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 $y$?
Continuity of inverse mapping of a continuous function
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real-analysis
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1Continuity is only defined for functions, which $f^{-1}$ will not be unless $f$ is a bijection. – 2011-03-27
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1And even if $f$ is a continuous bijection, we don't necessarily have $f^{-1}$ - see [here](http://en.wikipedia.org/wiki/Homeomorphism#Notes) – 2011-03-27
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0Yes but I meant some property of $f^{-1}$ as a correspondence. – 2011-03-27
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Well, benyond being bijective, $f$ must send open sets to open sets.
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0Suppose $y_n$ 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"? – 2011-03-27
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0@Thales Assuming $f^{-1}$ continous, yes. – 2011-03-28