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Let $p$ be a primes that does not divide $N$, then $T_p$ defined an endomorphism $J_0(N)\to J_0(N)$. what is $T_p^\vee$? In other words, we naturally have $J_0(N)^\vee \xrightarrow{T_p^\vee} J_0(N)^\vee$, and therefore $J_0(N) \xrightarrow{T_p^\vee} J_0(N)$. So, what is this dual map?

The reason that I ask this question is the following.

Let $Y_0(N)= \Gamma_0(N)\backslash \mathbb{H}$ be the usual modular curve of level $N$. Given $p\nmid N$, we have two maps from $Y_0(pN)$ to $Y(N)$. Let's call them $B_1$ and $B_p$. By definition, $B_1$ is induced by the identity map of the upper half plane, and $B_p$ is induced by $z\mapsto pz$. Now $B_1$ and $B_p$ uniquely determines maps from $X_0(pN)$ to $X_0(N)$. With an abuse of notation, I am going to denote the map between the Jacobian of the modular curves $J_0(pN)\to J_0(N)$ induced by Albanese functoraility still by $B_1$ or $B_p$, and denote the maps induced by the Picard functoriality by $B_1^\vee$ or $B_p^\vee$.

I was able to show that $B_1\circ B_p^\vee$ coincide with the Hecke operator $T_p$. This can be explained by the following fact. If we let $\alpha =\left(\begin{smallmatrix} 1 & 0 \\ 0 & p \end{smallmatrix} \right)$. Then $$\alpha^{-1}\Gamma_0(N)\alpha \cap \Gamma_0(N)= \alpha^{-1}\Gamma_0(pN) \alpha.$$

However, I am having trouble to figure out what is $B_p\circ B_1^\vee$.

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