1. ## Composition Function Proof

I am stuck on a composition function Proof.

Let f:A-->B and g:B-->D. Prove that g 0 f is onto and g is 1-1, then f must be onto.

I have ( and am not sure if it is correct...)

Let f:A-->B and g:B-->D
Let all d be and element of D and there exists and a in A such that f(a)=b.
Then (g o f)(a)=g(f(a))
=g(b)
=d
Therefore, g o f is onto

Where I am a little confused is it looks like to me is that f:A-->B would not only have to be onto but also one to one...

2. Originally Posted by Pi R Squared
Let f:A-->B and g:B-->D. Prove that gof is onto and g is 1-1, then f must be onto.
Note what is to be proven.

If $b\in B$ then $g(b)\in D$.
Because $g \circ f$ is onto $\left( {\exists a \in A} \right)\left[ {g \circ f(a) = g(f(a)) = g(b)} \right]$.

You know that $g$ is one-to-one. So Finish.