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Duality
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Tate-Shafarevich group and $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \mathrm{Sha}(E/L)$

$\DeclareMathOperator\Sha{Sha}\DeclareMathOperator\Gal{Gal}$Let $L/K$ be a quadratic extension of number field $K$. Let $\sigma$ be a generator of $\Gal(L/K)$.

Let $E/K$ be an elliptic curve defined over $K$ and $\Sha(E/K)$ be its Tate-Shafarevich group. $\Gal(L/K)$ acts on $\Sha(E/L)$ naturally (cf. How Galois group acts on Tate-Shafarevich group?).

There is a canonical isomorphism $\tau: E(L)\cong E_D(L)$ .

On the other hand , there exists an isomorphism $\phi: \Sha(E/L)\cong \Sha(E_D/L)$ induced by $\tau$ though I cannot write down the map between them.

What I want to prove is

$\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$

$\sigma \tau=-\tau \sigma$ holds because we can calculate its coordinates explicitly, but I cannot calculate both $\sigma \phi$ and $\phi \sigma$, so I'm having difficulty to prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$. How can I overcome this trouble and prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$ ?

Thank you in advance.

Duality
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