$1$)Let $A\subseteq\mathbb{R}^n$ be an open set and $f:A\rightarrow \mathbb{R}^n$ a continuously differentiable $1-1$ function such that $det f'(x)\neq 0$ for all $x$. Show that $f(A)$ is an open set and $f^{-1}: f(A )\rightarrow A$ is differ entiable. Show also that $f(B)$ is open for any open set $B\subseteq A$.
$2$) Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is $\mathbb{C}^{\infty}$, show that $f$ can not ne $1-1$
Could any one tell me whether my solutions are correct or not?
$1$)For every $y\in f(A)$,there is an $x\in A$, By Inverse function theorem, there is an open set $U\subsetneq A$ and open set $V\subsetneq \mathbb{R}^n$ such that $x\in U$ and $f(U)=V$, since clearly $y\in V$ $f(A)$ is open(Can I say this?) Furthermore $f^{-1}:V\rightarrow U$ is differentiable( by IFT), this implies $f^{-1}$ is differentiable at $y$ and as $y$ was chosen arbitrarily from $f(A)$ so $f^{-1}:f(A)\rightarrow A$ is differentiable.
$2$)I show the result is true even if $f$ is only defined in a non-empty open subset of$\mathbb{R}^2$, we know that $f$ is not constant in any open set,So, suppose we have $D_1f(x_0,y_0)\neq 0$ so there is a nbd $U$ of $(x_0,y_0$) where $D_1f(x,y)\neq 0\forall (x,y)\in U$ The function $g:U\rightarrow \mathbb{R}^2$ defined by $g(x,y)=(f(x,y),y)$ satisfies $det g'(x,y)\neq 0\forall (x,y)\in U$ , here $f$ and $g$ are one-to-one and We can apply the result of problem $1$, The inverse function is clearly of the form $(h(x,y),y)$ and hence $(f(h(x,y),y),y)=(x,y)\forall (x,y)\in V=\{(f(x,y),y):(x,y)\in U\}$, Now $V$ is open but each horizontal line intersects $U$ atmost once since $f$ is one-one, This is a contradiction since $U$ is non-empty and open.