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.
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- $\begingroup$ What does "bicontinuous" mean? $\endgroup$Chris Eagle– Chris Eagle2011-09-30 16:06:48 +00:00Commented Sep 30, 2011 at 16:06
- 6$\begingroup$ @Chris: Probably that the inverse is continuous. $\endgroup$Ted– Ted2011-09-30 16:10:05 +00:00Commented Sep 30, 2011 at 16:10
- 3$\begingroup$ Bicontinuous is a standard definition in some topology texts. See Gamelin and Greene $\endgroup$JeremyKun– JeremyKun2011-10-02 21:22:19 +00:00Commented Oct 2, 2011 at 21:22
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6 Answers
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10 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? $\endgroup$A. Thomas Yerger– A. Thomas Yerger2014-09-29 15:18:24 +00:00Commented Sep 29, 2014 at 15:18
- $\begingroup$ @hermes: is it not? The point $x=2$ is excluded from the domain $\endgroup$Giuseppe Negro– Giuseppe Negro2016-05-09 09:09:58 +00:00Commented May 9, 2016 at 9:09
- $\begingroup$ @hermes: That interval is excluded from the domain, too. $\endgroup$Giuseppe Negro– Giuseppe Negro2016-05-09 09:54:14 +00:00Commented May 9, 2016 at 9:54
- 3$\begingroup$ @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. $\endgroup$Giuseppe Negro– Giuseppe Negro2016-05-09 20:47:55 +00:00Commented May 9, 2016 at 20:47
- 1$\begingroup$ @CloudJR: It is not. As I said, the target space in your example is (0,1), which is not compact. $\endgroup$Giuseppe Negro– Giuseppe Negro2018-10-06 16:47:05 +00:00Commented Oct 6, 2018 at 16:47
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13 Define $f: [0,1) \cup [2,3] \rightarrow [0,2]$ by
$$f(x)=\begin{cases} x & x \in [0,1) \\ x-1 & x \in [2,3] \end{cases}$$
- 3$\begingroup$ I don't understand how the function is continuous since it has a big jump from 1 to 2 in its domain. $\endgroup$user2277550– user22775502015-07-17 06:49:11 +00:00Commented Jul 17, 2015 at 6:49
- 11$\begingroup$ @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. $\endgroup$Freyja– Freyja2015-07-21 16:51:55 +00:00Commented Jul 21, 2015 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). $\endgroup$Freyja– Freyja2015-07-21 16:53:01 +00:00Commented Jul 21, 2015 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? $\endgroup$peter.petrov– peter.petrov2020-04-06 13:21:16 +00:00Commented Apr 6, 2020 at 13:21
- 2$\begingroup$ 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. $\endgroup$peter.petrov– peter.petrov2020-04-06 13:42:35 +00:00Commented Apr 6, 2020 at 13:42
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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.
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2 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.
- 1$\begingroup$ If the inverse map also goes to $\mathbb{Q}$, then how is it discontinuous? $\endgroup$isomorphismes– isomorphismes2011-09-30 16:23:37 +00:00Commented Sep 30, 2011 at 16:23
- 3$\begingroup$ 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– Zhen Lin2011-09-30 17:34:16 +00:00Commented Sep 30, 2011 at 17:34
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1) Take any topological space,
2) Obtain another space by refining its topology,
3) ...
4) PROFIT!
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!
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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 $a$.
