= Solution
Let $D=\dim\mathcal H_A$ and introduce an isomorphic reference system $R$. Choose a basis $\{|j\rangle\}_{j=1}^D$ and the <maximally entangled state>
$$
|\Omega\rangle_{RA}=\frac1{\sqrt D}\sum_{j=1}^D|j\rangle_R|j\rangle_A.
$$
Complete positivity makes
$$
\omega_{RA}=(\operatorname{id}_R\otimes\Phi)(|\Omega\rangle\langle\Omega|)
$$
a positive <density operator>. This is the normalized <Choi matrix>, with the reference placed first. Its <spectral decomposition> can be written
$$
\omega_{RA}=\sum_{\ell=1}^{r}\lambda_\ell|v_\ell\rangle\langle v_\ell|,
\qquad \lambda_\ell>0.
$$
The number $r$ is at most $D^2$.
Here an <index state> is a chosen basis vector $|j\rangle_R$ in the reference. The corresponding <relative state> of $|v_\ell\rangle$ is the generally unnormalized vector
$$
|\eta_{\ell j}\rangle_A=(\langle j|_R\otimes I_A)|v_\ell\rangle_{RA}.
$$
Thus $|v_\ell\rangle=\sum_j|j\rangle_R|\eta_{\ell j}\rangle_A$. Define a <linear map> $A_\ell$ by its columns:
$$
A_\ell|j\rangle=\sqrt{D\lambda_\ell}\,|\eta_{\ell j}\rangle.
$$
Then $(I_R\otimes A_\ell)|\Omega\rangle=\sqrt{\lambda_\ell}|v_\ell\rangle$, so
$$
\omega_{RA}=\sum_\ell(I_R\otimes A_\ell)|\Omega\rangle\langle\Omega|(I_R\otimes A_\ell^\dagger).
$$
Expand both sides in reference <matrix units>. The block with reference indices $i,j$ on the left is $D^{-1}\Phi(|i\rangle\langle j|)$; on the right it is $D^{-1}\sum_\ell A_\ell|i\rangle\langle j|A_\ell^\dagger$. Since <matrix units> span all operators and $\Phi$ is linear,
$$
\boxed{\Phi(\rho)=\sum_\ell A_\ell\rho A_\ell^\dagger.}
$$
This constructs the <Kraus representation> explicitly from the relative states, rather than assuming it.
Finally, <trace> preservation implies for every <density operator> $\rho$ that
$$
\operatorname{Tr}\rho=\operatorname{Tr}\Phi(\rho)
=\operatorname{Tr}\left[\rho\sum_\ell A_\ell^\dagger A_\ell\right].
$$
Taking every rank-one <orthogonal projection> as $\rho$ forces the <Hermitian operator> in brackets to have the same quadratic form as the identity. Therefore
$$
\boxed{\sum_\ell A_\ell^\dagger A_\ell=I.}
$$
The argument is finite-dimensional, as appropriate to the quantum-information setting; a channel with a different output dimension has the same construction with rectangular <Kraus operators>.
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