Let $f: \mathbb{R} \rightarrow \mathbb{R}$. I recently came across the following property being used to solve some "coffin problems": $$f(x) = f^{-1}(x) \Rightarrow f(x) = x$$
The intuition given was that $f^{-1}$ is the reflection of $f$ about $y = x$, and hence $f$ and $f^{-1}$ must have the same intersection points as $f$ and $y=x$. (And this property has been used to simplify problems like $\frac{3x^3+1}{2} = (2x-1)^{\frac{1}{3}}$ to $\frac{3x^3+1}{2} = x$.)
But how do you prove this formally?
I tried but couldn't prove anything except $f(f(x)) = x$, which is obvious. If you need to make certain assumptions about f like surjectivity, injectivity, continuity, etc, go ahead but do state them clearly. Obviously the weakest assumptions are preferred.
Edit There is no implicit forall quantification over $x$ in any equation. The above property should be understood as: $$\{x \: | f(x) = f^{-1}(x) \} \subseteq \{x \: | f(x) = x \}$$
This fails for the simple function $f(x)=-x$. We have $f=f^{-1}$ (they are the same function), i.e., $f(x)=f^{-1}(x)$ for all $x\in {\Bbb R}$, yet $f(x)=x$ only for $x=0$.
More generally any invertible function satisfying $f(f(x))=x$ (of which were are uncountably many) has this property.