If a function has a inverse that is well defined is it a bijection If I have a function (binary relation), 
$f: X \to Y:x \mapsto y$
and I show that it is well defined 
and I show that its inverse is well defined.
Then have I shown that $f$ is a bijection? (That it is one to one, and onto)
By show that it is well defined I mean show that is is actually a function, not something masquerading as one.
So that $\forall a,b \in X, a=b \implies f(a)=f(b)$ 
and that $\forall x \in X,  \exists y \in Y$ such that  $f(x)=y$
Or are there invertable functions that are not bijections?
I am fairly sure there are not.
 A: I'll expand on my comment a little.
The usual definition of an invertible function is the following

A function $f\colon A\to B$ is invertible if and only if there exists a function $g\colon B\to A$ such that the compositions $f\circ g=\mbox{Id}_B$ and $g\circ f=\mbox{Id}_A$ hold, where $\mbox{Id}_X\colon X\to X$ is the identity function on $X$.

Recall that the definition of the identity function $\mbox{Id}_X$ is the (it turns out unique) function such that for all $\alpha\colon X\to Y$ and for all $\beta\colon Z\to X$, it holds that $\mbox{Id}_X\circ\alpha=\alpha$ and $\beta\circ\mbox{Id}_X=\beta$. This is equivalent to defining $\mbox{Id}_X(x)=x$ for all $x\in X$.
It's a useful exercise to show that if $f$ has an inverse $g$, then $g$ is unique, that is, if $h$ is also an inverse of $f$, then $h=g$ (here use the definition of the identity functions to prove this).
Now, how can we show that if $f\colon A\to B$ is invertible, then $f$ must be a bijection? Well contradiction is a reasonable method to attempt. Suppose $f$ has inverse $g$ but that $f$ is not surjective. That means there exists an element $b\in B$ such that there exists no $a\in A$ with $f(a)=b$. Now, consider what this says about the function $f\circ g$. Why can this function not be equal to the identity on $B$?
What happens if we suppose that $f$ has inverse $g$ but that $f$ is not injective? That means there exist $a,a'\in A$ with $a\neq a'$ and such that $f(a)=f(a')$. Can you see how this would lead to the fact that $a=a'$? (hint: $g\circ f=\mbox{Id}_A$).
That's all there is to it. It's also another useful exercise to show the converse of the above proposition, namely that if $f$ is a bijection, then $f$ is invertible, and hence these are actually equivalent statements. Here, a more direct approach is in order to show this. So given an element $b$ in the codomain, define where the inverse should map $b$, and show that this mapping is a well-defined function in the sense that every element in the codomain of $f$ has a unique element in the domain of $f$ to which it is mapped. Here you'll need to use the fact that $f$ is surjective and injective respectively.
