Is there any connection between the facts that $(ff^{-1})(x)=x$ and that $f$ and $f^{-1}$ are reflections of each other in the line $y=x$? Suppose $f$ is a real-valued function of a real variable and has inverse $f^{-1}$. Then


*

*$(ff^{-1})(x)=x.$

*Also, $f$ and $f^{-1}$ are reflections of each other in the line $y=x$.


Is there any connection between the above two facts?
 A: You choose any point (p, q) on the cartesian plane. Consider the line $y=x.$ Then find the image of (p, q) by reflecting with respect to line $y=x.$ Let ( m, n) be the image of (p, q). Then you'll find that (m, n) = (q, p). The equation (1) is implicitly telling this phenomenon. 
Let's do some high school coordinate geometry. If $ax + by +c =0$ is a given straight line, and $(p, q)$ be a point some given point that is not on the line, then the coordinate of the image of $(p,q)$ is given by $ \frac{h-p}{a} = \frac{k-q}{b} = \frac{-(ap +bq +c)}{a^2 +b^2}$ where $(h, k)$ is the coordinate of the image of (p, q) under reflection with respect to the given line. This algebra proves the fact that I stated.
Another way you might like to think: Let $r$ be a reflection operator with respect to some given line. Any reflection has order 2 i.e. $r^2(x) =x$ for any x. 
Since inverse function defined such a way that successively applying $f$ and then $f^{-1}$ return the given input i.e. $f^{-1}o f(x) =x.$ So, we see that $r^2$ and $f^{-1}o f$ have exact same effect from geometric point of view. ( 1) and (2) potray this fact.
