Let $\Omega$ be an open set of $\mathbb{R}^n$ and let $f:\Omega\to\mathbb{R}^n$ a function of $\mathscr{C}^r(r\in\mathbb{N}\cup\{\infty\})$. Let $a\in\Omega$ such that $f´(a):\mathbb{R}^n\to\mathbb{R}^n$ be a linear isomorphism and let $b=f(a)$. So there is an open neighbourhood $U$ of $a$ in $\Omega$, and there is an open neighbourhood $V$ of $b$ in $\mathbb{R}^n$ such that $f(U)=V$ and $f:U\to V$ is a diffeomorphism of $\mathscr{C}^r class$.
Proof $f$ is locally injective
Let $A=f´(a)$ and let $\lambda=\frac{1}{2||A^{-1}||}$
As $f´$ is continuous in $a$ there exists an open connected $U$ that contains $A$ and is contained in $\Omega$ such that $||f(x)-A||<\lambda\:\:\forall x\in U$ Given $y\in\mathbb{R}^n$ consider the function $\psi=\psi_y$ defined as:
$\psi_y:U\to\mathbb{R}^n\\x\to x+A^{-1}(y-f(x))$
Note $f(x)=y$ iff $\psi(x)=x$. For each $x\in\mathbb{R}^n$
$\psi´(x)=A^{-1}(A-f´(x))$ so that $||\psi´(x)||<\frac{1}{2}\:\:\forall x\in U$
By the intermdeite value theorem:
$||\psi(x_1)-\psi(x_2)||\leqslant\frac{1}{2}||x_1-x_2||\:\:\forall x_1,x_2\in U$.
By Banach fixed point Theorem we have $x\in U$ such that $\psi(x)=x$.
Questions:
1)I guess $\psi´(x)=A^{-1}(A-f´(x))$ is wrong once $(A^{-1}f(x))´=-A^{-2}f´(x)$, right?
2)Why does the author goes for this method to prove the function is locally injective? Why does not the author take the derivative as in the inverse theorem in one dimension? Why complicate so much?
Thanks in advance!