Is this a correct usage of the axiom of choice?
Let $f:B\rightarrow C$ and $g:A\rightarrow B$ be onto, then by definition: \begin{align*} &\forall c \in C, \exists b\in B,\text{ s.t. } f(b)=c \ \ \ \text{ and}\\ &\forall b \in B, \exists a\in A,\text{ s.t. } g(a)=b \end{align*} By definition of composition (from Hubbard p.15): \begin{align*} (f\circ g)(a)=f[g(a)] \end{align*} with domain $A$ and codomain $C$.
By the axiom of choice, every surjection has a right inverse. Since $f$ is surjective, there exists a $b$ such that $f^{-1}(b)=c$. Similarly, $g$ is also surjective, so there exists a $a$ such that $g^{-1}(b)=a$.
Therefore $\forall c \in C,\exists a\in A, \text{ s.t. } f[g(a)]=f(b)=c$.
The Axiom of Choice is needed if you wish to construct right inverse maps, i.e. maps $$ \phi_f: C\rightarrow B, \ \ \phi_g: B \rightarrow A$$ with the property that $$ f \circ \phi_f = 1_C, \ \ g \circ \phi_g = 1_B$$ This follows easily with the AoC but can in general not be done without. If you only want to find inverse images for specific elements this is just set-theory (no need for the AoC).