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Suppose $f$ is a conformal equivalence between two domains $D_1$ and $D_2$ in $\mathbb{C}$. Does this imply the existence of a map

$F_t(z): D_1 \times [0, a] \rightarrow \mathbb{C}$

such that each $F_t$ is conformal in $z$ and smooth in $t$, $F_0 = \text{id}$, and $F_{a} = f$? If not, does this hold if we make stronger assumptions, such as requiring that the boundary be a Jordan domain, etc.?

user897123
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    As it stands, the answer would be (in general) no. A conformal map $\mathbb{C}\to \mathbb{C}$ is of the form $z \mapsto cz+d$ where $c\neq 0$. Do you mean a homotopy $F\colon D_1\times [0,a] \to \mathbb{C}$ with $F_0 = \operatorname{id}_{D_1}$ and $F_a = f$? – Daniel Fischer Apr 22 '14 at 09:57
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    I suppose $D_1=D_2=\mathbb C\setminus{0}$ and $f(z)=\frac1z$ might pose problems: The two gaps $0$ and $\infty$ have to switch places, requiring $\infty$ in the image at intermediate $t$ values. – Hagen von Eitzen Apr 22 '14 at 10:15
  • Ah, yes, thanks for catching the typo. Fixed. – user897123 Apr 22 '14 at 10:53

1 Answers1

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Suppose that $D_1$ is the unit disk $D$. First, you can assume that $f(0)=0, f'(0)=1$. Now use the homotopy $f(tz)/t$ for $0<t\le 1$, extended by the identity to $t=0$.

This show that the space of univalent functions $D\to {\mathbb C}$ (normalized to have $f'(0)=1$) is contractible.

Via Riemann mapping theorem, the same applies to all simply connected domains $D_1$.

As for the general case, the map $z\mapsto z^{-1}$ with $D_1={\mathbb C}^\times$ is a counter-example.

Moishe Kohan
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