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Can a child of a conformal factor be conformal?
Let $\Omega \subset \mathbb R^2$ and let $f\in C^2$ be a conformal map, mapping $\Omega$ onto itself.
I would like to know whether, given any $\Omega_1 \subset \Omega$,
$$
f^{ -1}(\Omega_1) \subset \Omega
$$
is also conformal in the sense that
$$
f:f^{ -1}(\Omega_1) \to \Omega_1
$$
is conformal?
If not, is it true that $f$ is the restriction of a conformal map on some larger domain?
A:
Your problem has a positive answer.
Fix a conformal coordinate system $z=(u,v)$ on $\Omega$. Then, a priori, there exists $R,\epsilon>0$ so that $f(B(0,R))\subset\Omega^\epsilon$. In fact, fixing $z_0\in B(0,R)$, we have that $f_*
u_z=
u_{f(z)}$ for all $z\in B(z_0,\epsilon)$. So $f$ is an orientation-preserving diffeomorphism on $B(z_0,\epsilon)$ (and hence on $\Omega$).
Now suppose, for some $z_0\in B(0,R)$, that $f_*
u_z$ is $C^1$ in some neighborhood $U\subset\Omega^\epsilon$. Then we have that
$$
abla_z f\cdot \frac{\partial f_*
u_z}{\partial u} =
abla_z f\cdot \frac{\partial f_*
u_z}{\partial v} =
abla_z f\cdot f_*
u_z =1
$$
so that
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