Why does the following diagram commute?$\require{AMScd}$
\begin{CD}
 H^0(X,\mathscr{A}) \times \mathrm{Ext}^2\_X(\mathscr{A},\mu\_{\ell^n}) @>>> H^2(X,\mu\_{\ell^n}) \\ @VVV @| \\
 H^1(X,\mathscr{A}[\ell^n]) \times H^1(X,\mathscr{E}xt\_X^1(\mathscr{A},\mu\_{\ell^n})) @>>> H^2(X,\mu\_{\ell^n}).
\end{CD}
($\mathscr{A}/X$ is an Abelian scheme.)
The vertical homomorphism $\delta$ on the left is induced by the Kummer sequence
$$
 0 \to \mathscr{A}[\ell^n] \to \mathscr{A} \to \mathscr{A} \to 0.
$$


The upper row is the Yoneda pairing.


The lower row is induced by the Weil pairing
$$
 \mathscr{A}[\ell^n] \times \mathscr{A}^\vee[\ell^n] \to \mu\_{\ell^n}
$$
using $\mathscr{E}xt^1(\mathscr{A},\mu\_{\ell^n}) = \mathscr{H}om(\mathscr{A}[\ell^n],\mu\_{\ell^n}) = \mathscr{A}^\vee[\ell^n]$ by the long exact Ext sequence for the Kummer sequence
$0 \to \mathscr{A}[\ell^n] \to \mathscr{A} \to \mathscr{A} \to 0$:
$$
 0 = \mathscr{H}om\_X(\mathscr{A},\mu\_{\ell^n}) \to \mathscr{H}om\_X(\mathscr{A}[\ell^n],\mu\_{\ell^n}) \to \mathscr{E}xt^1\_X(\mathscr{A},\mu\_{\ell^n}) \stackrel{0}{\to} \mathscr{E}xt^1\_X(\mathscr{A},\mu\_{\ell^n}),
$$
the latter transition map being $0$ since $\ell^n$ kills $\mu\_{\ell^n}$. (Edit 2: Or, better,
$$
 0 = \mathscr{H}om\_X(\mathscr{A},\mathbf{G}\_m) \to \mathscr{E}xt^1\_X(\mathscr{A},\mu\_{\ell^n}) \to \mathscr{E}xt^1\_X(\mathscr{A},\mathbf{G}\_m) \stackrel{\ell^n}{\to} \mathscr{E}xt^1\_X(\mathscr{A},\mathbf{G}\_m)
$$
and using $\mathscr{E}xt^1\_X(\mathscr{A},\mathbf{G}\_m) = \mathscr{A}^\vee$, as pointed out in the comments.)


Finally, for the middle vertical arrow: The local-to-global Ext spectral sequence $H^p(X,\mathscr{E}xt\_X^q(\mathscr{A},\mu\_{\ell^n})) \Rightarrow \mathrm{Ext}\_X^{p+q}(\mathscr{A},\mu\_{\ell^n})$ gives us an injection (an edge morphism, see also [Natural morphism appearing in Grothendieck spectral sequence](https://mathoverflow.net/questions/28827/natural-morphism-appearing-in-grothendieck-spectral-sequence/159656#159656))
$$
 H^1(X,\mathscr{E}xt\_X^1(\mathscr{A},\mu\_{\ell^n})) \hookrightarrow \mathrm{Ext}\_X^2(\mathscr{A},\mu\_{\ell^n})
$$
since $\mathscr{H}om\_X(\mathscr{A},\mu\_{\ell^n}) = 0$. 


Edit: Differently stated, why does
\begin{CD}
 \mathrm{Ext}^2\_X(\mathscr{A},\mu\_{\ell^n}) @>>> \mathrm{Hom}(H^0(X,\mathscr{A}), H^2(X,\mu\_{\ell^n})) \\
 @AAA @A\delta^\*AA \\
 H^1(X,\mathscr{E}xt\_X^1(\mathscr{A},\mu\_{\ell^n})) @>>> \mathrm{Hom}(H^1(X,\mathscr{A}[\ell^n]), H^2(X,\mu\_{\ell^n}))
\end{CD}
commute? (the left vertical arrow a edge map of the spectral sequence and the right vertical arrow $\delta^\*$ with $\delta$ the connecting map of the Kummer sequence)



commutative diagram with Yoneda pairing, Weil pairing and edge morphism
1. Show that (in the lower diagram) $H^1(X,\mathscr{H}om(\mathscr{A}[\ell^n],\mu\_{\ell^n}))$ embeds via the first edge homomorphism into $\mathrm{Ext}^1(\mathscr{A}[\ell^n],\mu\_{\ell^n})$ (by the local-to-global Ext spectral sequence) and that the lower horizontal and left vertical arrows factor through this.
2. Now both paths map $e = (0 \to \mu\_{\ell^n} \to \mathscr{F} \to \mathscr{A}[\ell^n] \to 0) \in \mathrm{Ext}^1(\mathscr{A}[\ell^n],\mu\_{\ell^n})$ to $H^0(X,\mathscr{A}) \stackrel{\delta\_\mathrm{Kummer}}{\to} H^1(X,\mathscr{A}[\ell^n]) \stackrel{\delta\_e}{\to} H^2(X,\mu\_{\ell^n})$.


