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.
Best Answer
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.