= Solution
Let $Q\in E'[\widehat\phi]$. The compatibility of divisor classes with pullback identifies the class of
$$
\phi^*((Q)-(O))
$$
with $\widehat\phi(Q)=O$, so choose $g_Q\in K(E)^\times$ with
$$
\operatorname{div}(g_Q)=\phi^*((Q)-(O)).
$$
Since $[n]Q=\phi\widehat\phi(Q)=O$, choose $f_Q\in K(E')^\times$ with
$$
\operatorname{div}(f_Q)=n((Q)-(O)).
$$
The functions $f_Q\circ\phi$ and $g_Q^n$ have the same divisor. Their quotient is constant, and because $K$ is algebraically closed we may rescale $g_Q$ so that
$$
f_Q\circ\phi=g_Q^n.
$$
Thus define
$$
E'[\widehat\phi]\longrightarrow
\frac{K(E')^\times\cap K(E)^{\times n}}{K(E')^{\times n}},
\qquad Q\longmapsto[f_Q].
$$
Changing either function changes $f_Q$ only by an $n$th power of a constant. The divisor relation for $Q+R$ differs from the sum of those for $Q$ and $R$ by $n$ times a principal divisor, so the map is a homomorphism. If $[f_Q]$ is trivial, then $f_Q=h^n$ for $h\in K(E')^\times$, whence $\operatorname{div}(h)=(Q)-(O)$; the divisor-class isomorphism forces $Q=O$. The map is therefore well-defined and injective.
For $P\in E[\phi]$, translation by $P$ fixes $g_Q^n=f_Q\circ\phi$. Hence
$$
e_\phi(P,Q)=\frac{g_Q(X+P)}{g_Q(X)}\in\mu_n
$$
is independent of the auxiliary point $X$. This is the <Weil pairing> associated with $\phi$.
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