Why the graph of $\sin(\sin^{-1}(x))$ is the identity?

Question

I just made the following plots on Mathematica:

This may be really silly but I don't understand why the plot of $\sin(\sin^{-1}(x))$ is the identity. Isn't $\sin^{-1}(x)$ is defined only on the interval $[-1,1]$? How come it has values outside of that interval? Here on wikipedia, it says:

The trigonometric functions are periodic, and hence not injective, so strictly speaking, they do not have an inverse function. However, on each interval on which a trigonometric function is monotonic, one can define an inverse function, and this defines inverse trigonometric functions as multivalued functions. To define a true inverse function, one must restrict the domain to an interval where the function is monotonic, and is thus bijective from this interval to its image by the function. The common choice for this interval, called the set of principal values, is given in the following table.

So is Mathematica using more than one set of principal values? I'm confused.

Answer 1

The sine function maps the real axis onto the interval $[-1, 1]$, but it maps the complex plane onto itself. From $$ \sin(z) = \frac{e^{iz} - e^{-iz}}{2i} $$ it is not difficult to see that the equation $\sin(z) = w$ has (infinitely many) solutions for every $w \in \Bbb C$.

ArcSin[z] is defined for arbitrary complex values $z$ (the precise formula is given in the reference) and returns some (complex) number such that Sin[ArcSin[z]] == z.