TITLE: Proof that a bijection has unique two-sided inverse
QUESTION [6 upvotes]: For a bijection $\alpha:A\rightarrow B$ define a bijection $\beta: B\rightarrow A$ such that $\alpha \beta $ is the identity function $I:A\rightarrow A$ and $\beta\alpha $ is the identity function $I:B\rightarrow B$.

Prove that $\alpha\beta$ or $\beta\alpha $ determines $\beta $
  uniquely.

I am stonewalled here. I can understand the premise before the prove that, but I have no idea how to approach this. 

The motivation of the question in the book is to show that bijections have two sided inverses. Later questions ask to show that surjections have left inverses and injections have right inverses etc,

REPLY [4 votes]: In fact, we have the following:

Theorem. Let $f\colon A\to B$ be a function. The following are equivalent:

$f$ is one-to-one.
$f$ is left-cancellable: if $C$ is any set, and $g,h\colon C\to A$ are functions such that $f\circ g = f\circ h$, then $g=h$.

The following condition implies that $f$ is one-to-one:

$f$ has a left inverse, $h\colon B\to A$ such that $h\circ f=\mathrm{id}_A$.

If, moreover, $A\neq\emptyset$, then $f$ is one-to-one if and only if $f$ has an left inverse.

Dually:

Theorem. Let $f\colon A\to B$ be a function. The following are equivalent:

$f$ is onto $B$.
$f$ is right-cancellable: if $C$ is any set, and $g,h\colon B\to C$ are such that $g\circ f = h\circ f$, then $g=h$.

The following condition implies that $f$ if onto:

$f$ has a right inverse, $g\colon B\to A$ such that $f\circ g = \mathrm{id}_B$.

In addition, the Axiom of Choice is equivalent to "if $f$ is surjective, then $f$ has a right inverse."

Proposition. Let $f\colon A\to B$ be a function If $g$ is a left inverse of $f$ and $h$ is a right inverse of $f$, then $g=h$. In particular, a function is bijective if and only if it has a two-sided inverse.
Proof. $g = g\circ\mathrm{id}_B = g\circ(f\circ h) = (g\circ f)\circ h = \mathrm{id}_A\circ h = h.$ $\Box$
This is the same proof used to show that the left and right inverses of an element in a group must be equal, that a left and right multiplicative inverse in a ring must be equal, etc. 
The last proposition holds even without assuming the Axiom of Choice: the small missing piece would be to show that a bijective function always has a right inverse, but this is easily done even without AC.