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.


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.

  • $\begingroup$ "The composition of bijections is a bijection" is exactly the part that I was missing. That makes perfect sense in hindsight, but it didn't "click" until I read it explicitly. Thank you! $\endgroup$ – wz-billings Nov 30 '17 at 4:46

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.


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