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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.

  • 0
    What does "bicontinuous" mean?2011-09-30
  • 5
    @Chris: Probably that the inverse is continuous.2011-09-30
  • 1
    Bicontinuous is a standard definition in some topology texts. See Gamelin and Greene2011-10-02

6 Answers 6

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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!

54

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
    So could we say that $h$ has continuous inverse if both $X$ and $K$ are compact?2014-09-29
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    @hermes: is it not? The point $x=2$ is excluded from the domain2016-05-09
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    @hermes: That interval is excluded from the domain, too.2016-05-09
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    @hermes: Not in the example of my answer. There, the domain of $f$ is considered to be the topological space $[0, 1]\cup (2, 3]$.2016-05-09
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    @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](https://en.wikipedia.org/wiki/Bijection).2016-05-09
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    Thanks for ur help2018-10-06
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    you write in last paragraph as ** More generally, every bijective map h:X→K with X non-compact and K compact cannot have a continuous inverse.** But let f:(0,1)→(0,1) by f(x)=x .this is bicontinuous ...please consider my comment2018-10-06
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    @CloudJR: Yes. That's because $(0,1)$ is not compact.2018-10-06
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    @but the function i gave is bicontinuous but what you said is that every bijective map h:X→K with X non-compact and K compact cannot have a continuous inverse... Which is a contradiction (counter example) to your statement2018-10-06
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    @CloudJR: It is not. As I said, the target space in your example is (0,1), which is not compact.2018-10-06
42

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

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

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    I don't understand how the function is continuous since it has a big jump from 1 to 2 in its domain.2015-07-17
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    @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](https://en.wikipedia.org/wiki/Continuous_function#Weierstrass_definition_.28epsilon.E2.80.93delta.29_of_continuous_functions) 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.2015-07-21
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    @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*).2015-07-21
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    If I take the subset $(0.5,1.5)$ in the range, it's inverse image is the set $(.5,1)\bigcup [2,2.5)$. which is not open. Does this contradict the definition of continuity?2015-09-04
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    @julian.marr: that inverse image is open _when considered as a subset of the domain_.2015-09-04
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    Is that because there exists a neighborhood around 2 wntirely contained in $[2,2.5)$?2015-09-04
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    So this is becuase $(f^{-1})^{-1}([2,3])=[1,2]$ which isn't open in $[0,2]$?2016-06-20
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    @snulty: I wrote my comment in haste, and deleted it immediately.2016-06-20
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    @TonyK thank you for the response though :)2016-06-20
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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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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.

  • 0
    If the inverse map also goes to $\mathbb{Q}$, then how is it discontinuous?2011-09-30
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    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).2011-09-30
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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 $a$.