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:
Sergio Da Silva
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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. – Sergio Da Silva Apr 03 '14 at 01:42
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The reason I ask this question is because I am reading the proof of the following proposition: – velut luna Apr 03 '14 at 02:15
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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. – velut luna Apr 03 '14 at 02:19
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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. – Sergio Da Silva Apr 03 '14 at 02:26
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2I agree that the answer is no in general, but if $x_0$ is an interior point (of the domain of $f$), then could the answer be true? Let just assume $f : U \subset \mathbf R \to \mathbf R$ for simplicity. In the link you provided, the following example $$f(x)=\begin{cases} x & x \in [0,1) \ x-1 & x \in [2,3] \end{cases} $$ does not work because $x_0 = 2$ belongs to the boundary of the domain of $f$. – QA Ngô Jun 01 '22 at 02:32