Why does $\sin^{-1}(\sin(\pi))$ not equal $\pi$ And when does $\sin^{-1}(\sin(x)) = x$

Since, for any $y \in [-1,1]$, several different $x$ have $\sin(x)=y$, either $\sin^{-1}(y)$ is multivalued and so not a function, or you need to restrict its range by taking a principal value.  Usually the later choice is made, with $-\pi/2 \le \sin^{-1}(y) \le  \pi/2$. 
So in answer to your title question:  $  \sin^{-1} (\sin (\pi)) \not = \pi$ because $\pi \gt  \pi/2$ and in fact $\sin^{-1} (\sin (\pi)) = \sin^{-1} (0) = 0$.
$\sin^{-1} (\sin (x)) = x $ when $-\pi/2 \le x \le  \pi/2$.
