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The domain invariance theorem states that for an open set $U\subset \mathbb{R}^n$ and a continuous and injective mapping $f:U\to \mathbb{R}^n,$ the image $f(U)\subset \mathbb{R}^n$ is open. I've read that for a smooth functions the proof of the analogous statement is easy. Surely, when $f$ is continuously differentiable and for all $x\in U \ $ $\det f'(x)\ne 0$ the image $f(U)$ is open (a consequence of the inverse function theorem). But how to prove that $f(U)$ is open in the case when $f$ is continuously differentiable and injective? Is there an elementary proof?

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  • $\begingroup$ I don't recall ever seeing a proof apart from proofs of the full invariance of domain theorem for continuous functions. Where did you see this referred to as "easy"? I would love to see an easy proof! $\endgroup$ – Dan Ramras Feb 13 '14 at 18:10
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    $\begingroup$ For Completeness of the question, you should add links to the claims that the proof is simpler assuming smoothness. I suspect, the implicit (or explicit) assumption was that the derivative is invertible. $\endgroup$ – Moishe Kohan May 21 '20 at 21:12
  • $\begingroup$ There is a one-dimensional result that if $f$ is an injective differentiable function then $\{x: f'(x)=0\}$ has empty interior. The extends to higher dimensions. If you assume $C^1$, then the same set will be closed. $\endgroup$ – user123641 May 27 '20 at 15:18
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I don't know if what I'm going to state is an answer. But it reminds me that exercice: Let $f : U \subset \mathbb{R}^p \to V\subset \mathbb{R}^q$ $\mathcal{C}^1$ and injective. Prove that $p \leqslant q$ and that there exists an open dense subset $W \subset U$ on which $\mathrm{d}f(x)$ is injective for all $x$.

Thus, $f : U\subset \mathbb{R}^n \to \mathbb{R}^n$ is injective and $\mathcal{C}^1$, there would exist an open-dense subset of $U$ on which you could use the inverse function theorem. Can we then go further? I'm not sure this will help conclude.

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  • $\begingroup$ No, it does not help. $\endgroup$ – Moishe Kohan May 27 '20 at 15:36
  • $\begingroup$ An example to think about is the map $x\mapsto x^2$: It is open on ${\mathbb R} -\{0\}$ (which is open and dense). Of course, in this example the map is not injective, but it shows that openness on an open and dense subset is not nearly enough. $\endgroup$ – Moishe Kohan May 27 '20 at 21:15
  • $\begingroup$ Yes, my proposal was more about what can be said about $f$ if moreover it is injective. Of course it is a crucial hyothesis. The exercice stated above show that we can say many things on an open dense subset, and we just have to focus on the complementary! By the way I don't now how to begin on this complementary, but with the injective and $\mathcal{C}^1$ assumption.. Maybe there's something to look at $\endgroup$ – Didier May 27 '20 at 22:12
  • $\begingroup$ @DIdier_ Thanks. Do you know how to prove the statement in the exercise? $\endgroup$ – Asaf Shachar May 28 '20 at 7:15
  • $\begingroup$ Yes. It uses the constant rank theorem. The idea is to define $m = \max \mathrm{rank}\mathrm{d}u$, $W = \{ x ~|~ \mathrm{rank}\mathrm{d}u(x) = m \}$. Show $W$ is open, then with the constant rank theorem, $u$ is, in some coordinates, just the projection onto the first $m$ coefficients. The injective asumptions show that $m = p$ and that it should be lesser than $q$. Moreover, you can show that the complement of $W$ is of empty interior by re-doing this: if not, you would find an open ball where $u$ is in local coordinates a projection onto less than $p-1$ coordinates, thus non-injective. $\endgroup$ – Didier May 28 '20 at 7:21

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