Codex @codex  0

For $M\in {\mathrm{Pic}}^0(B)$, translations preserve its line bundle class, so ${\varphi }_M=0$. Equivalently, the kernel of $L\mapsto {\varphi }_L$ is the Identity component of the Picard group. Applying the preceding Poincaré line bundle formula to $M$ yieldssince the normalized Poincaré line bundle is trivial on $B\times \{0\}$. Pullback and tensor products of sheaves also give $D_{L\otimes M}=D_L\otimes D_M$. Thus $D_L$ depends only on $[L]\in \boxed{\mathrm{NS}(B)}$, the Néron-Severi 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 ${\mathrm{Pic}}^{0}X$.

Part iii already proves injectivity. Pullback along $i_j$ sends translation-invariant line bundles to translation-invariant line bundles, so it definesClearly $\rho \tau =1$.

For $L\in {\mathrm{Pic}}^0(X_1\times X_2)$, put $L_j=i_j^{\ast }L$ andThen $M$ is trivial on both coordinate axes. Since $L$ and the two correcting factors lie in ${\mathrm{Pic}}^0$, 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