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Consider the following statement of the inverse function theorem:

Let $f : W \to \mathbb{R}^n$ be a $C^1$ function from an open subset $W \subseteq \mathbb R^n$ and let $p \in W$. $f$ has a local $C^1$ inverse if $Df(p)$ is invertible.

This is effectively what one sees in Rudin's Principals of Mathematical Analysis. The question is: can the "if" can be replaced by "iff"? It seems like the answer is obviously "yes": just apply the chain rule to $f \circ g = 1$ and $g \circ f = 1$. Despite this, I can't find a statement that has the inverse function theorem in the "iff" form, so I feel like I must be making some kind of dumb mistake.

For example this question has someone saying it only holds for "low dimensions" and someone else saying that it suffices to require the inverse to be surjective, which doesn't make sense to me given the straightforward proof I gave above.

What am I missing?

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  • $\begingroup$ What you said is certainly true, the "only if" is trivially true by the chain rule, while the "if" is the non-trivial part (and what's usually called the IFT). I'm still reading through the link to see exactly what they're saying. $\endgroup$ – peek-a-boo Sep 5 at 16:23
  • $\begingroup$ @peek-a-boo What is going on in the linked question then? $\endgroup$ – Keefer Rowan Sep 5 at 16:24
  • $\begingroup$ The person who got the accepted answer changed the question. $\endgroup$ – Stephen Montgomery-Smith Sep 5 at 16:33
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    $\begingroup$ It seems that in the link, they're assuming fewer regularity on the inverse function. Here you said "local $C^1$ inverse", but in the link of the accepted answer it says "$f$ invertible and $C^1$ smooth" but makes no further assumptions on the regularity of the inverse. See also the abstract of this paper where they impose a certain Lipschitz condition on the inverse... but this is really different from what you're asking (because if you assume differentiability of the inverse this is trivial using the chain rule). $\endgroup$ – peek-a-boo Sep 5 at 16:34
  • $\begingroup$ That makes sense. $\endgroup$ – Keefer Rowan Sep 5 at 16:43

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