We are given the functions $f:A\rightarrow B$ and $g:B\rightarrow A$ with the property that $g\circ f=i_A$.
I have so far shown that $f$ is injective and $g$ is surjective. Now, I am stuck with trying to show that if $f$ is surjective, then $g$ is injective.
I think an easy approach to this question would be with a proof by contrapositive.
Help is really much appreciated.
Let $b_1,b_2\in B$ such that $g(b_1)=g(b_2)$, since $f$ is surjective we have:
$$ \exists\, a_1,a_2\in A \ \text{such that}\ f(a_1)=b_1 \ \ f(a_2)=b_2 $$
this implies that
$$ g(f(a_1))=g(b_1)=g(b_2)=g(f(a_2)) $$
so, since $\ g\circ f=id_{A}$ we have
$$ a_1=g(f(a_1))=g(f(a_2))=a_2 $$
so $a_1=a_2$ and finally
$$ b_1=f(a_1)=f(a_2)=b_2 $$
so $b_1=b_2$.
So far so good: from $g \circ f = i_A$, you can prove $f$ is injective and $g$ is surjective. Unfortunately, this doesn't necessarily imply $f \circ g = i_B$ (see my answer here).
Assuming $g \circ f = i_A$, then assuming $f$ is surjective or $g$ is injective will imply that $f$ or $g$ is invertible, and that each is the other's inverse. So no, you won't be able to prove these things without further assumption.
For example, is $A$ and/or $B$ finite with the same number of elements?
