What are some examples of functions which are continuous, but whose inverse is not continuous?

nb: I changed the question after a few comments, so some of the below no longer make sense. Sorry.

  • $\begingroup$ What does "bicontinuous" mean? $\endgroup$ Sep 30 '11 at 16:06
  • 5
    $\begingroup$ @Chris: Probably that the inverse is continuous. $\endgroup$
    – Ted
    Sep 30 '11 at 16:10
  • 2
    $\begingroup$ Bicontinuous is a standard definition in some topology texts. See Gamelin and Greene $\endgroup$
    – JeremyKun
    Oct 2 '11 at 21:22

A bijective map that is continuous but with non-continuous inverse is the following parametrization of the unit circle $\mathbb{S}^1$:

$$f: \colon [0, 2\pi) \to \mathbb{S}^1, \qquad f(\theta)=e^{i \theta}.$$

This map cannot have continuous inverse, because $\mathbb{S}^1$ is compact, while $[0, 2\pi)$ is not. Indeed, $f^{-1}$ jumps abruptly from $2\pi$ to $0$ when we travel round the unit circle.

Another example, somewhat similar in nature, is the map $g\colon [0,1] \cup (2, 3] \to [0, 2]$ defined by

$$g(x)=\begin{cases} x & 0 \le x \le 1 \\ x-1 & 2 < x \le 3 \end{cases}$$

The inverse map is $$g^{-1}(y)=\begin{cases} y & 0 \le y \le 1 \\ y+1 & 1 < y \le 2\end{cases}$$

and it is not continuous because of a jump at $y=1$. Note that, again, the range of $g$ is compact while the domain is not.

More generally, every bijective map $h\colon X \to K$ with $X$ non-compact and $K$ compact cannot have a continuous inverse.

  • 1
    $\begingroup$ So could we say that $h$ has continuous inverse if both $X$ and $K$ are compact? $\endgroup$ Sep 29 '14 at 15:18
  • $\begingroup$ @hermes: is it not? The point $x=2$ is excluded from the domain $\endgroup$ May 9 '16 at 9:09
  • $\begingroup$ @hermes: That interval is excluded from the domain, too. $\endgroup$ May 9 '16 at 9:54
  • $\begingroup$ @hermes: Not in the example of my answer. There, the domain of $f$ is considered to be the topological space $[0, 1]\cup (2, 3]$. $\endgroup$ May 9 '16 at 15:36
  • 2
    $\begingroup$ @hermes: I respectfully claim that you are wrong. A bijection is a one-one onto map between two sets: $f\colon A\to B$. In my example, the set $A$ is $[0, 1]\cup (2, 3]$ and the set $B$ is $[0,2]$. The assignment I gave above is a bijection of $A$ onto $B$. Please consult Wikipedia. $\endgroup$ May 9 '16 at 20:47

Define $f: [0,1) \cup [2,3] \rightarrow [0,2]$ by

$$f(x)=\begin{cases} x & x \in [0,1) \\ x-1 & x \in [2,3] \end{cases}$$

  • 1
    $\begingroup$ I don't understand how the function is continuous since it has a big jump from 1 to 2 in its domain. $\endgroup$ Jul 17 '15 at 6:49
  • 9
    $\begingroup$ @user2277550 You agree that f is continous on (0,1) and on (2,3), right? So, what about 0, 2 and 3? The Weierstrass definition of continuity says that f is continuous at c if for any ε > 0 there exists a δ > 0 such that $|x - c| < \delta \implies |f(x) - f(c)| < \varepsilon$ for any x in the domain. Note the last part. For c = 0 and c = 2, we thus only have to check that f is right continuous, and for c = 1 we check that it's left continiuous. $\endgroup$
    – Frxstrem
    Jul 21 '15 at 16:51
  • 3
    $\begingroup$ @user2277550 (This is because the numbers just to the left of 0 and 2 and the number just to the right of 3 are not in the domain of f). $\endgroup$
    – Frxstrem
    Jul 21 '15 at 16:53
  • 1
    $\begingroup$ @Frxstrem "and for $c = 1$ we check that it's left continiuous" ... Didn't you mean $c = 3$ here? After all $c=1$ is not in the domain of $f$ so... I think it does not make sense to ask if $f$ is continuous at $c=1$. Could someone confirm this? $\endgroup$ Apr 6 '20 at 13:21
  • 1
    $\begingroup$ OK, thanks, I just needed a confirmation because it is a highly upvoted comment, so I wanted to make sure I am not missing something subtle here. $\endgroup$ Apr 6 '20 at 13:42

Let $X$ be a set and $\tau_1,\tau_2$ two topologies on $X$ with $\tau_2\subsetneq\tau_1$. Then the identity function from the topological space $(X,\tau_1)$ to $(X,\tau_2)$ is a continuous bijection but the inverse function (the identity function from $(X,\tau_2)$ to $(X,\tau_1)$) is not continuous.


Let $\rm X$ be the set of rational numbers with the discrete topology. Then the identity map $\rm X\to \mathbb{Q} $ is bijective and continuous, with discontinuous inverse.

  • $\begingroup$ If the inverse map also goes to $\mathbb{Q}$, then how is it discontinuous? $\endgroup$ Sep 30 '11 at 16:23
  • 1
    $\begingroup$ A bijective continuous map is a homeomorphism if and only if it is also an open map. A singleton is open in $X$, but not in $\mathbb{Q}$ (under the metric topology). $\endgroup$
    – Zhen Lin
    Sep 30 '11 at 17:34

1) Take any topological space,
2) Obtain another space by refining its topology,
3) ...

In fact, consider the forgetful functor $F: \mathbf{Top} \to \mathbf{Set}$. For any set $S$ the continuous functions of the form $f: X \to Y$ such that $FX = FY = S$ and $Ff = 1_S$ form a linear order on the set of all topologies on $S$, and this order is in fact inverse to the usual one (formed by set inclusion of topologies).

For example, extending the answer by Marco, consider a simple curve $\gamma: I \to M$ on some manifold with the finite number of self-intersections. For each intersection, remove all corresponding points from $I$ except for one. Voila :) UPD: actually, you can remove all corresponding points, period!


Take a "8"-shaped plane curve $\mathcal C \subset \mathbb R^2$ endowed with the subspace topology. Let $\phi: \mathbb R \to \mathcal C$ be a continuous injective parametrization of $\mathcal C$. The inverse function $\phi^{-1}: \mathcal C\to\mathbb R$ cannot be continuous because $\phi^{-1}((a,+\infty))$ is not an open set for some $a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.