Intuition for why inverse of strictly increasing function is also strictly increasing?

Question

Is there any intuition (ideally graphical) of why the inverse of a strictly increasing function is also strictly increasing?

I was hoping that if we have a strictly increasing function, then its reflection in the line $y=x$ will "obviously" also be strictly increasing. But that this is so isn't obvious to me. (Perhaps someone can persuade me otherwise.)

Answer 1

It is very intuitive indeed. Function $f$ map $x$ to $f(x)$ and its inverse map $f(x)$ to $x$. In strictly increasing function, higher $x$ are mapped to higher $f(x)$ and vice versa, higher $f(x)$ is mapped to higher $x$

Answer 2

If $f$ is a strictly increasing function, then

$x_1<x_2$ and $f(x_1)<f(x_2)$

is always true.

To paraphrase the above statement and to make is so obvious

$f^{-1}(y_1)<f^{-1}(y_2)$ and $y_1<y_2$

Where $y = f(x)$

NO difficult theorem or reasoning included. The equations are not even that hard. I believe this is as easy as it can get.

Answer 3

Let $f:X\rightarrow Y$ with the property that $x<x‘ \Rightarrow f(x)<f(x‘)$. Assume that there are $x_1,x_2$ with $f(x_1) < f(x_2)$ and $x_1\geq x_2$. But then $f(x_2) \leq f(x_1) < f(x_2)$. A contradiction. Hence the inverse function must be (if it exists) strictly monotone as well.

Answer 4

Let $f$ be a strictly increasing function. Consider two points $A$ and $B$ on the curve $y=f(x)$.

Let $A \equiv (x_1 , f(x_1))$ and

$B \equiv (x_1+k , f(x_1 +k))$ for some $k \gt 0$.

Clearly, $x_1+k \gt x_1$ and $f(x_1+k) \gt f(x_1)$

Now, $f^{-1}(x)$ will be its reflection about $y=x$. So, the points after reflection are

$A' \equiv (f(x_1),x_1)$

$B' \equiv (f(x_1+k),x_1)$

Here obviously, $f(x_1+k) \gt f(x_1)$ and $x_1+k \gt x_1$. Thus, $f^{-1}$ must also be a strictly increasing function given $f$ is strictly increasing.