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

1 Answer 1

up vote 0 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. –  user21725 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

Your Answer

 
discard

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