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Let $f: X\rightarrow Y$ be a map between Hausdorff spaces.

Is there an example, s.t. Y is compact, $f$ continuous, bijective, but $f$ is not a homeomorphism.

If $X$ is compact instead of $Y$, then it is clear (the answer is no)

but is $f^{-1}(Y)$ not compact ?

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    $\begingroup$ Let $X$ and $Y$ have the same underlying set, but different topologies. $\endgroup$ – Daniel Fischer Feb 9 '14 at 14:49
  • $\begingroup$ @DanielFischer Yes i got it. but if we consider $f^{-1}(Y)$, is this not a contradiction ? $\endgroup$ – derivative Feb 9 '14 at 14:53
  • $\begingroup$ A contradiction to what? $f^{-1}(Y)=X$ for any continous $f\colon X\to Y$. $\endgroup$ – Hagen von Eitzen Feb 9 '14 at 14:54
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    $\begingroup$ $f^{-1}$ has no reason a priori to be continuous. $\endgroup$ – Daniel Fischer Feb 9 '14 at 14:54
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For example, consider the set $\Bbb N\cup\{\infty\}$ with the discrete topology for $X$, and with the natural topology induced by the order for $Y$ (where neighborhoods of $\infty$ are the cofinite sets). This $Y$ is compact, but $X$ is not.

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  • $\begingroup$ yes, clear. i thought too quickly.(i assumed closed+hausdorff = compact) $\endgroup$ – derivative Feb 9 '14 at 15:10
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I think the classic example of $f : [0,1) \to S^1$ given by $f(x) = e^{2 \pi i x}$ satisfies all your criteria.

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