A question on an argument in Woronowicz’s paper on the compact quantum group $ {\text{SU}_{q}}(2) $ 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.
 A: As Jan suggests, there's a typo in the definition of $f$. It should read
$$
f(t) = \begin{cases}\frac{1}{\sqrt{t}} & (t \ge 1 - q^2)\\ 0 & (t = 0).\end{cases}
$$
Since $\alpha \alpha^* \ge 1-q^2$ (this follows from usual realization as in p. 123), this will imply that $a^*$ is an isometry. And $\alpha^* \alpha + \beta^* \beta = 1$ implies that the eigenspace of $\beta^* \beta$ for the eigenvalue $1$ is precisely the kernel of $\alpha$, which implies $a^* a + b^* b = 1$.
Morally, at $q=0$ you have the Toeplitz algebra as the 'quantum limit' of the deformation quantization of the $2$-dimensional symplectic leaves
$$
\left\{\begin{pmatrix}\bar{z} \lambda & \sqrt{1-|z|^2}\lambda\\-\sqrt{1-|z|^2}\bar{\lambda}&z\bar{\lambda}\end{pmatrix} \mid |z| < 1 \right\} \quad (|\lambda|=1)
$$ 
inside $SU(2)$ together with boundary $U(1)$, and $C(SU_q(2))$ is an elaborate patching of these. The $q$-disc algebra generated by an operator $Z_q$ satisfying $\|Z_q\|=1$ and $1-Z_q^* Z_q = q^2 (1-Z_qZ_q^*)$ interpolates the Toeplitz algebra and the usual function algebra on the closed disc, but these algebras are in fact isomorphic to each other for $|q|<1$.
