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I need an example of a continuous bijection $f:X \to Y$, where $X$ is NOT compact and $Y$ is Hausdorff, such that $f$ is not a homeomorphism. (It is easy to show that if $X$ is compact, then $f$ is necessarily a homeomorphism)

Any help is appreciated, Thanks !

the8thone
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    Glue 2 end of a string together is a continuous bijection, cut the glued part away is not. – Gina Jul 27 '14 at 01:35

2 Answers2

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Using the usual topologies $f:[0,2\pi)\to S^1$ given by $f(x)=e^{ix}$, half open interval to a circle. Continuous, bijective, inverse is discontinuous at $1$, has to break the circle somewhere.

Conifold
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  • I thought about this example , but I was not sure if it works. you mean $f$ is not an open map ? Oh I see ! The image of $[0,\epsilon)$ is not open ... Thanks ! – the8thone Jul 27 '14 at 00:44
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$X=\mathbb{R}$ with discrete topology and $Y=\mathbb{R}$ with the usual topology, and $f=id$. $f$ is continuous bijective and $f^{-1}=id$ is not continuous.

Hamou
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