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

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 ?

• Let $X$ and $Y$ have the same underlying set, but different topologies. – Daniel Fischer Feb 9 '14 at 14:49
• @DanielFischer Yes i got it. but if we consider $f^{-1}(Y)$, is this not a contradiction ? – derivative Feb 9 '14 at 14:53
• A contradiction to what? $f^{-1}(Y)=X$ for any continous $f\colon X\to Y$. – Hagen von Eitzen Feb 9 '14 at 14:54
• $f^{-1}$ has no reason a priori to be continuous. – Daniel Fischer Feb 9 '14 at 14:54

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.
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.