Since i could not find the reference book, below the following earlier version of it is used:
CQGRC.pdf - Compact Quantum Groups and Their Representation Categories - Incomplete preliminary version as of January 17, 2014, by Sergey Neshveyev and Lars Tuset.
In [CQGRC], it is Proposition: 1.7.2, CQGRC, page 21 claiming that
$f_z$ has the property (i) of being a is a homomorphism $\Bbb C[G]\to\Bbb C$.
and the proof of (i) claims the relation
$$
\tag{$\dagger$}
\hat\Delta(\rho^z)=\rho^z\otimes \rho^z\ .
$$
So let us comment first on relation $(\dagger)$ above, which is the question.
For the convenience of the reader, we unpack the notations one by one, and describe
the framework:
Framework:
$G=(A,\Delta)$ is the given $C^*$-algebraic CQG. The underlying algebra is in an alternative notation $A=C(G)$.
(I will use preferably $A$ below, since it is simpler to type.)
We need the space of matrix coefficients of $A$, denoted alternatively by $A_0=\Bbb C[G]$.
The functionals $f_z$ are linear functionals $f_z:A_0\to \Bbb C$, and we write $A_0'$ for this space of functionals.
So for each complex $z$ we have $f_z\in A_0'$. To put the hands on some $f_z$ we need to know the structure of $A_0$.
What is $A_0$? The first proposition in CQGRC, §1.6, is:
For a compact quantum group $G=(A,\Delta)$ denote by $A_0=\Bbb C[G]\subset C(G)=A$ the linear span of
matrix coefficients of all finite dimensional (co)representations of $G$, it is a dense subspace of $A$ by CQGRC, Corollary 1.5.6.
So let $(U_\alpha)$ be such a maximal
family of irreducible, unitary, mutually inequivalent representations of $G=(A,\delta)$. The index $\alpha$ runs in some index set
$\Lambda$, that we will often omit.
Here,
$$
U_\alpha = \sum m^\alpha_{ij}\otimes u^\alpha_{ij}\in B(H_\alpha)\otimes A\ .
$$
The matrices $m^\alpha_{ij}$ are the elementary matrices, the canonical basis of the matrix space $B(H_\alpha)$.
For an irreducible, unitary representation $U\in B(H) \otimes A_0$ its contragredient cousin is $U^c$.
CQGRC, Proposition 1.3.13 shows that $U$ and $U^{cc}$ are equivalent.
The space of intertwiners between $U$, and $U^{cc}$, in notation $\operatorname{Mor}(U,U^{cc})\operatorname{Mor}(U,U)$ is thus one-dimensional
(Schur),
and we have even an explicitly constructed intertwiner $j(Q_r)\in \operatorname{Mor}(U, U^{cc})$.
Here $Q_r$ is a positive, invertible operator in $B(\bar H)$, and $j$ is the antimorphism on $B(H)$.
We rescale $Q_r$ by a positive scalar $\lambda_U$, so that the obtained invertible, positive operator $\rho_U:=\lambda_U\; j(Q_r)$
is normed by the condition:
$$
\operatorname{Trace}(\rho_U) =
\operatorname{Trace}(\rho_U^{-1}) \ .
$$
From now on, we can forget about $j(Q_r)$, and $\lambda_U$, and need only the information that the space of intertwiners is one-dimensional,
generated by an invertible, positive operator $\rho_U$.
Since $\rho_U>0$, and for each $z$ the map $a\to a^z:=e^z\log a$ is analytic on $(0,\infty)$, we have an analytic functional calculus
defining the positive operator
$$(\rho_U)^z=:\rho_U^z\in B(H)\ .$$
Now, $f_z$ is introduced in CQGRC, Definition 1.7.1, implicitly. We do not have a formula to evaluate
$f_z$ on an elements from $A_0$, instead, we use the matrix coefficient spaces of individual irreducible representations.
Note that for different (inequivalent) unitary (co)representations $U_\alpha$, $U_\beta$ the corresponding
spaces of matrix coefficients
$\mathscr C(\alpha):=\operatorname{Span}(U_\alpha):=\operatorname{Span}(u_{\alpha,ij})$ and
$\mathscr C(\beta):=\operatorname{Span}(U_\beta):=\operatorname{Span}(u_{\beta,kl})$
generated by their entries are linearly independent (orthogonal). Moreover, this independence holds also for the coefficients
of $U_\alpha$, so that extending $f_z$ is uniquely defined on $\mathscr C(\alpha)$ by the relation:
$$
%\Big(\operatorname{id}_{B(H_{\alpha})}\otimes f_z\Big)\;\Big(U_\alpha\Big) \ =\ \rho_{\alpha}^z\in B(H_{\alpha})_{>0}\ .
(\operatorname{id}\otimes f_z)(U_\alpha)\ =\ \rho_{\alpha}^z\in B(H_{\alpha})\ .
$$
We denoted by $\rho_\alpha$ the positive operator $\rho_{H_\alpha}$ constructed above.
(Unpacking, if $m_{\alpha,ij}$ is the canonical basis of $B(H_\alpha)$, and $U_\alpha =\sum m^\alpha_{ij}\otimes u^\alpha_{ij}$ to obtain
$f_z(u^\alpha_{ij})$ we compute $\rho_\alpha$, then its $z$-power by analytic functional calculus, then we write this result in the $m^\alpha$--basis and take the
coefficient in $m^\alpha_{ij}$.
So far we have the $\rho_U$ objects. Then the $\rho_U^z$ objects. Using them we get $f_z\in A_0'$.
We forget about them, and need in fact only $f_1$.
So we need in fact only $\rho_U$ to know $f_1$ on the vector space spanned by matrix coefficients of $U$.
And now, the text also uses the letter $\rho$ for $f_1$.
(This may be nice a posteriori. But at this point, let us be more careful.)
So i will slightly change notation and use instead:
$$ \varrho:=f_1\in A_0'=\mathscr U(G)\overset{\Phi=\Phi_G}\longrightarrow \prod_\alpha B(H_\alpha)\ .
$$
The text identifies $\varrho$ with its image $\Phi(\varrho)$ in the product of matrix spaces.
Let us compute it explicitly, using $\pi_{U_\alpha}$ as in CQGRC, page 20:
$$
\begin{aligned}
\Phi(\varrho) &=(\ \Phi(\varrho)_\alpha\ )_\alpha \ ,\\
\Phi(\varrho)_\alpha &:= \pi_{U_\alpha}(\varrho)\\
&:=(\operatorname{id}\otimes\varrho)(U_\alpha)\\
&:=(\operatorname{id}\otimes f_1)(U_\alpha)\\
&:=\rho_{U_\alpha}^1=\rho_{U_\alpha}=\rho_\alpha\ .
\end{aligned}
$$
In other words, $\Phi(\varrho)$ is an element in $\prod B(H_\alpha)$ which has as $\alpha$-component the
positive operator $\rho_\alpha\in B(H_\alpha)$. I will write $\rho$ for this family, $\rho=(\rho_\alpha)=\Phi(\varrho)$.
(At any rate, we can now imagine why the notation $\rho$ was kept for $\varrho$.)
What is $\varrho^z\in \mathscr U(G)=A_0'$? It is by definition the functional which is mapped by
$\Phi$ into $\rho^z:=(\rho_\alpha^z)_\alpha$.
So identities involving $\varrho^z$ should be rather moved from the $\mathscr U$-spaces to the matrix spaces.
Let us recall $\hat\Delta$ is as a map $\mathscr{U}(G)\to \mathscr{U}(G\times G)$, which is
determined by its evaluation at some functional $\omega \in A_0' =\mathscr{U}(G)$. The image lies in $\mathscr{U}(G\times G)$.
(It can be bigger than $A_0'\otimes A_0'$.) The functional $\hat \Delta(\omega)\in \mathscr{U}(G\times G)$ is determined
by evaluation on elements of the shape $a\otimes b\in A_0\otimes A_0$. So $\hat\Delta$ is determined by the double evaluation
$$
\hat\Delta(\omega)\ a\otimes b :=\omega(ab)\ .
$$
Our $\omega$ of interest is $\rho^z$.
Recall that we also have a map needed in the sequel (and also displayed vertically),
$$
\require{AMScd}
\begin{CD}
\mathscr U(G) @. \omega\\
@V \Phi_G V V @VVV\\
\prod_{\alpha} B(H_\alpha) @. (\ (\operatorname{id}_{H_\alpha}\otimes\omega)(U_\alpha)\ )_\alpha
\end{CD}
$$
The same $\Phi$-mapping can be written also for $G\times G$, the corresponding irreducible representations are parametrized by tuples $(\alpha,\beta)$,
and are of the shape $U_\alpha\odot U_\beta\in B(H_{(\alpha,\beta)})\otimes A_0\otimes A_0$, where $B(H_{(\alpha,\beta)}):=B(H_\alpha)\otimes B(H_\beta)$
it the algebraic tensor product of the two matrix spaces.
(The composition with the product $A_0\otimes A_0\to A_0$ gives rise to the representation denoted by $U_\alpha\odot U_\beta$ in CQGRC.)
Then consider the diagram, which is for a general $\omega$ not commutative:
$$
\require{AMScd}
\begin{CD}
\omega @>\hat\Delta_G>> \hat\Delta(\omega)\\
\\
\mathscr U(G) @>\hat\Delta_G>> \mathscr U(G\times G)\\
@V \Phi_G V V (??) @VV \Phi_{G\times G} V\\
\prod_\alpha B(H_\alpha) @>>\underline\Delta > \prod_{(\alpha,\beta)} B(H_\alpha) \otimes B(H_\beta)\\
\\
(w_\alpha)_{\alpha\in\Lambda} @>>\underline\Delta > (w_\alpha\otimes w_\beta)_{(\alpha,\beta)\in\Lambda\times\Lambda}
\end{CD}
$$
However for the special value $\omega=\varrho$, which is a "group-like" element,
$\hat\Delta(\varrho)=\varrho\otimes \varrho$ we have the commutativity
$$
\require{AMScd}
\begin{CD}
\varrho @>\hat\Delta >> \hat\Delta(\varrho)\\
@V \Phi_G V V @VV \Phi_{G\times G} V\\
\rho=(\rho_\alpha) @>>\underline\Delta > \rho\otimes \rho =(\rho_\alpha\otimes\rho_\beta)_{(\alpha,\beta)}
\end{CD}
$$
In fact, this group-like property of $\varrho$ is checked in the $\rho$-world,
CQGRC, Theorem 1.4.8,
$$
\rho_{U_\alpha\times U_\beta}=\rho_{U_\alpha}\otimes \rho_{U_\beta}\ .
$$
(This is as stated an equality in the category of representations for $G$.
The part from $A$ involved in what we need below,
when $U_\alpha\times U_\beta$ occurs,
is evaluated to a factor.)
So we compute the component $(\alpha,\beta)$ of $\hat\Delta$ evaluated in $\varrho$.
We write explicitly at some point:
- $U_\alpha =\sum m^\alpha_{ij}\otimes u^\alpha_{ij}\in B(H_\alpha)\otimes A_0$, representation of $G$,
- $U_\beta =\sum m^\beta_{kl}\otimes u^\beta_{kl}\in B(H_\beta)\otimes A_0$, representation of $G$, so that
- $U_\alpha\odot U_\beta= U_{(\alpha,\beta)}=\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes u^\alpha_{ij}\otimes u^\beta_{kl}\in B(H_\alpha)\otimes B(H_\beta)\otimes A_0\otimes A_0$
is the corresponding tensor product representation of $G\times G$. (Not for $G$, when the times notation is used.)
$$
\begin{aligned}
(\ \Phi_{G\times G}\ \hat\Delta(\varrho)\ )_{(\alpha,\beta)}
&:=
\big(\ \operatorname{id}_{B(H_{(\alpha,\beta)})}\otimes\hat\Delta(\varrho)\ \Big)(U_{(\alpha,\beta)})
\\
&=
\big(\
\operatorname{id}_{B(H_{\alpha})}\otimes
\operatorname{id}_{B(H_{\beta})}\otimes
\hat\Delta(\varrho)\ \Big)
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes u^\alpha_{ij}\otimes u^\beta_{kl}
\\
&=
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes \hat\Delta(\varrho)(u^\alpha_{ij}\otimes u^\beta_{kl})
\\
&=
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes \hat\Delta(\varrho)(u^\alpha_{ij}\otimes u^\beta_{kl})
\\
&=
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes \underbrace{(\varrho\otimes\varrho)(u^\alpha_{ij}\otimes u^\beta_{kl})}_{\in\Bbb C}
\\
&=
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\cdot \varrho(u^\alpha_{ij})\cdot \varrho(u^\beta_{kl})
\\
&=
\sum m^\alpha_{ij}\cdot \varrho(u^\alpha_{ij})\otimes \sum m^\beta_{kl}\cdot \varrho(u^\beta_{kl})
\\
&=
(\operatorname{id}\otimes\varrho)\left(\sum m^\alpha_{ij}\otimes u^\alpha_{ij}\right)
\otimes
(\operatorname{id}\otimes\varrho)\left(\sum m^\beta_{kl}\otimes u^\beta_{kl}\right)
\\
&=
(\operatorname{id}\otimes\varrho)(U_\alpha)
\otimes
(\operatorname{id}\otimes\varrho)(U_\beta)
\\
&=\rho_\alpha\otimes\rho_\beta
\\
&=\underline\Delta(\rho)_{(\alpha,\beta)}
\\
&=(\ \underline\Delta(\Phi(\varrho)\ )_{(\alpha,\beta)}
\ .\qquad\text{ So:}
\\[2mm]
\Phi_{G\times G}\; \hat\Delta(\varrho)
&=\underline\Delta\; \Phi(\varrho)\ .
\end{aligned}
$$
As a final word, a way of giving a sense to the boxed identity from the question,
$$
\hat\Delta(\varrho^z) =\varrho^z\otimes\varrho^z\ ,
$$
which is related to the upper horizontal arrow in the above diagrams,
is by moving it downwards via the $\Phi$ arrows to the lower horizontal arrow,
which is a map $\underline\Delta$ clearly compatible with the functional calculus,
$$
\underline\Delta(\rho^z)_{(\alpha,\beta)}
=
\rho_\alpha^z\otimes\rho_\beta^z
=
(\rho_\alpha\otimes\rho_\beta)^z
=
(\ \underline\Delta(\rho)\ )^z
\ .
$$
The definition of $\varrho^z$ is by taking $\rho^z$ from the L.-most.H.S. and pushing it via $\Phi^{-1}$
into $\mathscr U(G)$. It may be then useful in the vertical $G\times G$-arrow to write
$\varrho^z\otimes\varrho^z$ as a product of $\varrho^z\otimes\epsilon$ and $\epsilon\otimes\varrho^z$,
then go down via $\Phi$ to get by definition of $\varrho$ the commuting operators
$\rho^z\otimes 1$ and $1\otimes \rho^z$, and here we have
$$
(\rho\otimes\rho)^z =
(\ (\rho\otimes 1)\;(1\otimes\rho)\ )^z =
(\rho\otimes 1)^z\;(1\otimes\rho)^z\ .
$$
The notations in CQGRC were my biggest problems. I hope the above
answer - a general nonsense categorial translation - hits the wound point.
