= Solution
A <holomorphic function> with $f'(z)\ne0$ at every point is locally a <conformal map>. Write $a=z_0$. Since $|a|<1$ and $|z|<1$, $1-\overline az\ne0$, and
$$
\boxed{f'(z)=\frac{|a|^2-1}{(1-\overline az)^2}\ne0.}
$$
Thus the specified <Möbius transformation> is conformal throughout the disk. Expanding the two squared absolute values gives
$$
\boxed{1-|f(z)|^2=\frac{(1-|a|^2)(1-|z|^2)}{|1-\overline az|^2}>0,}
$$
so $f$ maps the <open unit disk> into itself. Solving $w=(a-z)/(1-\overline az)$ gives $z=(a-w)/(1-\overline aw)=f(w)$. Equivalently $f(f(z))=z$. Hence $f$ is its own inverse, proving both <injectivity> and <surjectivity> on the disk. Nonvanishing derivative establishes the local angle-preserving property; the explicit inverse supplies the separate global bijection claim.
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