Prove that $``ff^{-1}(x)$ $=$ $x$ $=$ $f^{-1}f(x)"$ $\implies$ $``$the graph of $f$ and $f^{-1}$ are reflections of each other in the line $y = x"$.

Question

According to the Cambridge International AS & A Level Pure Mathematics $1$ book $(2019$ edition, page $48)$,

The graph of $f$ and $f^{-1}$ are reflections of each other in the line $y = x$.
This is because $ff^{-1}(x)=x=f^{-1}f(x)$.
When a function $f$ is self-inverse, the graph of $f$ will be symmetrical about the line $y=x$.

But how does $``ff^{-1}(x)$ $=$ $x$ $=$ $f^{-1}f(x)"$ imply $``$the graph of $f$ and $f^{-1}$ are reflections of each other in the line $y = x"$? Can someone prove it?

Answer 1

The reflection of the point $(\hat x, \hat y)$ in the line $y=x$ is $(\hat y, \hat x)$

So you simply show that for each $\hat x \in \mathcal D(f), \hat y \in \mathcal R(f)$, we have: $$f(\hat x)=\hat y\iff f^{-1}(\hat y)=\hat x$$

Can you show this?