Let $D$ be the closed unit disk in $\mathbb{C}$ and let $f:D\to D$ be a function such that:
- $f$ is equal to the identity function $\mathrm{Id}$ specifically on the unit circle ($\partial D$)
- $f$ is continuous on $D$
- $f\circ f=\mathrm{Id}$ on $D$
How do we show that $f=\mathrm{Id}$ on all of $D$?