TITLE: Poincaré line bundle
QUESTION [8 upvotes]: I am being stuck by the proof of the existence of Poincaré line bundle of complex torus in Griffiths-Harris. Here is the question: 
Let $M$ be a complex torus and $M'$ be the complex torus dual to $M$ (the one consists of all holomorphic line bundle whose chern class is zero). By Kunneth's formula and the definition of $M',$  one knows that the identity element $I$ of 
  $$ Hom(H^1(M,\mathbb{Z}),H^1(M,\mathbb{Z}))\equiv H^1(M,\mathbb{Z})\otimes H^1(M',\mathbb{Z})$$
belongs to $H^2(M\times M',\mathbb{Z})$ and is of type (1,1). Hence, there exists a line bundle $P$ over $M\times M'$ having $I$ as its chern class. For each $\xi$ in $M'$, let $P_{\xi}$ be the restriction of $P$ on  $M\times \{\xi\}$, clearly it is in $M'$. Hence we get a mapping $\Phi: M' \rightarrow M'$ that send $\xi$ to $P_{\xi}.$
Why the induced mapping of $\Phi$ on the first homology is the identity?
Thank you so much.

REPLY [2 votes]: [Not exactly an answer. Still, it may be helpful, I hope.]
I think there are two obscure points in this proof: apart from the homology 
question, why the map is holomorphic? (Is it really obvious?). 
Of course, all this can be settled. However, I prefer a more explicit 
construction of the Poincare bundle, which can be found, e.g., in 
Hindry&Silverman, "Diophantine Geometry", Exercise A.5.6. 
(I like it, but even if you won't, it is definitely worth reading).