Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there an example of a continuous invertible map $f:X\to Y$ between topological spaces $(X,T_X)$ and $(Y,T_Y)$ such that $f$ is continuous, but its inverse $f^{-1}$ is not continuous?

share|improve this question

4 Answers 4

It's easy to cook up small examples, but my favorite is the map $f\colon [0, 2\pi) \to S^1$, $f(t) = (\cos t, \sin t)$ because you can sort of see where things go wrong. To show that $f$ cannot be a homeomorphism, note that $S^1$ is compact.

For contrast, here's a useful lemma: if $f\colon X \to Y$ is a continuous bijection, $X$ is compact, and $Y$ is Hausdorff then $f$ is a homeomorphism.

share|improve this answer
2  
Boom! This is the canonical example. –  ncmathsadist May 25 '12 at 1:55

The simplest example is surely the identity function from $\{ 0,1 \}$ with the discrete topology to $\{ 0,1 \}$ with the trivial topology.

share|improve this answer

Using discrete metric is the easiest examplethat you can think of a continuous biyection that is not a homeomorphism but is some artificial. Hardly expect that a case like this occurs during the solution of a problem of analysis or geometry.

An example more natural is the following. Let $M=[-1,0]\cup(1,\infty)$ and $N=[0,\infty)$ and let $f:M\to N$ defined by $f(x)=x^2$. You can check easly that $f$ is continuous bijection (the graph has 2 separated parts but you don't surprise with this fact because domain has two disjoint parts). But $f^{-1}$ is discontinuous at $y=1$ and domain of inverse has only one part, but graph has two:

Function $f(x)=x^2$

share|improve this answer

Let $\,R_1\,$ be the reals with the discrete topology, and $\,R_2\,$ the reals with the usual euclidean toplogy. Then, the map $\,f: R_1\to R_2\,$ defined by $\,f(x):=x\,,\,\forall x\in R_1\,$ is continuous and bijective, but the inverse map isn't continuous.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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