}\times B$; the Seesaw theorem makes their quotient trivial. Here ${\varphi }_L:B\to \hat{B}$ is the homomorphism associated to a line bundle on an abelian variety, and $\hat{B}$ is the dual abelian variety. Pull back by $(f,g)$ to obtain the required identityThe choice of fiber trivialization is immaterial to this identity in the Picard group.

For a line bundle $\mathcal{L}$ on $X$, define the homomorphism associated to a line bundle on an abelian varietyThe translation identity from part (i) givesso ${\varphi }_{\mathcal{L}}$ is a homomorphism.

Suppose $\mathcal{M}\simeq T_y^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{\vee }$ lies in its image. Thenwhich is trivial by the same translation identity. Hence ${\varphi }_{\mathcal{M}}=0$; the image of every ${\varphi }_{\mathcal{L}}$ lies in the Identity component of the Picard group ${\mathrm{Pic}}^{0}X$.

For a line bundle $L$ on $X$, define the homomorphism associated to a line bundle on an abelian varietyThe Theorem of the square givesso ${\varphi }_L$ is a homomorphism. Pullback distributes over the tensor product of sheaves, and thereforeIterating the homomorphism law in $x$ gives

Suppose $L^{\otimes n}\in {\mathrm{Pic}}^0(X)$. Then ${\varphi }_L(nx)$ is trivial for every $x$. The multiplication-by-n morphism on an abelian variety is surjective, so ${\varphi }_L$ is trivial and $L\in {\mathrm{Pic}}^0(X)$. Thus the Néron-Severi groupis torsion-free.

Finally, put $M={\varphi }_L(x)$. For every $y$,which is trivial by the Theorem of the square. Hence

Define the homomorphism associated to a line bundle on an abelian varietyThe Theorem of the square gives ${\varphi }_L(x+y)={\varphi }_L(x)\otimes {\varphi }_L(y)$, so this is a homomorphism.

Tensor products satisfy ${\varphi }_{L\otimes M}={\varphi }_L\otimes {\varphi }_M$ and ${\varphi }_{L^{-1}}={\varphi }_L^{-1}$. Thereforeis a subgroup of the Picard group. If $M={\varphi }_L(x)$, translations commute and the theorem of the square givesfor every $y$. Hence $M\in {\mathrm{Pic}}^{0}X$, proving $\mathrm{im}{\varphi }_L\subseteq {\mathrm{Pic}}^{0}X$.

The normalized universal line bundle restricts to the class $\xi$ on $A\times \{\xi \}$ and is trivialized on both zero sections. For a line bundle $L$ on an abelian variety, its homomorphism associated to a line bundle on an abelian variety satisfiesas Picard group classes. A choice of trivialization of $L_0$ normalizes the right side; uniqueness then follows from the Seesaw theorem.

For $T_q(x)=x+q$, the homomorphism associated to a line bundle on an abelian variety gives ${\lambda }_{\Theta }(q)=T_q^{\ast }\mathcal{O}(\Theta )\otimes \mathcal{O}(\Theta {)}^{-1}$. Restrict to the negative Abel map of an algebraic curve. The restriction of the theta divisor to an Abel curve identifies this restriction with the degree-zero class $q$. Pullback of algebraically trivial line bundles does not change when the embedding is translated. Thus pullback $\rho :\widehat{J(C)}\to J(C)$ satisfies $\rho {\lambda }_{\Theta }=1$. Equal dimensions and properness make ${\lambda }_{\Theta }$ an isomorphism, including on schemes, rather than only on closed points.