Proof: If $fg$ and $gh$ are bijections, then $f,g,h$ are all bijections also. Note: my textbook uses the notation $gf$ to mean the composition $g$ of $f$, i.e. $gf=g(f(x))$.
Let $A$ be a nonempty set, and let $f,g,h \in \mathbb{F}(A),$ the set of all functions mapping from $A \to A$. If $fg$ and $gh$ are bijections, then $f,g,h$ are all bijections as well.
So far, all I have is:


*

*Since $fg$ is a bijection, $f$ is surjective and $g$ is injective.

*Since $gh$ is a bijection, $g$ is surjective and $h$ is injective.

*$g$ is a bijection follows immidiately.


The first two things are direct consequences of a proof on an earlier assignment, so I am allowed to apply them here without explanation, and then the third is just the definition of a bijection.
I know that since $g$ is bijective, $g^{-1}$ exists and is also a bijection, but I don't see why $f=fg^{-1}g$ makes $f$ an injection.
I assume that in the same way, I can show $h=g^{-1}gh$, and this implies the surjectivity of $h$.
An explanation of these last two elements of the proof would really help me out. In short, I think (correct me if I'm wrong) I understand how to do the proof, but I would like to understand why I'm doing it this way.
 A: Hint: Try to show that the composition of bijections is a bijection (or perhaps you already know this). Apply this to $f=(fg)\circ g^{-1}$, and similarly for $h$. 
Edit: It seems like you are looking for something more explicit. Let's show $f$ is injective. Suppose 
$$
f(a)=f(b),
$$
we wish to show $a=b$. As $g$ is a bijection, find $x,y$ with $g(x)=a$ and $g(y)=b$. Then $fg(x)=fg(y)$, by the choice of $x$ and $y$. But now $fg$ is a bijection, hence injective, and hence $x=y$. In particular, $a=g(x)=g(y)=b$, as desired. A similar argument works to show $h$ is surjective. 
A: The part your missing can be easily solved using the fact that the bijections are precisely the invertible functions. For example, you can compute the inverse
$$ \left( (fg) g^{-1} \right)^{-1} = g(fg)^{-1} $$
and since since $f = (fg)g^{-1}$, this gives the inverse of $f$.

You can also use this approach from the very beginning. By verifying:
$$ f(g (fg)^{-1}) = 1 $$
$$ (g (fg)^{-1}) f = g (fg)^{-1} f gh (gh)^{-1} = gh(gh)^{-1} = 1 $$
you show that $f$ is invertible, with inverse $g(fg)^{-1}$. And the rest follows from there, using a similar method to the previous section.
