Peter-Weyl theorem (compact quantum groups) I'm reading the paper Notes on compact quantum groups. In this paper, the following theorem is proven:

Question: Why is the marked equality true?
 A: The problem is actually in the line above.  The setup is that $u$ is the regular representation on $\newcommand{\mc}{\mathcal}\newcommand{\id}{\operatorname{id}}H$, so $u\in M(\mc B_0(H)\otimes A)$, and $v$ is on $K$, so $v\in M(\mc B_0(K)\otimes A)$.  Also $x$ should be in $\mc B_0(H,K)$.
Then the quote has $\xi_1\in H, \eta_1\in K$ and defined $x$ by $x(\xi) = \langle\xi,\xi_1\rangle \eta_1$ so $x\in\mc B_0(H,K)$.  On the next line we choose $\xi\in H, \eta\in K$ and it's claimed that
$$ (v^*(x\otimes 1)u)(\xi\otimes\eta) = (v^*(1\otimes a))(\eta_1\otimes\eta). $$
Here $a = (\omega_{\xi,\xi_1}\otimes\iota)u$.
This doesn't make sense as while $a\in M(A)$ and $u\in M(\mc B_0(H)\otimes A)$, we have that $\eta\in K$, and $K$ is just some auxiliary Hilbert space which has no relation to $A$.

I believe we should proceed as following.  By definition, $H$ is the GNS space with respect to the Haar state $h$ on $A$.  Let $\pi:A\rightarrow\mc B(H)$ be the $*$-representation, and let $\xi_0\in H$ be the cyclic vector.  Let $v_0 = (\id\otimes\pi)(v) \in M(\mc B_0(K)\otimes\pi(A)) \subseteq\mc B(K\otimes H)$, and similarly $u_0 = (\id\otimes\pi)(u)\in\mc B(H\otimes H)$.  Then for $\alpha\in H$, we have
$$ (v_0^*(x\otimes 1)u_0)(\xi\otimes\alpha) = (v_0^*(1\otimes \pi(a)))(\eta_1\otimes\alpha). $$
This follows as $(\omega_{\xi,\xi_1}\otimes\id)u_0 = \pi(a)$.
Notice that this equation is equality of vectors in $K\otimes H$.  Set $\alpha=\xi_0$ and apply $(\id\otimes\xi_0)$ to this to get
$$ (\id\otimes\omega_{\xi_0,\xi_0})(v_0^*(x\otimes 1)u_0) \xi
= (\id\otimes\omega_{\xi_0,\xi_0})(v_0^*(1\otimes \pi(a))) \eta_1. $$
However, $\omega_{\xi_0,\xi_0}\circ\pi = h$ and so
$$ y \xi = (\id\otimes h)(v^*(x\otimes 1)u) \xi
= (\id\otimes h)(v^*(1\otimes a)) \eta_1. $$
As $y=0$ this proves the equation which was causing trouble.
