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Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Let $f:D \to D$ be a smooth bijective map with everywhere invertible differential.

Is $f$ a diffeomorphism of the closed disk?

Here is what I know: The assumptions imply* that $f(D^o) \subseteq D^o$. Since $f$ is surjective, we also have $f(\partial D) \supseteq \partial D$.

The problem is that I am not sure that $f(\partial D) \subseteq \partial D$. If this was the case, then $f$ would be a diffeomorphism.

Note that I assume that $f$ is smooth on entire closed disk (with the boundary).

*See here.

Asaf Shachar
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2 Answers2

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Since $f:D\rightarrow D$ is a continuous bijection and $D$ is compact, we derive that $f$ is a homeomorphism. Let $g = f^{-1}$. By invariance of domain we derive that $g(D^0)\subseteq D^0$. Hence we have (as you noticed) $$\partial D\subseteq g(\partial D)$$ Thus $$f(\partial D) \subseteq f(g(\partial D)) = \partial D$$

Slup
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Here is an alternative to using invariance of domain.

We know that $f : D \to D$ must be a homeomorphism. Assume that $f(x) \in \partial D$ for some $x \in D^0$. Then $D \setminus \{x\}$ is homeomorphic to $D \setminus \{f(x)\}$. But $D \setminus \{f(x)\}$ is contractible and $D \setminus \{x\}$ is homotopy equivalent to the circle $S^1$. This is impossible.

Thus $f(D^0) \subset D^0$ and similarly $f^{-1}(D^0) \subset D^0$. Hence $f(D^0) = D^0$.

Paul Frost
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