Surjections and Injections Please examine following Theorem and the accompanying proof. I understand the idea behind the proof, I am just concerned that I might not have put it in the correct words.
Is the argument correct? If so can the write up be improved?

Given that $f:A\to B$ and that $g:B\to C$.  Prove that if $f$ is not onto and $g$ is one-to-one, then $g\circ f$ is not onto.
Proof. Assume that $f$ is not onto, $g$ is one-to-one and $g\circ f$ is onto. Since $f$ is not onto it follows that for some $x\in B$ it is that case that $$\forall a\in A(f(a)\neq x)\ (1)$$
  Since $g:B\to C$ it must be that for some $c\in C$, $g(x)=c$ and since $g$ is one-to-one, $$\forall y\in B(g(y)=c\implies y=x)\ (2)$$
  Since $g\circ f$ is onto it follows that for some $z\in A$, $g(f(z))=c$ but this implies the existence of some $\alpha\in B$ such that $(z,\alpha)\in f$ and $(\alpha,c)\in g$ but $g(\alpha)=c$ and $(2)$ entails that $\alpha=x$ but $(1)$ suggests that no element in $A$ has an image $x$ under $f$ thus no such $\alpha$ exists consequently $$\neg\exists a\in A(g(f(a))=c)$$ contradicting the assumption that $g\circ f$ is onto.
$\blacksquare$


The proof is correct. On the other hand you can avoid contradiction, by proving the contrapositive.

Proposition. Given the maps $f\colon A\to B$ and $g\colon B\to C$, if $g$ is one-to-one and $g\circ f$ is onto, then $f$ is onto.

Proof. Let $b\in B$. Since $g\circ f$ is onto, there exists $a\in A$ such that
$$
g(b)=g\circ f(a)=g(f(a))
$$
so $b=f(a)$ because $g$ is one-to-one. Therefore $f$ is onto.$\quad\square$
