Inverse of function is continuous or not?

Question

If $f:I\subseteq \Bbb R \rightarrow \Bbb R$ is injective and everywhere continuous in its domain then $f^{-1}$ is also continuous everywhere. Here $I$ cannot be discrete set and $f$ is single variable function The above statement is always correct or not? I thought it is always correct. Am I correct or not?

Answer 1

Not in general, no.

The mapping $$f:[0,2\pi)\to \{(x,y)| x^2+y^2=1\}$$

defined as

$$f(t) = (\sin t, \cos t)$$ is injective and continuous on $[0,2\pi)$, but $f^{-1}$ is not.

Answer 2

In general, no. Take$$\begin{array}{rccc}f\colon&\mathbb{Z}^+&\longrightarrow&\mathbb R\\&n&\mapsto&\begin{cases}0&\text{ if }n=0\\\frac1n&\text{ otherwise.}\end{cases}\end{array}$$Then $f$ is injective and continuous, but $f^{-1}$ is discontinuous at $0$.

Answer 3

A more theoretical answer.

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Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces and let $f:X\to Y$ be a function.

I preassume that in your question $f^{-1}$ stands for the inverse of $f$ purely as a morphism in category $\mathbf{Sets}$, i.e. a function $f^{-1}:Y\to X$ that satisfies $f\circ f^{-1}=\mathsf{id}_Y$ and $f^{-1}\circ f=\mathsf{id}_X$

Then let $X=Y\neq\varnothing$ let $\tau_X$ be the discrete topology and let $\tau_Y\neq\tau_X$.

Further let $f=\mathsf{id}:X\to Y$. It is evidently continuous and injective, but $f^{-1}=\mathsf{id}:Y\to X$ is not continuous.

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Things would be different if $f^{-1}$ stands for the inverse of $f$ as a morphism in category $\mathbf{Top}$.

In that case its existence allready implies its continuity.