= Solution
The quotient $q(w)=\phi(w)/w$ extends to a <holomorphic function> at zero with value one. Since $\phi$ 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> by
$$
h(z)=z\exp[\ell(z^2)/2].
$$
It satisfies $h(z)^2=\phi(z^2)$, $h(0)=0$, $h'(0)=1$ and $h(-z)=-h(z)$. To check <injectivity>, $h(z)=h(w)$ gives $\phi(z^2)=\phi(w^2)$, 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.
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