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Let $I\subseteq\mathbb{R}$ be an open interval, and $f:I\to\mathbb{R}$ an injective function. Let $a\in I$, and suppose that $f$ is continuous at $a$. Does it follow that $f^{-1}$ is continuous at $f(a)$ ?

I know that if $f$ is assumed to be continuous in the entire $I$, then $f^{-1}$ is continuous. (Here I use the fact that $f$ is strictly monotonic.) With local continuity, however, I cannot use monotonic properties.

ashpool
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    Check this: https://math.stackexchange.com/q/737465/42969 – Martin R Apr 02 '24 at 07:07
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    @MartinR Thanks for the link. It is interesting that it is not true in general, but here I'm assuming that the function is real-valued and defined on an interval, and none of the examples in the link seems to address this case. – ashpool Apr 02 '24 at 07:14

3 Answers3

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If you relax the condition that $I$ is open, then I believe this is a counterexample.

Let $f:[0,+\infty) $ be given by $$f(x) = \begin{cases} x & \text{$x \in \mathbb{Q}$} \\ -x & \text{$x \in \mathbb{I}\cap(0,1)$ }\\ \dfrac{1}{x} &\text{$x \in \mathbb{I}\cap(1,+\infty)$} \end{cases}$$

This function is injective and continuous at $x=0$ as $x,-x\to 0$, but its inverse is not continuous at $f(0)=0$, because if $c\in \mathbb{I}\cap (0,1)$ then $f^{-1}(c)=\dfrac{1}{c}\in (1,+\infty)$.

Julio Puerta
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A counterexample is as follows (now $I$ is open). Define $f:(-1,+\infty)\to\mathbb{R}$ given by:

$$f(x) = \begin{cases} x & \text{ $x\neq \dfrac{1}{n}, x\neq n$} \\ \dfrac{1}{2n} &\text{$x=\dfrac{1}{n}$}\\ \dfrac{1}{2n+1} &\text{$x=n$} \end{cases}$$

This function is injective (notice that points of the form $\dfrac{1}{n}$ map to even denominators and those of the form $n$ map to odd denominators), and continuous at $0$ as $x,\dfrac{1}{2n}\to 0$. Nonetheless, $f^{-1}$ is not continuous at $f(0)=0$ because $f^{-1}\left(\dfrac{1}{2n+1}\right)=n\not\to0.$

Julio Puerta
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  • Is $n$ a positive integer? Is $f(1)$ $1/2$ or $1/3$ ? – ashpool Apr 03 '24 at 01:59
  • I wonder if there are counterexamples when $I$ is an open interval of finite length or when $I=\mathbb{R}$. – ashpool Apr 03 '24 at 02:01
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    $n$ is a positive integer. You could include $1$ in any of the two definitions, both lead to a counterexample. The above function can be extended to $\mathbb{R}$, just include all missing numbers in the first definition of $f$. – Julio Puerta Apr 03 '24 at 05:21
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I think one needs the assumption that $f$ is injective.

Because otherwise $f(a)$ might several pre-images. For example, assume $f$ is not injection, and for $a\neq b$, we have $f(a)=f(b)$. Now consider open sets $U, V, W$ containing $a,b, f(a)$ respectively such that $U\cap V=\phi$. Since $f$ is continuous at $a$, we have $f(U) \subset W$.

In order to be $f^{-1}$ to be continuous at $f(a)$, one must have both $f^{-1}(W) \subset U$ and $f^{-1}(W) \subset V$, which is impossible by our assumption that $U \cap V=\phi$.

So injective property is necessary.

MAS
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  • I did assume injectivity. Please read the more detailed question posed below the title. – ashpool Apr 02 '24 at 07:43
  • @ashpool, aah I see you assumed injective function. Anyway, this is necessary condition but not sufficient. In other words the answer to your question is NO, in general. – MAS Apr 02 '24 at 08:29