 Solution

The quotient $q(w)=\varphi (w)/w$ extends to a holomorphic function at zero with value one. Since $\varphi$ is injective and vanishes at zero, $q$ has no zero in the unit disc. A holomorphic logarithm $\ell$ exists there, with $\ell (0)=0$. Define the odd square-root transform of a normalized univalent function byIt satisfies $h(z{)}^2=\varphi (z^2)$, $h(0)=0$, $h^{\prime }(0)=1$ and $h(-z)=-h(z)$. To check injectivity, $h(z)=h(w)$ gives $\varphi (z^2)=\varphi (w^2)$, hence $z^2=w^2$. If $w=-z$, oddness gives $h(z)=0$, so $z=w=0$. Thus the two arguments always coincide, and $h\in \mathcal{S}$. Choosing the square root by its derivative at zero removes its sign ambiguity.