Is a bijective function always invertible?

Solution 1:

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).

Solution 2:

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 $f$.