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.
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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.
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