If $f(x)$ is an invertible function, is then $f^{-1}(x)$ always invertible (consider $\sin(x)$)?

Question

So there is this question where it says:

If $f(x)$ is invertible, then is $f^{-1}(x)$ also invertible?

The answer may come to be: Yes, because if $f(x)$ is invertible, then the function is one-to-one, therefore the inverse, $f^{-1}(x)$, is also one-to-one. But why? Specifically, please consider $\sin(x)$, it has an inverse (with restricted domain) but if you take the inverse of the inverse of $\sin(x)$, then you would have to expand the domain. How would that work? Thanks.

NOTE: I am aware this question is similar to other questions, however this is specifically asked for $\sin(x)$ and not involving higher-level mathematical terms such as "bijective". Thanks.

Answer 1

The function $\sin(x)$ has an inverse on a restricted domain, $\arcsin(x)$. Technically it is not an inverse of the function $\sin(x)$ because the domain is restricted.

Remember a function definition has three elements to the definition, a domain, a function definition, and a co-domain. The sets must be the same for the function and its inverse. That is, The domain of the inverse function is the co-domain of the original function. The co-domain of the inverse function is the domain of the original function. If this doesn't hold then the inverse function is not a true inverse.

Answer 2

If the function $f:A\to B $ is one-to-one and onto, then there is a function $g:B\to A$ such that $gof(x) =x $ for all $x\in A$ and fog(x)=x for all $x\in B$

We call $g$ the inverse function of $ f$ and by the same token $f$ is the inverse function of $g.$