Spivak's proof of Inverse Function Theorem

$\lambda\colon \mathbb{R}^n\to\mathbb{R}^n$ 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 $z \in \lambda(W')$, $(f^{-1})'(z) = (g^{-1}\circ\lambda^{-1})'(z) = (g^{-1})'(\lambda^{-1}(z))\circ \lambda^{-1} = [g'(g^{-1}(\lambda^{-1}(z)))]^{-1}\circ \lambda^{-1} = [g'(f^{-1}(z))]^{-1}\circ \lambda^{-1} = [\lambda\circ g'(f^{-1}(z)]^{-1} = [f'(f^{-1}(z))]^{-1}$

Suppose the theorem holds for all functions $f$ such that $Df(a) = I$. What Spivak has shown is that for $\lambda = Df(a)$, $D(\lambda^{-1} \circ f(a)) = I$. By the assumption above, the theorem must be true for $\lambda^{-1}\circ f$. That means there exists $g$ which is the continuously differentiable inverse of $\lambda^{-1} \circ f$. This implies that $(g \circ \lambda^{-1}) \circ f = I$. Now $g \circ \lambda^{-1}$ is continuously differentiable, and so the theorem is true for $f$.