calculus - The intermediate value property of functions

The intermediate value property (IVP): A function has the intermediate value property on an interval $[a,b]$ if for all $x

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?

Answer

Take $f(x)=x^2$. Then $f$ is continuous, and thus it has the IVP property but

$$f^{-1}\big([1,4]\big)=[-2,-1]\cup[1,2].$$