I read Munkres' analysis on manifolds. In the book, with another condition:$\det Df(x)\ne 0, \ \forall x\in A$, it proves that $f(A)$ is open in $\mathbb R^n$.
However, I wonder whether there exists another way to show $f(A)$ is an open set in $\mathbb R$ by only using the two conditions without using $\det Df(x)\ne 0$.
I tried for some time, but it seems either it's too hard for me or with only two conditions, it's impossible to show $f(A)$ is open.
Any help on this? Thank you!
There is a much more general theorem known under the name "invariance of domain". It says that each continuous injection $f : A \to \mathbb R^n$ has an open image $f(A)$. See here.
Usually this theorem is proved by methods from algebraic topology - which is not elementary. There are also other proofs (see Is there some elementary proof of invariance of domain?), but they are also non-elementary.
If you only require $C^1$ without $\det Df(x) \ne 0$ for all $a \in A$, then I doubt that you can give an easy proof that $f(A)$ is open.
