1. ## Transitive Surjection proof

Hey all, can't figure out how to lay out the proof for the following problem:

If f: A -> B is surjective and g: B -> C is surjective then g o f : A -> C is surjective.

well by definition, for all b ∈ B there exists a ∈ A such that f(a) = b
and, for all c ∈ C there exists b ∈ B such that g(b) = c

Any help would be appreciated!

2. ## Re: Transitive Surjection proof

well you already have started the proof. first identify the goal precisely.
The goal is to prove that $\displaystyle g\circ f:A\to C$ is surjective. i.e. For all
$\displaystyle c\in C$ there exists $\displaystyle a\in A$ such that $\displaystyle (a,c)\in g\circ f$.
So first let $\displaystyle c\in C$ to be arbitrary. But as you pointed out , there exists
$\displaystyle b\in B$ such that $\displaystyle g(b)=c$ , which means that
$\displaystyle (b,c)\in g$. Again as you said, for all $\displaystyle b'\in B$ there exists
$\displaystyle a\in A$ such that $\displaystyle f(a)=b'$. Since $\displaystyle b\in B$, we have
$\displaystyle f(a)=b$, which means that $\displaystyle (a,b)\in f$.

So what can you conclude from this ?