Let $f: A\rightarrow A$. Prove that if $(f ◦ g)$ is surjective, then $f$ is surjective. Let $f: A\rightarrow A$. Prove that if $(f ◦ g)$ is surjective, then $f$ is surjective.
We know that $(f ◦ g)$ is surjective, so that means that $\forall x\in A, \exists a$ such that $f(g(a)) = x$
Let $y = g(a) \in A$
$\implies f(y) = f(g(a)) = x$
$\implies \forall x\in A, \exists y$ such that $f(y) = x$
$\implies f$ is surjective
Is this proof correct?
 A: You seem to have all the pieces, but they are arranged in a way that is making the reader do too much work.  Here is a more typical way of arranging the facts (with a lot of extra text along the way to support apprentice mathematicians).

Let $x\in A$ be given.  We intend to show that there is some $y\in A$ such that $f(y)=x$.  Since $f\circ g$ is surjective, we may choose $a\in A$ such that $(f\circ g)(a)=f(g(a))=x$.  Since $g(a)\in A$, we may take $y=g(a)$ to be the value we were seeking.  Since $x$ was arbitrarily chosen, $f$ is surjective.

For accessibility, I used the variable names you chose for your proof.  That said, it is more typical and readable for others to use $y$ for a member of a range and $x$ for a member of the domain (so that we would be able to say $f(x)=y$ instead of $f(y)=x$.  
A: Another proof: the image of $f$ restricted to $g(A)$ is $A$, so the image of $f$ contains $A$.  Since the codomain of $f$ is $A$, the image is a subset of $A$, so it follows that the image is $A$.
