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Suppose $X,Y$ are metric spaces, let $A \subset X$ be a bounded subset of $X$ and $f: A \to Y$ to be a continuous bjection. Prove or disprove that $f^{-1}$ is continuous.

Remark: If each closed subset of $A$ is compact then $f$ would map closed sets to closed sets, which would then imply the continuity of $f^{-1}$. But how to prove/disprove that $f$ is a closed map?

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    $\begingroup$ spelling- bijection $\endgroup$ – Lost1 Apr 12 '13 at 12:27
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Let $X = \mathbb{R}$ with discrete metric and let $Y = \mathbb{R}$ with the usual metric. All maps $X \to Y$ are continuous. No maps $Y \to X$ are continuous.

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    $\begingroup$ Constant maps $Y\longrightarrow X$ (and only them) are continuous. $\endgroup$ – Julien Apr 12 '13 at 12:45
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$X=\mathbb{R},A=[0,1),Y=S^1, f:x\mapsto e^{2\pi ix}$

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Let $X$ be a finite set with discrete topology $d(x,y)=1 \mathrm{if} x \neq y$ and$ =0 $ otherwise. And $Y$ be the same set with trivial topology $d(x,y)=0$ $A=X$ and $f$ is the identity function, then $f$ is continuous but $f^{-1}$ is surely not continuous.

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