Intertwining relations and operator decomposability The following proposition is made in a functional analysis paper without proof. I assume that the argument is a very simple one-liner, but I can't seem to come up with it. Any help would be appreciated.
Consider separable Hilbert spaces $\mathcal{H}$, $\mathcal{K}$ and $\mathcal{K}_1$, where the dimension of $\mathcal{K}_1$ is not lower than the dimension $\mathcal{K}$. If an isometry $U: \mathcal{H} \otimes \mathcal{K} \to \mathcal{H} \otimes \mathcal{K}_1$ satisfies the following relation
$$
 U \left( a \otimes 1_K \right) = \left( a \otimes 1_{{K}_1}\right)U,
$$
for all bounded operators $a \in \mathcal{B}(\mathcal{H})$, it is of the form $U = 1 \otimes U_1$ with $U_1: \mathcal{K} \to \mathcal{K}_1$ an isometry.
 A: Let $E$,  $F$,  and $G$  be
orthonormal bases for $H$, $K$,  and $K_1$,  respectively.
Fix any vector $e_0$ in $E$ and consider the operators
$$
  W:\xi \in K\mapsto e_0\otimes \xi \in H\otimes K,
  $$
and
$$
  W_1:\eta \in K_1\mapsto e_0\otimes \eta \in H\otimes K_1.
  $$
Finally, put $U_1=W_1^*UW$, so that $U_1$ is an operator from $K$ to $K_1$.
We then  claim that $U=1\otimes U_1$.
To see this
notice that, given $e_1,e_2\in E$, $f\in  F$, and $g\in G$, we have that
$$
  \big \langle (1\otimes U_1)(e_1\otimes f),e_2\otimes g\big \rangle  =
  \big \langle e_1\otimes U_1(f),e_2\otimes g\big \rangle ,
  \tag 1
  $$
which clearly vanishes when $e_1\neq e_2$. Otherwise we have that (1) equals
$$
  \big \langle U_1(f),g\big \rangle  =
  \big \langle UW(f),W_1(g)\big \rangle  =
  \big \langle U(e_0\otimes f),e_0\otimes g\big \rangle .
  $$
In order to prove the claim it is enough to show that
$\big \langle U(e_1\otimes f), e_2\otimes g\big \rangle $ coincides with (1).
Assuming first that $e_1=e_2$, let $\sigma $ be any permutation of $E$
such that $\sigma (e_0)=e_1$, and consider the unitary operator $a$ on $H$ determined by its values on $E$ by
$$
  a(e) = \sigma (e),\quad\forall e\in  E.
  $$
Then,
$$
  \big \langle U(e_1\otimes f), e_2\otimes g\big \rangle  =
  \big \langle (a\otimes 1)^{-1} U(a\otimes 1)(e_1\otimes f), e_2\otimes g\big \rangle  =$$$$=
  \big \langle U(\sigma (e_1)\otimes f), \sigma (e_2)\otimes g\big \rangle  =
  \big \langle U(e_0\otimes f), e_0\otimes g\big \rangle .
  $$
In case $e_1\neq e_2$, let $a$ be the orthogonal projection from $H$ to ${\mathbb C}e_1$, so
$$
  \big \langle U(e_1\otimes f), e_2\otimes g\big \rangle  =
  \big \langle U(p\otimes 1)(e_1\otimes f), e_2\otimes g\big \rangle  = $$ $$ =
  \big \langle (p\otimes 1)U(e_1\otimes f), e_2\otimes g\big \rangle  =
  \big \langle U(e_1\otimes f), (p\otimes 1)(e_2\otimes g)\big \rangle  = 0.
  $$
This concludes the proof.
