I am having trouble with Spivak's proof of the Inverse Function Theorem in his Calculus on Manifolds:
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. If the theorem is true for $\lambda^{-1}\circ f$, it is clearly true for $f$…
How is the theorem true for $f$ if it is true for $\lambda^{-1}\circ f$?
Best Answer
$\lambda\colon \mathbb{R}^n\to\mathbb{R}^n$ is a bijection and $\lambda$ and $\lambda^{-1}$ are both continuously differentiable. Note that $\lambda'(z) = \lambda$ for all $z \in \mathbb{R}$.
Let $g = \lambda^{-1}\circ f$. Suppose the theorem is true for $g$. Then there is an open set $V'$ containing $a$ and an open set $W'$ containing $g(a)$ such that $g:V'\to W'$ has a continuous inverse $g^{-1}:W'\to V'$ which is differentiable and for all $y\in W'$ satisfies $$(g^{-1})'(y) = [g'(g^{-1}(y))]^{-1}$$ Then $\lambda(W')$ is open and $f = \lambda\circ g:V'\to \lambda(W')$ has a continuous inverse $g^{-1}\circ \lambda^{-1}:\lambda(W')\to V'$.
By the chain rule, for all $z \in \lambda(W')$, $(f^{-1})'(z) = (g^{-1}\circ\lambda^{-1})'(z) = (g^{-1})'(\lambda^{-1}(z))\circ \lambda^{-1} = [g'(g^{-1}(\lambda^{-1}(z)))]^{-1}\circ \lambda^{-1} = [g'(f^{-1}(z))]^{-1}\circ \lambda^{-1} = [\lambda\circ g'(f^{-1}(z)]^{-1} = [f'(f^{-1}(z))]^{-1}$