Prove that any central function of $\mathrm{SU}_{2}$ is uniquely determined by its restriction to the following subgroup.

Question

The question is given below:

3. Prove that any central functon $f$ on $\mathrm{SU}_2$ is uniquely determined by its restriction to the subgroup $$ \mathbf{T} = \biggl\{ A(z) = \begin{pmatrix} z & 0 \\ 0 & z^{-1} \end{pmatrix} \,\Bigg|\, z ∈ \mathbf{C}, |z| = 1 \biggr\} \,, $$ and that $f(A(z)) = f(A(z^{-1}))$.

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But I do not know how to solve it, could anyone give me a hint please?

EDIT:

Definition. A function $f$ on the group $G$ with the property that $$ f(g x g^{-1}) = f(x) \qquad \text{for all $g, x ∈ G$} $$ is called a central function.

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Answer 1

What does it mean to restrict a function: For a function $f:X\rightarrow Y$, the restriction of $f$ to a subset $U\subset X$ is a function $g:U\rightarrow Y$ such that $g(u)=f(u)$ for all $u\in U$. Extending $g$ to $X$ means that we construct a new function $g':X\rightarrow Y$ such that $g'(u)=g(u)$ for all $u\in U$.

Now back to the initial problem:

Remark: Any matrix $A\in SU(2)$ can be written as $\begin{pmatrix}z & -\overline w \\ w & \overline z\end{pmatrix}$ for some $w,z\in\mathbb C$ with $|w|^2+|z|^2=1$. Moreover, every matrix of this form is diagonalizable, that is, there exists $B\in SU(2)$ such that $BAB^{-1}$ is a diagonal matrix.

The $f(A(z))=f(A(z^{-1}))$ part follows from $A(z^{-1})=BA(z)B^{-1}$, where $B=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$.

Uniqueness: If $f$ is a central function and $A\in SU(2)$, then $$f(A)=f(BAB^{-1})$$ where $B\in SU(2)$ such that $BAB^{-1}$ is diagonal.

Claim: For any matrix $C\in SU(2)$, $CAC^{-1}=A(z)$ is diagonal if and only if $C=B$ or $C=B^{-1}$, where $B$ is as in the remark.

The eigenvalues of a matrix do not change under unitary change of basis, and since $BAB^{-1}$ has eigenvalues $z$ and $\overline z$ (because diagonal), $A$ has the same eigenvalues. Hence $CAC^{-1}$ is diagonal if and only if $CAC^{-1}=A(z)$ or $A(z^{-1})$.
But this is equivalent to $C=B$ or $C=B^{-1}$.