Why is arcsin a function and not a relation since $\arcsin(\sin(\frac{3\pi}{4})) = \frac{\pi}{4}$? Since $\sin(\frac{\pi}{4})$ and $\sin(\frac{3\pi}{4})$ are both $\frac{\sqrt{2}}{2}$, shouldn't $\arcsin(\frac{\sqrt{2}}{2})$ map to both $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ and therefore not be a function?
when I input $\arcsin(\frac{\sqrt{2}}{2})$ into Wolfram, I get back $\frac{\pi}{4}$ only, but isn't the point of an inverse function that $f^{-1}(f(x)) = x$?
In this case however, $\arcsin(\sin(\frac{3\pi}{4})) = \frac{\pi}{4}$
What am I missing?
 A: The important thing to realise here is that a function is not just a rule.  To specify a function properly you must also define the domain and codomain.  For example, the sine function is
$$\sin:{\Bbb R}\to{\Bbb R}\quad\hbox{where}\quad
  \sin x= \langle\hbox{insert your favourite definition}\rangle\ .$$
This function is not one-to-one (injective) and therefore has no inverse.
Here is a different function:
$$f:\Bigl[-\frac\pi2,\frac\pi2\Bigr]\to[-1,1]\quad\hbox{where}\quad
  f(x)=\sin x=\langle\hbox{same definition as above}\rangle\ .$$
This function is one-to-one and onto and therefore has an inverse.  Its inverse is commonly denoted $\sin^{-1}$ or $\arcsin$.  Notice however that this terminology is not strictly accurate: $\arcsin$ is not the inverse of $\sin$, it is the inverse of $f$, which is a different function.
Hope this clears things up.
A: Recall that $\arcsin: [-1,1] \mapsto [-\pi/2,\pi2/]$, i.e., the range of $\arcsin$ is $[-\pi/2,\pi/2]$.
The function $\sin$ on the other hand is a non-injective function that maps the real line to $[-1,1]$. Defining an inverse for a non-injective functions is not possible. Hence, one route that is often taken is to restrict the domain of the original function such that the function on the new restricted domain is injective.
Hence, the function $\arcsin$ is the inverse of $\sin$ only when the domain of the function $\sin$ is restricted to $[-\pi/2,\pi/2]$.

EDIT
What mathematica does is the following. It first evaluate $\sin(3\pi/4)$, which is $1/\sqrt{2}$. Now it goes and asks the function $\arcsin$, what $\arcsin(1/\sqrt2)$ is, which is now $\pi/4$.
A: As others have pointed out, $\arcsin x$ has the restricted domain $[-1,1]$ and range $[\frac{-\pi}{2},\frac{\pi}{2}]$. This restriction is a choice, a definition.
What's really happening is that $\arcsin x$ is multi-valued. We're only selecting what is called the principal branch, and we're doing this for convenience.
Since the trigonometric functions are periodic, a single-valued function that acts as an inverse doesn't exist. We elect to "chop up" the inversion relation into sections - the branches. In the case of $\arcsin x$ we focus on returning $\sin x$ input values $[\frac{-\pi}{2},\frac{\pi}{2}]$. This branch is the principal branch - the one we've elected to use most often. Why? Well, consider that $\sin x$ cycles through all possible outputs when we give it $[\frac{-\pi}{2},\frac{\pi}{2}]$. Also, it's "nicely packaged" - there's a distinct symmetry and accessibility in using $[\frac{-\pi}{2},\frac{\pi}{2}]$.
When evaluating a query such as yours, we compensate by finding an equivalent result in the principal branch.
