to show that class $\mathbb S$ is preserved under square-root transformation, i.e. $g(z)=\sqrt{f(z^2)} \in \mathbb S$

Question

$\mathbb D$={${z \in \mathbb C / \mid z \mid < 1 }$}

$\mathbb S$ be the class of all functions $f$ on $\mathbb D$ such that $f$ is univalent (analytic + injective) and $f(0)=0$,$f'(0)=1$.

I want to show that class $\mathbb S$ is preserved under square-root transformation, i.e. $g(z)=\sqrt{f(z^2)} \in \mathbb S$.

I have the following logic but don't know how worth it is.

I know the proof that if $f \in \mathbb S$, then there exist an odd function $h \in \mathbb S$, such that $h^2(z)=f(z^2)$.

So, We have $g(z)=\sqrt{h^2(z)}$, then using any one of the branch, we have $g(z)=h(z)$, and as $h \in \mathbb S$, we have $g \in \mathbb S$. Is the argument ok?