Suppose g: A-->C and h: B-->C.
Prove: If h is bijective then there exists a function f: A-->B such that g=h(composed with)f.
You know that $\displaystyle h^{ - 1} :C \mapsto B$. WHY?
Consider the mapping $\displaystyle h^{ - 1} \circ g:A \mapsto B$.
Does that work?
Now, I have left alot for you to explain.