Math for Non-Geeks/ Continuity of the inverse function

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In this chapter we will introduce a theorem that gives a sufficient condition for the continuity of the inverse function of a bijective function. In mathematical literature this theorem is often referred to as the "Inverse function Theorem". What is astounding about this result is that the inverse function of a discontinuous function can actually be continuous.

Motivation

The continuous and invertible function $f$ with the domain $[-1,0]\cup (1,2]$

First we want to consider the most general condition possible for when a bijective function $f:D\to W$ with $D,W\subseteq ℝ$ has a continuous inverse function. The first ansatz that we naturally wan to investigate is the continuity of $f$ itself. We might spontaneously assume that the continuity of the inverse function $f^{-1}$ follows directly from the continuity of $f$. This is not necessarily the case, as is shown in the following example:

The discontinuous inverse function $g=f^{-1}$

The function is continuous since it's continuous on both $[-1,0]$ and $(1,2]$. Furthermore it is strictly monotonically increasing, so it's injective. The image of $f$ is $[-1,1]$ and therefore $f:([-1,0]\cup (1,2])\to [-1,1]$ is surjective. All together, $f$ is bijective and therefore invertible. The inverse mapping $f^{-1}:[-1,1]\to ([-1,0]\cup (1,2])$ has functional values:

The inverse function is not continuous since it has a jump discontinuity at $y=0$. So it can definitely be the case that a continuous function has a discontinuous inverse.

But now another thing is apparent: the inverse function $f^{-1}=g$ is also invertible with the inverse function $g^{-1}={\left(f^{-1}\right)}^{-1}=f$. This means, however, that a discontinuous function (like $g$) can have a continuous inverse function. Hence, we posit:

The problem lies in the domain of $f$. Namely, this is $[-1,0]\cup (1,2]$, i.e. not an interval. The domain is not connected and has a "gap." How does this "gap" affect the inverse function $f^{-1}$?

Since the domain of $f$ corresponds to what should be the codomain of $f^{-1}$, the inverse function also has this "gap" in its codomain. Similarly, the codomain of $f$ corresponds to the domain of $f^{-1}$. It is for this reason that $f^{-1}$ can not be continuous. The domain of $f$ is an interval and therefore connected, while its codomain contains a gap. Since $f^{-1}$ is surjective and must attain every value in what should be its codomain, the graph of $f^{-1}$ must lend itself to a jump point somewhere. This is why $f^{-1}$ is not continuous.

Thus, we must require that, by assumption, the domain of $f$ is an interval to ensure that we don't encounter any problems. Usually this requirement is sufficient to guarantee the continuity of $f^{-1}$. We can prove this using the $ϵ$-$\delta$ characterization of continuity. With the help of Zwischenwertsatzes we can futhermore conclude that the codomain of $f$ - and therefore the domain of $f^{-1}$ - is an interval.

Exercises

Exercise 2

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