# Inverse Function Theorem, Spivak's Proof

I'm having a lot of trouble following the proof of the following theorem. This is from Spivak's 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 let $$\det f'(a) \not= 0$$. 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 $$(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}.$$

The proof starts off with the following.

Proof. Let $$\lambda$$ be the linear transformation $$Df(a)$$. Then $$\lambda$$ is non-singular, since $$\det f'(a) \not= 0$$. 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 $$f$$. Therefore we may assume at the outset that $$\lambda$$ is the identity.

The problem I'm having is that I don't quite see where he's trying to go with this argument. I saw another post (Spivak's proof of Inverse Function Theorem) that explained the statement

If the theorem is true for $$\lambda^{-1}\circ f$$, it is clearly true for $$f$$.

but exactly why is it necessary for this proof? Also, when it says

Therefore we may assume at the outset that $$\lambda$$ is the identity.

why are we assuming that $$\lambda$$ is the identity?

• It will simplify the proof if we assume $Df (a)$ is the identity. And, once we prove the theorem in the special case where $Df (a)$ is the identity, the more general case where $Df (a)$ is not the identity will be an easy corollary. Jul 17, 2015 at 1:50

I'm guessing Spivak is making the assumption that $\lambda = I$ because it simplifies the notation greatly. You could try recreating the proof without this assumption, with $\lambda$ cropping up everywhere, and you should be able to see why Spivak has done this.