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Possible Duplicate:
Are continuous self-bijections of connected spaces homeomorphisms?

I know that a function $f\colon X\to Y$ need not be a homeomorphism if it is:

  1. A continuous bijection (e.g. $f:[0,2\pi)\to S^1$),
  2. An open bijection (e.g. inverse of above function),
  3. A bijection between homeomorphic spaces (e.g. $x\mapsto -x$ on $\mathbb R$ with lower limit topology),

but what if $X,Y$ are already known to be homeomorphic, and $f$ is a continuous bijection. Must f then be a homeomorphism, or does there exist a counterexample?

I have given this some thought, but I have so far been unsuccessful at generating either a proof or a counterexample. Also, my searches have turned up unsuccessful.

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    I think it would be hard to beat [Mike's unzipped pants space](http://math.stackexchange.com/a/20932/822) for simplicity, whether you care about connectedness or not. If you really want a *disconnected* example, add an isolated point.2012-01-13

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