In the book of Analysis on Manifolds by Munkres, at page 71, it is asked that
Let $A$ be open in $\mathbb{R}^n $; let $f : A \to \mathbb{R}^n $ be of class $C^r$; assume $Df(x)$ is non-singular for $x\in A$. Show that even if $f$ is not one-to-one on $A$, the set $B = f(A)$ is open in $\mathbb{R}^n $.
I have argued that:
By the inverse function theorem, for $x \in A$, $\exists U_x$ s.t $U_x \subseteq A$ and $U_x$ is open, and $f$ is one-to-one on $U_x$.Moreover, we have $\cup_{x\in A} U_x = B $. Since $B$ is the union of the open sets, $B$ is also open.
Is there any problem in this argument ? Could someone provide me a feedback, please.
It is not quite correct. By the inverse function theorem, for each $x\in A$ there is a neighborhood $U_x \subset A$ of $x$ such that $f(U_x)$ is open and $f:U_x \to f(U_x)$ is diffeomorphism. Then $B= \cup_x f(U_x)$ is open because each $f(U_x)$ is open.
Notice that it is not enough for $f$ to be one to one at $U_x$, you need $f(U_x)$ to be open.