Problem statement, as written:
Let $f\colon C\rightarrow A$ and $g\colon B\rightarrow A$ be functions, and suppose $g$ is bijective. Prove that there exists $h\colon C\rightarrow B$ such that $f=g\circ h$ if and only if $ran(f) \subseteq ran(g)$. Prove that $h$ is unique.
[Throughout, $I_S$ represents the identity function on the set $S$.]
Proof: Since $g$ is bijective, it is invertible; hence, $g^{-1}\colon A\rightarrow B$. Let $h\colon C\rightarrow B$ be defined by $h=\{(c,b)|c\in C$ and $b=g^{-1}(f(c))\}$.
To see that $h$ satisfies the definition of function, we show that 1) $dom(h) = C$, 2) $ran(h) \subseteq B$, and 3) $(x,y_1),(x,y_2) \in h \rightarrow y_1 = y_2$. 1) is trivial to show, as $dom(h) = C$ by how $h$ is defined. To show 2), let $b_0\in ran(h)$. Then by the definition of $h$, $b_0 = g^{-1}(f(c))$, for some $c\in C$. Since $f\colon C\rightarrow A$, $f(c)\in A$. In turn, $g^{-1}\colon A\rightarrow B$, so $b_0 = g^{-1}(f(c))\in B$. Thus, $ran(h)\subseteq B$. To prove 3), let $(x,y_1),(x,y_2) \in h$. Then, again, by the definition of $h$, $y_1 = g^{-1}(f(x))$ and $y_2 = g^{-1}(f(x))$. Since $g^{-1}$ is well-defined, $g^{-1}(f(x))$ is unique; hence, $y_1=y_2$.
Next, we show that $f=g\circ h$ if and only if $ran(f) \subseteq ran(g)$. $(\rightarrow)$ Suppose $f=g\circ h$. Let $a_0\in ran(f)$. Then $a_0=f(c)$ for some $c\in C$. Since $f=g\circ h$ by hypothesis, $(c,a_0)\in g\circ h$. This means $(c,b)\in h$ and $(b,a_0)\in g$ for some $b\in B$. Applying the definition of $h$, $b=g^{-1}(f(c))$. In consequence, $a_0 = g(g^{-1}(f(c)))$; so $a_0 \in ran(g)$. $(\leftarrow)$ Now suppose $ran(f) \subseteq ran(g)$. To see that $f=g\circ h$, let $c\in C$. Then $(g\circ h)(c) = g(h(c)) = g(g^{-1}(f(c))) = I_A(f(c)) = f(c)$.
Lastly, to see that $h$ is unique, let $l\colon C\rightarrow B$ be any other function such that $f=g\circ l$ if and only if $ran(f) \subseteq ran(g)$. Let $a_0\in ran(f)\subseteq A$; since $g\colon B\rightarrow A$ is, by hypothesis, bijective (and hence, surjective), $a_0 = g(b)$ for some $b\in B$. Thus, $a_0\in ran(g)$, so $ran(f)\subseteq ran(g)$. This implies, by hypothesis, that $f=g\circ l$ and $f=g\circ h$; hence $g\circ l = g\circ h$. Then $g^{-1}\circ (g\circ l)$ = $g^{-1}\circ (g\circ h)$. On one hand, $g^{-1}\circ (g\circ l) = (g^{-1}\circ g)\circ l= I_B\circ l = l$. On the other hand, $g^{-1}\circ (g\circ h) = (g^{-1}\circ g)\circ h = I_B\circ h = h$. Thus, $l=h$. Hence, $h$ is unique. Q.E.D.