Let $ q \in [0,1) $. The compact quantum group $ {\text{SU}_{q}}(2) $ is defined to be the universal unital $ C^{*} $-algebra that is generated by two elements $ \alpha $ and $ \beta $ satisfying the following five relations:
\begin{align}
\alpha^{*} \alpha + \beta^{*} \beta & = 1, \\
\alpha \alpha^{*} + q^{2} \beta \beta^{*} & = 1, \\
\beta^{*} \beta & = \beta \beta^{*}, \\
\alpha \beta & = q \beta \alpha, \\
\alpha \beta^{*} & = q \beta^{*} \alpha.
\end{align}
In his paper *Twisted $ \text{SU}(2) $ Group. An Example of a Non-Commutative Differential Calculus*, Woronowicz proves that there is a $ * $-isomorphism between $ {\text{SU}_{q}}(2) $ and $ {\text{SU}_{0}}(2) $. Part of his proof proceeds as follows. He begins with the claim that
\begin{align}
\sigma(\alpha^{*} \alpha)
& \subseteq \left\{ 0,1 - q^{2},1 - q^{4},1 - q^{6},\ldots,1 \right\} ~ \text{and} \\
\sigma(\beta^{*} \beta)
& \subseteq \left\{ 0,\ldots,q^{6},q^{4},q^{2},1 \right\}.
\end{align}
Next, he chooses arbitrary continuous functions $ f,g: [0,1] \to \Bbb{R} $ such that:

- $ f(0) = 0 $ and $ f(t) = 1 $ for all $ t \in \left[ 1 - q^{2},1 \right] $.
- $ g(1) = 1 $ and $ g(t) = 0 $ for all $ t \in \left[ 0,q^{2} \right] $.

Letting $ a = \alpha ~ f(\alpha^{*} \alpha) $ and $ b = \beta ~ g(\beta^{*} \beta) $, he then says that $ (a,b) $ is a generating pair for $ {\text{SU}_{0}}(2) $, i.e., \begin{align} a^{*} a + b^{*} b & = 1, \\ a a^{*} & = 1, \\ b^{*} b & = b b^{*}, \\ a b & = 0, \\ a b^{*} & = 0. \end{align}

The problem is, none of this seems to work. For example, let us try to verify that $ a a^{*} = 1 $. Observe that \begin{align} a a^{*} & = [\alpha ~ f(\alpha^{*} \alpha)] [\alpha ~ f(\alpha^{*} \alpha)]^{*} \\ & = [\alpha ~ f(\alpha^{*} \alpha)] [f(\alpha^{*} \alpha)^{*} ~ \alpha^{*}] \\ & = [\alpha ~ f(\alpha^{*} \alpha)] \left[ \overline{f}(\alpha^{*} \alpha) ~ \alpha^{*} \right] \qquad (\text{By the continuous functional calculus.}) \\ & = [\alpha ~ f(\alpha^{*} \alpha)] [f(\alpha^{*} \alpha) ~ \alpha^{*}] \qquad \left( \text{As $ \overline{f} = f $.} \right) \\ & = [f(\alpha \alpha^{*}) ~ \alpha] [\alpha^{*} ~ f(\alpha \alpha^{*})] \qquad (\text{As $ \alpha ~ p(\alpha^{*} \alpha) = p(\alpha \alpha^{*}) ~ \alpha $ for every polynomial $ p $.}) \\ & = [f(\alpha \alpha^{*})] (\alpha \alpha^{*}) [f(\alpha \alpha^{*})] \\ & = h(\alpha \alpha^{*}), \end{align} where $ h: [0,1] \to \Bbb{R} $ is defined by $ h(t) \stackrel{\text{df}}{=} t [f(t)]^{2} $ for all $ t \in [0,1] $. However, $$ \sigma(\alpha \alpha^{*}) \cup \{ 0 \} = \sigma(\alpha^{*} \alpha) \cup \{ 0 \}, $$ so $ h|_{\sigma(\alpha \alpha^{*})} $ is the identity function on $ \sigma(\alpha \alpha^{*}) $, which yields $ h(\alpha \alpha^{*}) = \alpha \alpha^{*} $ instead of $ h(\alpha \alpha^{*}) = 1 $.

Question.For those who are familiar with this topic, am I hallucinating, or is something not right here?

**Note:** Woronowicz’s argument is given on Page 179, in Section A2, of his paper.