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

So what can you conclude from this ?