Function composition and bijection Let $E, F, G$ be sets and $f : E \to F$, $g : F \to G$, $h : G \to E$. Suppose


*

*$h \circ g \circ f$ is surjective;

*$g \circ f \circ h$ and $f \circ h \circ g$ are injective.


I have already proven that h is bijective. Now I want to prove that $g \circ f$ is bijective. How can I do that ?  
Thank you
 A: We will use the following property of functions from which the proofs are elementary:


*

*A function $F:X\to Y$ is injective iff there exists $G:Y\to X$ such that $G\circ F=\mathrm{id}_X$;

*A function $F:X\to Y$ is surjective iff there exists $G:Y\to X$ such that $F\circ G=\mathrm{id}_Y$;

*A function $F:X\to Y$ is bijective iff there exists $G:Y\to X$ such that $G\circ F=\mathrm{id}_X$ and $F\circ G=\mathrm{id}_Y$.

*For $F:X\to Y$, if there exists a surjection $G:Z\to X$ such that $F\circ G$ is injective, then $F$ is injective.

*For $F:X\to Y$, if there exists an injectition $G:Y\to Z$ such that $G\circ F$ is surjective, then $F$ is surjective.

*A function which is injective & surjective is bijective.




Let $E, F, G$ be sets and $f:E\to F$, $g:F\to G$ and $h:G\to E$. Suppose that $h\circ g\circ f$ is surjective and $g\circ f\circ h$, $f\circ h\circ g$ are injective.

By the property given, it is immediate that 


*

*$h$, $h\circ g$, $f\circ h$ are injective;

*$h$, $h\circ g$ are surjective;


by associativity of composition of functions. Therefore, like you showed, $h$ is bijective. Also, we see that $h\circ g$ is bijective.  
Appliying the 4th and 5th property, we have $(g\circ f)\circ h$ injective with $h$ surjective, which implies $g\circ f$ injective. In a similar manner, $h\circ(g\circ f)$ is surjective with $h$ injective, which implies $g\circ f$ surjective. By our 6th proprety, $g\circ f$ is bijective.
