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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?

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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.

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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.

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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 $f(x)=x^2$

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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.

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