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$1$)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.

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up vote 2 down vote accepted

1) The statement "$f^{-1}\colon f(A)\to A$ is differentiable" is false. For example, take $n=2$ and $f(x,y)=(e^x\cos y, e^x \sin y)$. This map satisfies all the assumptions with $A=\mathbb R^2$, and we have $f(A)=\mathbb R^2$. Yet, there is no inverse $f^{-1}\colon f(A)\to A$.

What is true is that for sufficiently small neighborhoods there is an inverse, and this inverse is differentiable. Your proof is fine.

2) is fine.

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