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I'm given that:=
$E\subseteq \mathbb R^n$ be open and $f:E\to \mathbb R^n$ be a $C^1$ map . Suppose that for some $a\in E$ , the linear map $f'(a)$ is invertible ,and $b=f(a)$ .Then :=
I've to show that :=

There are open set $U$ and $V$ in $\mathbb R^n$ such that $a\in U,b\in V$ and $f|_{U}$ is one-one and onto $V$ i.e. $f(U)=V$ .

The proof in my notes involves

$1.)$ first proving $f:U\to \mathbb R^n$ is $1$-$1$ .

which I understood.. I can't understand the motive of the remaining part of the proof whose outline I'm presenting below:

$2.)$ Outline of remaining part of proof:
Let $V=f(U)$ then $b=f(a)\in V$ .Then we show that $V$ is open . Let $y_0=f(x_0)\in V $ and let $r\gt 0$ is such that $B=N(x_0,r)\subseteq U$ with $\overline B\subseteq U$.
We show that :
$N(y_0,\epsilon r)\in V$ and so $y_0$ is interior point of $V$ .Hence,then $V$ is open..

I'll be obliged if there is someone kind enough to explain me the motive behind $2.)$ and also I can't understand why nowhere in the proof did we prove $f:U\to \mathbb R^n$ is onto , whereas it was proved in the beginning of the proof that it is $1$-$1$..

Thanks in advance for any help...

coool
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1 Answers1

1

We don't prove that $f:U\to \Bbb R^n$ is onto, because it is false. Consider $\sin x$ with $U$ being a neighbourhood of $x=0$.

In the first step you prove that there exists an open set $U$ on which the function $f$ is injective.

Then you recall a basic theorem that for any injective function with arrival set equal to its range read ($f: U\to V$) is invertible.

In the second step you prove that the image $V=f(U)$ of this set is also open.

The motivation behind the whole theorem is to prove that if the differential of a function is invertible at some point (i.e. its a matrix of non-zero determinant), then the function itself is invertible is the small neighbourhood of that point.

TZakrevskiy
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  • as you said $f:U\to \Bbb R^n$ is onto is wrong...but the statement of my theorem says that $f|_{U}$ is one-one and onto. – coool Nov 05 '14 at 08:39
  • @coool the statement say that $f\big|_U$ is one-to-one and (onto $V$). The notion of onto depends on the arrival set. $\sin x:[0,\pi/2]\to [0,1]$ is onto, but $\sin x:[0,\pi/2]\to \Bbb R$ is not onto. – TZakrevskiy Nov 05 '14 at 08:41
  • I've another doubt that in second step showing that $V=f(U)$ .I can't understand how showing that Let $y_0=f(x_0)\in V$ and let $r\gt 0$ is such that $B=N(x_0,r)\subseteq U$ with $\overline B$⊆$U$. We show that : $N(y_0,\epsilon r)\in V$ and so $y_0$ is interior point of $V$ . proves $V$ is open.. I can't understand this proving that $V$ is open ...Please help,.. – coool Nov 05 '14 at 08:46
  • This is a definition of an open set. If for any point $x$ in the set $X$ there exists a ball $N(x,r)$ of radius $r$ centered in $x$ such that $N(x,r)\subset X$, then $X$ is called an open set. – TZakrevskiy Nov 05 '14 at 08:54
  • just one last doubt before accepting answer : why did we choose $N(y_0,\epsilon r) $ to belong to $V$ .. and what is $\overline B$ in $\overline B\subseteq U$.. – coool Nov 05 '14 at 08:58
  • Erm, as per above - to show that the set is open. $\bar B$ seems to be the closure of a set $B$. – TZakrevskiy Nov 05 '14 at 09:03
  • but why only $\epsilon r$ ?in $N(y_0,\epsilon r)$ – coool Nov 05 '14 at 09:04
  • Well, given that you control $\epsilon$, you can say $\epsilon r$ is an arbitrary positive number. – TZakrevskiy Nov 05 '14 at 09:07