Another question about continuity and inverse maps

Question

In my studies I came across the reversibility of strictly monotonic functions.

The sentence states[S]:

[S] The function $f$ is on any intervall $I$ strictly growing. Then there exists the inverse $f^{-1}$ on $f(I)$, which is also strictly growing and continuous.

If $f$ is also continuous, so $f(I)$ has to be an intervall with (sometimes even infinitly many) boundries, $\inf f$ and $\sup f$. In this case $f(I)$ is exactly than left or right closed, when $I$ is left or right closed.

This also holds in the same way for strictly decreasing.

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I had to show in an exercise that with $f$ being strictly monotonic on an intervall $I$. Which implies $f^{-1}$ being also strictly monotonic, that its wrong to assume $f$ being continuous.

Thing is: [S] states, that $f^{-1}$ is continuous and also strictly monotonic on $f(I)$

But if we now call $g:=f^{-1}$ than there exists $g^{-1}$ since $g$ also fullfills the requirements. And $g^{-1}=f$

But than $g^{-1}=f$ should be also strictly monotonic (which is indeed the case) and continuous on $g(f(I))=g(g^{-1}(I))=I$

But Im $100$% sure im pretty damn wrong somewhere..

Okey my questions more precisesly:

[S] is writen in my textbook as a lemma. And it doenst require $f$ which [S] refers to, to be continuous. And I cant think of why when $f$ isnt continuous and strictly (for example) growing on an intervall $I$, why $f^{-1}$ will also strictly grow but be continuous on the given intervall $f(I)$.. I mean as you can see above, thats the way how I think about it right now, and if I would be correct (which cant be true) than $f$ has to be continuous in order for $f^{-1}$ to be continuous (and ofc $f$ and $f^{-1}$ both are strictly (for example) growing)

Can someone help? :)

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Picture of exercise and of [S]: (as wished in the comments :))

Answer 1

I think your intuitions will become better as you see more examples.

My translation of your "S", where I have stuck in a phrase in square brackets.

37.1 Inversion theorem for strictly monotone functions: Let the function $f$ be strictly increasing on a given interval $I$, which may be finite or infinite. Then its inverse function $f^{-1}$, defined on $f(I)$, is strictly monotone and continuous [on $f(I)$]...

For example: this $f$ is strictly monotone increasing function on $I=\mathbb R$: $$f(x)=\begin{cases} x&\text{ for } x\le0\\x+1&\text{ for } x>0.\end{cases}$$ It has a jump discontinuity at $x=0$. In this case $f(I)=(-\infty,0]\cup(1,\infty)$, and the inverse function $g$ given, for $y\in f(I)$, by $$ g(y)=\begin{cases}y&y\le0\\y-1&y>1.\end{cases}$$ As advertised, this is continuous and strictly increasing on the disconnected set $f(I)$. There is an extension of $g$ to all of $\mathbb R$, namely $$ \hat g(y)=\begin{cases}y&y\le0\\0&0<y\le 1\\y-1&y>1.\end{cases}$$ This function is continuous on all of $\mathbb R$ but not strictly monotone there. A careless reader of MSE (such as myself), or of the textbook, might assume $f^{-1}$ is this $\hat g$; but your textbook author really wants $g$.

The question posed in the second snapshot is

3. Suppose the function $f$ is strictly monotone on the interval $I$. Then, as we know, $f^{-1}$ is strictly monotone. Why is it an incorrect application of the inversion theorem to conclude that the inverse of $f^{-1}$, which is $f$, is also continuous?

The answer is, $f^{-1}$ is defined on $f(I)$, which might not be an interval, but a disjoint union of intervals, as in the example I gave above. The inversion theorem applies only to functions defined on an interval, but not to functions defined on disjoint unions of intervals.