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The intermediate value property (IVP): A function has the intermediate value property on an interval $[a,b]$ if for all $x<y$ in $[a,b]$ and all $K$ with $f(x) < K < f(y)$ there exists $c \in (x,y)$ with $f(c) = K$.

I think that

$f$ has IVP if and only if $f^{-1}$ maps connected sets to connected sets.

Please, can you tell me if what I think is correct?

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  • $\begingroup$ BTW the functions with the intermediate value property are also often called Darboux functions (because of Darboux's theorem). $\endgroup$ – Martin Sleziak Apr 12 '15 at 10:20
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Take $f(x)=x^2$. Then $f$ is continuous, and thus it has the IVP property but $$f^{-1}\big([1,4]\big)=[-2,-1]\cup[1,2].$$

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  • $\begingroup$ Oh, very good, thank you! Now I see it also is false with $f(x)=|x|$. $\endgroup$ – blue Jan 22 '14 at 15:36

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