If $f$ is invertible is $f\left(f^{-1}(x)\right)=x$ and $f^{-1}(f(x))=x$ always True? If $f$ is invertible, is $f\left(f^{-1}(x)\right)=x$ and $f^{-1}(f(x))=x$ always True?
I have doubt in the second one since $$\sin^{-1}(\sin x)=x$$ is not always true.
Any comments on this?
 A: Yes, it is always true. It turns out that $\sin$ is not invertible. The function which is denoted by $\sin^{-1}$ (or, more generally, by $\arcsin$) is the inverse of the restriction of $\sin$ to $\left[-\frac\pi2,\frac\pi2\right]$. And so$$\left(\forall x\in\left[-\frac\pi2,\frac\pi2\right]\right):\arcsin\bigl(\sin(x)\bigr)=x$$and$$\bigl(\forall x\in[-1,1]\bigr):\sin\bigl(\arcsin(x)\bigr)=x.$$
A: It is always true.  Technically a function is not invertible except on certain intervals.  $f:X\to Y$ is invertibel iff $f$ is a bijection between $X$ and $Y$. A bijection is into(injective) and onto(surjective). Into means that for every $a$ and $b$ in $X$, $f(a)=f(b)$ necessarily implies $a=b$. $f$ is onto iff for every $y \in Y\, \exists x \in X$ that $f(x)=y$. 
So it follows that for any $b \in Y$, there exists a unique $f^{-1}(b)$ in X. So $f(f^{-1}(b))=b$.
The reason this fails for $\sin$ is that sine is only bijective for $X=[-\pi/2, \pi/2]$ and $Y=[-1,1]$ (up to periodicity). And if you don't have a  bijection, you do not have an inverse. 
