# Extending a diffeomorphism onto a larger set

Let $$K\subset\mathbb{R}^d$$ be compact such that $$K^\circ\neq\emptyset$$. Let $$U\subset\mathbb{R}^d$$ bounded and open ($$U$$ could e.g. be chosen to be the open unit ball). Let $$Φ: K^\circ\to U$$ be a diffeomorphism.

Can I find a set $$V \supset K$$ open and a diffeomorphism $$\Psi: V \to \Psi(V)$$ such that $$\Psi|_{K^\circ} = \Phi$$? The range of $$\Psi$$ is of no great importance to me, I only care that it's a proper superset of $$U$$.

A similar question was asked at Extending a diffeomorphism outside a compact set where a counterexample was given for non-contractable sets. However, the question asked for an extension onto the whole space $$\mathbb{R}^d$$, whereas I'm only interested to extend $$\Phi$$ a little bit outside of $$K$$.

The answer is, in full generality, no. I will use Riemann's uniformisation Theorem:

If $$U$$ and $$V$$ are non-empty simply connected open subsets of $$\mathbb{C}$$, different than $$\mathbb{C}$$, then there exists a biholomorphism $$\varphi\colon U \to V$$. In particular, such a biholomorphism is a diffeomorphism.

Consider $$\Delta = \left\{ z \in \mathbb{C} \mid |z|<1 \right\}$$ and $$U$$ the interior of a Von Koch curve. Then $$\Delta$$ and $$U$$ are non-empty simply connected open subsets of $$\mathbb{C}$$: there exists a diffeomorphism $$\varphi\colon \Delta \to U$$.

Suppose by contradiction that $$\varphi$$ extends to $$\psi \colon V \to \psi(V)$$ as a diffeomorphism with $$\overline\Delta \subset V$$. Then $$\psi(\partial\Delta) = \partial \psi(\Delta) = \partial U$$ and $$\psi$$ maps the unit circle to the Von Koch curve. But a diffeomorphism sends smooth curves to smooth curves, which is a contradiction. It follows that $$\psi$$ does not exist.

Comment: there is no reason for a smooth / conformal / holomorphic map to extend to the boundary of a domain, as shows the above example. We indeed could have chosen a square, heptagone or any non-smooth curve instead of the Von Koch curve, but this example shows how pathological it can be (and in fact, it can be even worse).

• In your definition of $\Delta$, did you intend a strict inequality $<$ instead of equality? Commented Aug 25, 2021 at 9:00
• @peek-a-boo Yes of course. Thank you for pointing that out! Commented Aug 25, 2021 at 9:02
• Thanks for the counter example! Would the result hold, if we assumed more regularity of $K$? Maybe $C^1$ boundary or just "$K$ such that $\Phi$ extends to $\partial K$ in a suitable sense". Commented Aug 25, 2021 at 9:12
• @CallMeStag To be fair, I don't know. Extending conformal maps to the boundary is a whole field of research in complex geometry / CR geometry, and what I remember from a course in geometry is that "we don't know much things about extending maps to boundaries". Commented Aug 25, 2021 at 9:15
• All right, thanks anyway! Commented Aug 25, 2021 at 9:16