Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

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 $y$?

share|improve this question
1  
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
1  
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

1 Answer

up vote 0 down vote accepted

Well, benyond being bijective, $f$ must send open sets to open sets.

share|improve this answer
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.