# Domain invariance for smooth functions

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?

• 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! Feb 13, 2014 at 18:10
• 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. May 21, 2020 at 21:12
• 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.
– user123641
May 27, 2020 at 15:18

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.

• No, it does not help. May 27, 2020 at 15:36
• 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. May 27, 2020 at 21:15
• 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 May 27, 2020 at 22:12
• @DIdier_ Thanks. Do you know how to prove the statement in the exercise? May 28, 2020 at 7:15
• 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. May 28, 2020 at 7:21