# Is monotonicity a necessary condition for the inverse function theorem?

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?

• 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. – Conifold Jun 2 at 4:09
• @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? – D. Soul Jun 2 at 4:20
• 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. – Conifold Jun 2 at 4:24
• @Conifold Thank you! – D. Soul Jun 2 at 7:11

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.
• 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? – D. Soul Jun 2 at 4:21
• @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. – Clarinetist Jun 2 at 4:23
• @D.Soul over the interval $f(I)$ of course, and not $\mathbb{R}$. – Clarinetist Jun 2 at 4:26