Does continuity of an invertible function imply continuity of its inverse?

[quasar987]

My textbook always says things like "[...] such that f is invertible, continuous, and f^-1 is continuous also" and " [...] such that g is invertible, smooth, and g^-1 is smooth also"

But doesn't it follow from "f is conitnuous" that "f^-1 is continuous", and same thing for smooth?

 

[StatusX]

No. The standard counterexample is the map from [0,2pi) to the circle (the natural one), which is a bijection, and is continuous since, roughly speaking, small changes in x give small changes in f(x). But in the circle, a small change about f(0) gives a very large change in the inverse image, which will suddenly jimp from just above 0 to just below 2pi.

There are theorems that the continuity of f^-1 follows from that of f under certain additional assumptions, like if the domain is compact and the range is Hausdorff, as is the case, for example, with all functions from a closed interval to R, ie, the functions you usually graph.

 

[matt grime]

Of course it doesn't follow. Exercise, think of lots of counter example (hint, the inveser function theorem. If f is invertible and f'(0)=0, then what about the derivative of the inverse of f at zero?)