A textbook I was reading, Introduction To Real Analysis By Robert G. Bartle (page 169) states that the inverse theorem is defined as:

Let $I$ be an interval in $\mathbb{R}$ and let $f: I \rightarrow \mathbb{R}$ be strictly monotone and continuous on $I$. Let $J := f(I)$ and let $g: J \rightarrow \mathbb{R}$ be the strictly montone and continuous function inverse to $f$. If $f$ is differentiable at $c \in I$ and $f'(c) \neq 0$, then $g$ is differentiable at $d := f(c)$ and $$ g'(d) = \frac{1}{f'(c)} = \frac{1}{f'(g(d))} $$

However, in the proof given using Caratheodory's theorem in the book, the fact of the functions $f$ or $g$ being monotone was not used. Also, it seems a bit strange that only this should apply to only monotone functions. I can understand that if $f$ is monotone, then $g$ is monotone by continuous inverse theorem. But is this really necessary for the inverse function theorem to be used?

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    $\begingroup$ Yes, it is necessary. A continuous function which has an inverse is strictly monotone, so without monotonicity the inverse is undefined. You can generalize by selecting a branch of (multivalued) inverse in a neighborhood of $d$, but this directly reduces to this version by selecting a small enough interval around $c$ where $f$ is strictly monotone. $\endgroup$ – Conifold Jun 2 at 4:09
  • $\begingroup$ @Conifold So in the cases where $f$ is not strictly montone, if I would like to find the inverse of a point $c$, I need to restrict the interval of $f$ to a neighbourhood of $c$ where $f$ is strictly montone? But wouldn't this be a bit strange cause we know that non monotonic functions e.g. $f(x) = \sin(x)$ is not strictly monotone but yet we know that its inverse $g(y) = \arcsin(y)$ exists? $\endgroup$ – D. Soul Jun 2 at 4:20
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    $\begingroup$ The $\arcsin(y)$ is not the inverse of $\sin(x)$, it is only the inverse of $\sin(x)$ restricted to $[-\pi/2,\pi/2]$, where it is strictly monotone. This is the "small enough interval" around $0$, say. $\endgroup$ – Conifold Jun 2 at 4:24
  • $\begingroup$ @Conifold Thank you! $\endgroup$ – D. Soul Jun 2 at 7:11

You must have a different edition of Bartle's text than I do (I have the 3rd edition, p. 164) - Bartle says here, in defining the inverse of a function:

If $f$ is a continuous strictly monotone function on an interval $I$, then its inverse function $g = f^{-1}$ is defined on the interval $J = f(I)$ and satisfies the relation $g(f(x)) = x$ for $x \in I$.

What this means, then, is that the inverse of $f$ may only be defined if $f$ is strictly monotone and continuous. You thus cannot claim the existence of the function $g$ in the claim you've provided above unless $f$ is strictly monotone and continuous.

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  • $\begingroup$ Hi Thanks! But what about functions like $f(x) = \sin (x)$ where it is not monotone but yet we know the inverse $g(y) = \arcsin(y)$ exists? $\endgroup$ – D. Soul Jun 2 at 4:21
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    $\begingroup$ @D.Soul Per the definition, choose an interval $I$ in which $f$ is strictly monotone. For the $\sin$ example, by convention, we choose the interval $I = [-\pi/2, \pi/2]$, in which $f$ is continuous and strictly monotone, hence the inverse exists. $\endgroup$ – Clarinetist Jun 2 at 4:23
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    $\begingroup$ @D.Soul over the interval $f(I)$ of course, and not $\mathbb{R}$. $\endgroup$ – Clarinetist Jun 2 at 4:26
  • $\begingroup$ Thank you for your help! $\endgroup$ – D. Soul Jun 2 at 7:11

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