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In C.H. Edwards's Advanced Calculus of Several Variables we find

Corollary III 2.8 Let $f:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be $\mathcal{C}^{1}$ at $a$. If $df_{a}:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is one-to-one, then $f$ itself is one-to-one on some neighborhood of $a$.

Which says to me: If the differential of a mapping is non-singular at a point, then the mapping is invertible near that point. That is, a one-to-one $\mathcal{C}^{1}$ function on some neighborhood is invertible on some neighborhood contained in the first neighborhood.

Then, after much further development, we encounter the inverse mapping theorem:

Theorem III 3.3 Suppose that the mapping $f:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is $\mathcal{C}^{1}$ in a neighborhood $\mathcal{W}$ of the point $a$, with the matrix $f^{\prime}\left(a\right)$ being nonsingular. Then $f$ is locally invertible--- there exist neighborhoods of $a$ $\mathcal{U}\subset\mathcal{W}$ and $\mathcal{V}$ of $b=f\left(a\right)$, and a one-to-one mapping $g:\mathcal{V}\to\mathcal{W}$ such that

$$ g\left(f\left(x\right)\right)=x\text{ for }x\in\mathcal{U}, $$

and

$$ f\left(g\left(y\right)\right)=y\text{ for }y\in\mathcal{V}. $$

In particular, the local inverse of $g$ is the limit of the sequence $\left\{ g_{k}\right\} _{0}^{\infty}$ of successive approximations defined inductively by

$$ g_{0}\left(y\right)=a,g_{k+1}\left(y\right)=g_{k}\left(y\right)-f^{\prime}\left(a\right)^{-1}\left[f\left(g_{k}\left(y\right)\right)-y\right] $$

for $y\in\mathcal{V}.$

The contraction mapping aspect is probably valuable, but it seems to me, the essential invertibility part of the inverse mapping theorem has already been established in Corollary III 2.8. Other than the fact that the inverse mapping theorem states its hypothesis as $f$ is $\mathcal{C}^1$ in a neighborhood of $a$, as opposed to only at $a$, I don't see a major difference. To me, saying $f^{\prime}\left(a\right)$ is non-singular is just fancy talk for saying $df$ is one-to-one. For example, Nering defines a non-singular linear transformation as one which is invertible. The associated matrix is also shown to have a non-zero determinant. And one-to-one and onto implies invertibility. I'm pretty sure the property of being $\mathcal{C}^1$ at $a$ ensures me that the mapping is onto in some neighborhood of $a$.

Is my understanding of this correct? In other words, does Corollary 2.8 establish the invertibility of $f$ near $a$ based on the hypotheses of Theorem 3.3?

Steven Thomas Hatton
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  • I think there is a subtle difference. In the corollary, there is no guarantee about what the image of $f$ looks like. If you take a neighbourhood $U$ of $a$, you don't know whether $f(U)$ is/contains a neighbourhood of $f(a)$ or not (i.e. you don't know whether $f(U)$ is an open set or not), even though $f$ is bijective between $U$ and $f(U)$. – user1551 Oct 28 '19 at 18:13
  • For $f$ to be differentiable at $a$ doesn't it need to be continuous near $a$? – Steven Thomas Hatton Oct 28 '19 at 18:39

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One very important part of the inverse mapping theorem is the continuous differentiability of $f^{-1}$ in a neighbourhood of $f(a)$, so that $f$ is a local diffeomorphism. This is probably the reason why you need differentiability in a neighbourhood of $a$; only if you know that $f$ is $C^1$ near $a$, you can follow that $f^{-1}$ is $C^1$ near $f(a)$, too.

Im wondering why this isn't part of the theorem in your book.

Zimbo
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  • What you seem to be saying is the distinction between $\mathcal{C}^1$ at $a$, and $\mathcal{C}^1$ near $a$ is significant. But that that begs the question as to why. – Steven Thomas Hatton Oct 28 '19 at 18:34
  • You need $C^1$ near $a$ to know that $f^{-1}$ is $C^1$ near $f(a)$, too. – Zimbo Oct 28 '19 at 18:46
  • If you add that to your answer I will accept it. I'm still left wondering if there is not a more direct way of establishing this than the very arcane contraction mapping proof given by Edwards. It's instructive to study the proof, but it seems as though the essentials are obscured by the details. Perhaps I should read Spivak's Calculus on Manifolds. – Steven Thomas Hatton Oct 28 '19 at 18:55