I'd like an example of a function $f:(a,b)\to\mathbb R$ and a point $c\in(a,b)$ such that:

  • $f$ is invertible.
  • $f$ is continuous at $c$.
  • $f^{-1}$ is discontinuous at $f(c)$.

Motivation: There is a calculus book that states the following.

Let $f$ be an invertible function defined on an interval $I$. If $f$ is differentiable at $c\in I$ and $f'(c)\neq 0$, then $f^{-1}$ is differentiable at $f(c)$.

In the proof, the continuity of $f^{-1}$ at $f(c)$ is essential. Usually, the said essential fact is an hypothesis (if the domain is not an interval) or it is implied by the hypothesis that $f$ is continuous in a neighborhood of $c$ (if the domain is an interval). But in the said book, both hipothesis are missing and the fact is justified as follows:

As $f$ is differentiable at $c$, $f$ is continuous at $c$. Therefore, $f^{-1}$ is continuous at $f(c)$.

I suspect continuity at $c$ does not imply continuity of the inverse at $f(c)$ due to the following facts:

  1. It seems it is not a common result in analysis books.
  2. In the usual proofs that the inverse of a continuos map (on an interval or on a compact set) is continous, in order to prove that the inverse is continuous at a given point, we need the continuity of $f$ in the whole domain.
  3. In more recent editions of the said book, the statement was modified (now, it is supposed that $f$ is differentiable in a neighborhood of $c$, which implies what is needed).

However, I do not have a counterexample.

  • $\begingroup$ Does this help? math.stackexchange.com/questions/362592/… $\endgroup$
    – Andrew
    Feb 20 at 23:14
  • $\begingroup$ this is the one variable case, so if f is continuous its inverse is continuous too. Look this https://proofwiki.org/wiki/Continuous_Inverse_Theorem $\endgroup$
    – tac
    Feb 20 at 23:19
  • $\begingroup$ @AndrewZhang My question is different. I'm talkin about continuity at a single point of an interval. Thus, $f$ and $f^{-1}$ cannot be continuous (in the whole domain). $\endgroup$
    – Pedro
    Feb 20 at 23:21
  • $\begingroup$ @tac I'm not talking about continuous functions (in the entire interval). See my previous comment. Also, I referred to the linked result in item 2 of my question. $\endgroup$
    – Pedro
    Feb 20 at 23:22
  • 1
    $\begingroup$ @Taladris Not even. That's true if $f$ is continuous on an interval. But it can be defined on an interval and continuous at just one point. See my answer $\endgroup$
    – Didier
    Feb 20 at 23:44

2 Answers 2


Here is an example of such a function. It comes from the book Les contre-exemples en mathématiques by Bertrand Hauchecorne (more precisely, section 8.22, page 150, in the second edition).

Consider $g\colon \Bbb R_+ \to \Bbb R_+$ defined by $$ g(x) = \begin{cases} \frac{n}{2} & \text{if} \quad x=n \quad \text{is an even integer},\\ \frac{1}{n+2} & \text{if} \quad x=n \quad \text{is an odd integer},\\ \frac{1}{2(n-1)} & \text{if} \quad x=\frac{1}{n} \quad \text{with} \quad n \geqslant 2 \quad \text{integer},\\ x & \text{in any other case}. \end{cases} $$ Now, consider $f\colon \Bbb R \to \Bbb R$ defined as $$ f(x) = \begin{cases} g(x) & \text{if} \quad x \geqslant 0,\\ -g(-x) & \text{if} \quad x < 0. \end{cases} $$ One shows with a bit of effort that $f$ is a bijection.

Note that $|f(x)| \leqslant |x|$ for all $x$, so that $f$ is continuous at $0$ with $f(0)=0$. However, for all $n \geqslant 1$, $f(2n-1) = \frac{1}{2n+1}$ shows that $f^{-1}\left(\frac{1}{2n+1}\right) = 2n-1 \underset{n \to \infty}{\longrightarrow} \infty$, so that $f^{-1}$ is not continuous at $0$.

  • 1
    $\begingroup$ Amazing source. $\endgroup$
    – Pedro
    Feb 20 at 23:47
  • 1
    $\begingroup$ @Pedro I was advised to read this book while I was in my first or second year as a student, and that's one of the best advices I've been given ever since. $\endgroup$
    – Didier
    Feb 20 at 23:50
  • 2
    $\begingroup$ (+1) Et est-ce qu'il existe un contre-exemple avec en plus $f'(0)$ existe et $\ne0?$ $\endgroup$ Feb 21 at 0:16
  • 1
    $\begingroup$ @AnneBauval Très bonne question. À cette heure-ci, je ne saurais répondre $\endgroup$
    – Didier
    Feb 21 at 0:23
  • 3
    $\begingroup$ @AnneBauval: Oui ! See the answer that I've posted. $\endgroup$ Feb 21 at 10:33

Here's a bijection $f$ from the interval $(-2,2)$ to the interval $(-\frac32,\frac32)$, such that $f$ is continuous at $0$, and in fact even differentiable there with $f'(0)=1$, but $f^{-1}$ is discontinuous at $f(0)=0$.

Define $f$ on the interval $(-1,1)$ by $$ f(0) = 0 $$ and $$ f(x) = \left( 1 + \frac{1}{n} \right) x ,\quad\text{for}\quad \frac{1}{n} \le |x| < \frac{1}{n-1} \quad (n = 2,3,\dots) . $$ For $1 \le |x| < 2$, define $f$ in a piecewise linear way, such that it assumes precisely those values that it previously jumped over, as indicated by this picture:

  • $\begingroup$ Very nice. The picture is illuminating. $\endgroup$
    – Pedro
    Feb 21 at 11:23
  • $\begingroup$ What a nice example! $\endgroup$
    – Didier
    Feb 21 at 12:06

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