I learn from SGA4$\frac{1}{2}$ Dualité Proposition 3.4 that Tr($\phi$(u)$\cup x$)=$\phi (x)$,where $\phi$ is a homomorphism $\phi$: $H^1_c(X,\mu_n) \to \mathbb{Z}/n$ and u is the (1,1) part of diagonal class $H^2_{\Delta}(X\times \bar{X},j_!\mu_n)$ which can be proved agrees with "de son image réciproque par $f_0$:$X\to Pic_D^0(\bar{X})$"in $H^1_c(X,Pic_D^0(\bar{X})）$ . Maybe we can assume that $\phi$ is $\frac{g(X+S)}{g(X)}$ where g is the function associate with the point $T\in A[n]$ used to define Weil-pairing.My question: How to prove $\phi(u)$ is the divisor corresponding to T and the agreement of the pairing？
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Let $X$ be a smooth projective curve over field $k$. Let $a\in A[n], \mathcal{L}\in A^\vee[n]$.
Recall the definition of Weil pairing (Mumford's Abelian varieties, IV.20): since $\mathcal{L}\in\mathrm{Pic}^0(A)$, we know $n^*\mathcal{L}=\mathcal{L}^{\otimes n}=\mathcal{O}_A$, so after pulling back by $n\colon A\to A$,the line bundle $\mathcal{L}$ becomes trivial, thus corresponds to a factor of automorphy $\chi\colon A[n]\to\mu_n$, the Weil pairing is defined by $e_n(a,\mathcal{L})=\chi(a)$.
To show the Weil pairing coincides with cup product, using SGA41/2 (6.2.2.3)(Duality, Proposition 3.4), it suffices to show the composition $$A^\vee[n]\to H^1(A[n],\mu_n)\to A^\vee[n]$$ is identity. The first map is taking factor of automorphy of the $n$-torsion line bundle (or equivalently, the $\mu_n$-torsor), the second map is collapsing the $A[n]$-torsor $$0\to A[n]\to A\overset{n}{\to} A\to 0$$ to a $\mu_n$-torsor. When we collapse such a torsor, the $n^{2g}$ fibers correspond to $A[n]$ are identified, the scaling is given by a homormophism $A[n]\to\mu_n$. Think this through one see the composition is identity.