2-11 Theorem (Inverse Function Theorem). Suppose that $f: \mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable in an open set containing $a$, and det $f'(a)\neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f:V\to W$ has a continuous inverse $f^{-1}:W\to V$ which is differentiable and for all $y\in W$ satisfies $$(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}$$
Proof. Let $\lambda$ be the linear transformation $Df(a)$. Then $\lambda$ is non-singular, since det $f'(a)\neq 0$. Now $D(\lambda^{-1}\circ f)(a) = D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$ is the identity linear transformation.
why is $D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$ ?
Best Answer
That is because the derivative of a linear transformation $L : V \subseteq \mathbb{R}^n \to W \subseteq \mathbb{R}^m$ is the linear transformation itself. In your case, $L$ is $\lambda^{-1}$. This may be a little counterintuitive but it follows directly from the definition:
The derivative of $L$ at $x \in V$, if you can find it, is some linear transformation $DL_x : V \to W$ that must satisfy: $$ \lim_{h \to \overrightarrow{\mathbf{0}}}\Big| \frac{L(x + h) - L(x) - DL_x(h)}{h} \Big| = 0 $$ But $L$ is linear so $L(x + h) - L(x) = L(x) + L(h) - L(x) = L(h)$. So in fact you are just looking for a linear transformation $DL_x$ such that $$ \lim_{h \to \overrightarrow{\mathbf{0}}}\Big| \frac{L(h) - DL_x(h)}{h} \Big| = 0 $$ Well, $L$ itself is a linear transformation. So what if you do the obvious thing and let $DL_x$ equal $L$? Then the condition above is indeed satisfied because the numerator becomes just $\overrightarrow{\mathbf{0}}$. So defining $DL_x = L$ works and we have discovered that the derivative $DL_x$ of a linear transformation $L$, at any arbitrary point $x$, is $L$ itself.