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If a function $y = f(x)$ is continuous at $x_0$. Suppose the function is invertible in a neigborhood of $x_0$. Is it true that the inverse function must be continuous at $y_0 = f(x_0)$?

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The answer is no in general (of course if $f$ is also differentiable there, then you could employ the inverse function theorem). Here is a link with many examples of continuous functions with inverses which are not continuous:

Functions which are Continuous, but not Bicontinuous

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  • $\begingroup$ I should mention that many examples are of functions which are continuous everywhere, and invertible everywhere. You can simply restrict to a neighborhood of a correctly chosen point to get the local example you want. $\endgroup$ – Sergio Da Silva Apr 3 '14 at 1:42
  • $\begingroup$ The reason I ask this question is because I am reading the proof of the following proposition: $\endgroup$ – velut luna Apr 3 '14 at 2:15
  • $\begingroup$ The reason I asked this question is because I am reading the proof of the following proposition: "Suppose a continuous function is invertible near x_0 and is differentiable at x_0. If f'(x_0) is nonzero, then the inverse function is also differentiable at y_0 = f(x_0), with derivative 1/f'(x_0).". I wonder why it is necessary to specify that the function is continuous, because differentiability already imply continuity at x_0. $\endgroup$ – velut luna Apr 3 '14 at 2:19
  • $\begingroup$ Yes, this is a good observation. It would be nice to have the inverse function of a continuous function automatically continuous, but alas, it's not. $\endgroup$ – Sergio Da Silva Apr 3 '14 at 2:26

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