This is the theorem:
9.24 Theorem Suppose $f$ is a continuously differentiable map of an open set $E
\subseteq \mathbb{R}^n$ into $\mathbb{R}^n$, $f'(a)$ is invertible for
some $a \in E$ and $b= f(a)$. Then(a) There exists open sets $U,V$ in $\mathbb{R}^n$ with $a\in U, b\in V$ such that $f: U \to V$ is a bijection.
(b)$f^{-1}$ is continuously differentiable.
From the second part of the proof:
Pick $y, y+k\in V$. Then there exist $x,x+h\in U$ so that $y=f(x),y+k=f(x+h)$. Then
$$f^{-1}(y+k)-f^{-1}(y)-Tk=h-Tk=-T(f(x+h)-f(x)-f'(x)h)$$, where $T= [f']^{-1}$
I fail to attain the second equality. For me
$$
\begin{align}
h-Tk
&= TT^{-1}h-Tk \\
&= T(f'(h))-T(f(x+h)-f(x)) \\
&= -T(f(x+h)-f(x)-f'(h))\\
&\neq -T(f(x+h)-f(x)-f'(x)h)
\end{align}
$$
Can anyone explain to me why the second equality hold?
Best Answer
@Sakamoto
The error is $T^{-1}h=f'(h)$.
Observe that $T^{-1}$ stands for the linear operator with matrix $f'(x)$ (the Jacobian matrix at $x$) and therefore $$ T^{-1}(h)=f'(x)h $$ Rudin's notation here is not well chosen in my opinion. The derivative is taken at the base point $x$ and acts on the tangent vectors $h$. It should be $T=[f'(x)]^{-1}$, etc. Hope this helps.