- #1

psie

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- TL;DR Summary
- I have two questions on Spivak's proof of the inverse function theorem in Calculus on Manifolds.

I know of a thread on this site with a similar question, but no definite answer. I will not state the whole proof, as it is quite long.

1. Why can we assume ##f## to have the identity map as its derivative? I understand how if the theorem is true for ##g = \lambda^{-1} \circ f##, then its true for ##f## and I also understand that ##Dg(a)=\mathrm{id}##, where ##\mathrm{id}## is the identity map. If ##\lambda=\mathrm{id}##, then simply ##g=f## and this is how the proof proceeds. But what when ##\lambda\neq \mathrm{id}##?

2. Why is $$|x_1-x_2|-|f(x_1)-f(x_2)|\leq |f(x_1)-x_1-(f(x_2)-x_2)|,$$ true? I'm tempted to say it is the reverse triangle inequality, but shouldn't there be an extra pair of absolute value signs around the LHS of this inequality?

. 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 $$(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}$$2-11 Theorem (Inverse Function Theorem). 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##. Therefor we may assume at the outset that ##\lambda## is the identity.Proof

... ... ...

Note that (3) and Lemma 2-10 applied to ##g(x)=f(x)-x## imply for ##x_1,x_2\in U## that $$|f(x_1)-x_1-(f(x_2)-x_2)|\leq\frac12|x_1-x_2|.$$ Since \begin{align*}

|x_1-x_2|-|f(x_1)-f(x_2)|&\leq |f(x_1)-x_1-(f(x_2)-x_2)| \\ &\leq\frac12 |x_1-x_2|,

\end{align*}

... ... ...

1. Why can we assume ##f## to have the identity map as its derivative? I understand how if the theorem is true for ##g = \lambda^{-1} \circ f##, then its true for ##f## and I also understand that ##Dg(a)=\mathrm{id}##, where ##\mathrm{id}## is the identity map. If ##\lambda=\mathrm{id}##, then simply ##g=f## and this is how the proof proceeds. But what when ##\lambda\neq \mathrm{id}##?

2. Why is $$|x_1-x_2|-|f(x_1)-f(x_2)|\leq |f(x_1)-x_1-(f(x_2)-x_2)|,$$ true? I'm tempted to say it is the reverse triangle inequality, but shouldn't there be an extra pair of absolute value signs around the LHS of this inequality?