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Apologize if this is a dumb and/or well-known question.

Let $n\geq 1$ be a positive integer, $f: \mathbf{D}^n\rightarrow\mathbf{D}^n$ be a continuous injection that is identity on the boundary $\mathbf{S}^{n-1}=\partial \mathbf{D}^n$. Must $f$ be a homeomorphism?

I am aware of this question, from which it follows that $f$ is surjective, hence bijective since $f$ is assumed to be injective. But I do not see a way of showing that $f^{-1}$ is continuous.

aaa acb
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  • Isn't an embedding by definition a homeomorphism from $D^n$ to $f(D^n)$? In that case you would need to show $D^n\subset f(D^n)$. Maybe via contradiction? (Suppose not, ...) – user408858 Apr 25 '24 at 02:36
  • Ah that’s right (embedding = homeomorphism onto image with induced topology)! Let me change embedding to continuous injection. Thank you @user408858. – aaa acb Apr 25 '24 at 02:45

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This boils down to (a special case of) the following basic result:

Proposition Let $f : X \to Y$ be a continuous bijection where $X$ is compact and $Y$ is Hausdorff. Then $f$ is a homeomorphism.

Proof. Let $C \subseteq X$ be closed. Since $X$ is compact, $C$ is compact, and so $f(C)$ is compact. Since $Y$ is Hausdorff, $f(C)$ is closed. This shows that $f$ is a closed map, and since $f$ is a bijection, we conclude that $f$ is an open map. Now $f$ is a continuous open bijection, a.k.a. a homeomorphism. $\square$

diracdeltafunk
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