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Let me consider a continuous function $y=f(x)$ for $x \in [0,1]$. Now consider its inverse $f^{-1}(y)= \{x:f(x)=y \}$. How can I characterize continuity property of $f^{-1}(y)$ in terms of $y$?

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Continuity is only defined for functions, which $f^{-1}$ will not be unless $f$ is a bijection. – Zev Chonoles Mar 27 '11 at 15:24
And even if $f$ is a continuous bijection, we don't necessarily have $f^{-1}$ - see here – Zev Chonoles Mar 27 '11 at 15:29
Yes but I meant some property of $f^{-1}$ as a correspondence. – Thales Mar 27 '11 at 15:52
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Well, benyond being bijective, $f$ must send open sets to open sets.

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Suppose $y_n$ converges to $y$. Then for $x_n$ with $f(x_n)=y_n$, its limit $x$ has $f(x)=y$, so can we say that $f^{-1}(y)$ and $f^{-1}(y_n)$ are "close"? – Thales Mar 27 '11 at 16:08
@Thales Assuming $f^{-1}$ continous, yes. – Júlio César Mar 28 '11 at 15:43

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