Let $S$ be the space of univalent (i.e. injective) mappings from the disk $\mathbb{D}$ to the plane $\mathbb{C}$ normalized so that $f(0) =0$ and $f'(0)=1$. So

$$f(z) = z+a_2z^2+a_3z^3+\cdots.$$

I'd like to show that if $f,g \in S$ and $f(\mathbb{D}) = g(\mathbb{D})$ then $f=g$.

Attempt: I'm aware the Schwarz lemma implies this. Let $U = f(\mathbb{D})= g(\mathbb{D})$ and let $F: U \to \mathbb{D}$ be a Riemann conformal mapping with $F(0)=0$. Then the Schwarz lemma applies to $F\circ f$ and $F\circ g$. But other than seeing that $F'(0)\leq 1$, I do not see what this gets me. Any help would be appreciated.


If $f(\mathbb{D}) = g(\mathbb{D})$ then $h = g^{-1} \circ f$ is a holomorphic function from $\mathbb{D}$ into $\mathbb{D}$ with $h(0) = 0$ and $h'(0) = 1$. Now use the Schwarz Lemma to conclude that $h$ is the identity.


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