# To prove $h \circ g \circ f$ is bijective.

Given are 3 functions $f,g,h$ of afinite Set $P$, and also given is that $g \circ f$ and $h \circ f$ are bijective.

Need to prove $h \circ g \circ f$ is bijective.

I know that as $g \circ f$ and $h \circ f$ are bijective, so both are injective as well as surjective. So

If $g \circ f$ is surjective then $g$ is surjective

If $h \circ g$ is injective then $g$ is injective

So it means $g$ is bijective

But using this information along with the information that the set is finite, how will I be able to conclude that $h \circ g \circ f$ is bijective.

Can anybody provide a hint to tackle this problem.

• Can you show that $f$ and $h$ are bijections as well? Then $h \circ g \circ f$ is the composition of bijections... Nov 2 '15 at 21:58
• With what you've given, you can show quite easily that $h\circ g\circ f$ is surjective. A surjective map from a finite set to itself can be shown to be bijective Nov 2 '15 at 21:59
• @TheRob you are correct but then how to proceed to show that $h \circ g \circ f$ is surjective Nov 2 '15 at 22:03
• @MichaelBiro we can only show for $g$, as $g$ is common in both $g \circ f$ and $h \circ g$ Nov 2 '15 at 22:05
• You're given $h\circ f$ bijective, so $h$ is surjective, $g\circ f$ is bijective so it's definitely surjective. The composition of surjective maps is surjective, so $h\circ g\circ f$ is surjective Nov 2 '15 at 22:07

Use that for a function on a finite set $P$, being an injection, a surjection or a bijection are all equivalent.
$f$ is already injective (from $g \circ f)$, you already know $g$ is too and $h$ is surjective. So all are bijections...
$g\circ f$ bijective implies $f$ is injective and $g$ surjective. Similarly, $h$ is surjective.
Furthermore, for a map from a finite set into itself, injective $\iff$ surjective $\iff$ bijective. Hence $f, g,h$ are all bijective. So $h\circ g\circ f$ is bijective.