Thread: Function Composition

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.

Originally Posted by GoldendoodleMom

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 . WHY?
Consider the mapping .
Does that work?
Now, I have left alot for you to explain.