Given the continuous curve $y=\sin \left(x\right)$ restricted to domain $[-\frac{\pi }{2},\frac{\pi }{2}]$ and codomain $[-1,1]$, evaluate whether the restricted function has an inverse on the codomain, and if so, what is an explicit form for that inverse.

No, sinusoidal functions never have inverses on any interval because they oscillate; ${\sin }^{-1}$ is not a true inverse

Yes, the inverse exists and is $f^{-1}\left(y\right)=\sin \left(y\right)$ because sine is its own inverse when restricted to any symmetric interval

Yes, the restriction is one-to-one and has inverse $f^{-1}\left(y\right)={\sin }^{-1}\left(y\right)$ with domain $[-1,1]$ and range $[-\frac{\pi }{2},\frac{\pi }{2}]$

No, because the codomain $[-1,1]$ is too large for an inverse; you must restrict the codomain to $\left\{0\right\}$ to have an inverse