Take the 2-minute tour ×
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.

The Inverse Function Theorem states sufficient conditions for a function to have a continuous inverse.

When, if it all, are these conditions necessary conditions? Is there a nice counterexample?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Consider $f(x)=x^3$, which has zero derivative at $x=0$.

share|improve this answer
    
That's embarrassingly easy. I suppose this didn't occur to me because in one-dimension I have seen the IFT stated as saying that if a function is continuous and strictly monotonic (on some interval), then it is invertible. This seems to me to be a necessary condition as well. –  Eric Gregor Jan 25 '13 at 2:11
    
@EricGregor Yes, that is a necessary condition. Somewhere on this site it is proven that if $f$ is not monotonic we either have a "vee" or "wedge", that is some $x<y<z$ with $f(x)\le f(y)\ge f(z)$ or $f(x)\ge f(y)\le f(z)$. If we take $c$ between $\min\{f(x),f(z)\}$ (or $\max\{f(x),f(z)\}$ in the second case) and $f(y)$ then by the IVT we see that $f$ achieves $c$ on $(x,y)$ and $(y,z)$, so $f$ is not invertible. –  Alex Becker Jan 25 '13 at 2:20
add comment

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.