Is a bijective function always invertible? I know that in order for a function to be invertible, it must be bijective, but does that mean that all bijective functions are invertible?
 A: A function is invertible if and only if it is injective (one-to-one, or "passes the horizontal line test" in the parlance of precalculus classes).  A bijective function is both injective and surjective, thus it is (at the very least) injective.  Hence every bijection is invertible.
As pointed out by M. Winter, the converse is not true.  For example, the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x + \lfloor x \rfloor$ is injective (and therefore invertible), but not surjective.
The graph of $f$.
A: I don't think "invertible" is a formal term with a precise definition.
Most texts take it to by synonymous with "injective" so: bijective = injective & surjective $\implies$ injective = invertible.  But invertible = injective $\not \implies$ injective & surjective = bijective.
But some texts take it to be synonymous with "bijective".
The main idea is that for $f:A\to B$ to be "invertible" then for any $y \in B$ the set of all $\{x_i\} \subset A$ such that $f(x_i) = y$ should be "invertible" that is there should only be one such, not multiple, $x$ so that $f(x) = y$. 
[ The set  $\{x_i| x\in A, f(x_i) = y\}$ is denoted $f^{-1}(y)$.  In general this is a multivalued set.  But it could be an emptyset.  Or a single valued set.  If it's a single valued set for all $y \in {f(x)|x \in A} = f(A)$, then we can say that $f^{-1}(x)$ is a well-defined function $f^{-1}:f(A) \rightarrow A$.]
That is the very definition of "$f$ is injective".  So no matter how we define "invertible" is must imply that $f$ is injective. 
On the other hand it's ambiguous whether it's possible that there are no $x\in X$ so that $f(x) = y$.  (In other words $y \in B$ but $y \not \in f(A)$; or in other words $f^{1}(y) =\emptyset$.)  It could be argued it's not invertible if $f^{-1}(x) = \emptyset$ rather than a single element.  Or it could be argued that it doesn't matter and we are only interested in $y \in f(A)$.
At any rate "invertible means that for all $y \in f(A)$ then $f^{-1}(y)$ is a single element; not a multiple valued set.  In other words "invertible" $\implies$ "injective".   
A: Yes. For every bijective function $f$ there is a unique function $f^{-1}$ so that $f(f^{-1}(x))=f^{-1}(f(x))=x$, or in other words: $f\circ f^{-1}=f^{-1}\circ f=\mathrm{id}$. Note that this is not hundred percent precise. For this we have to state the domain and codomain of all involved functions and have to be careful to state which set $x$ belongs to.
However, your initial statement is wrong. There are invertible functions that are not bijective, e.g. $\exp:\Bbb R\to\Bbb R$, which fails to be surjective (hence not bijective).
