Given A to C, and B to C with known complexities, what can be said about A to B?
We can have sets $A, B, C$ with linear-time computable maps $f : A \to C$ and $g : B \to C$ such that there exists a map $h : A \to B$ with $f = g \circ h$, but the needed time complexity/Turing degree for $h$ is as high as you want.

Proof: Pick a map $H : \Sigma^* \to \Sigma^*$ which is hard in whatever sense you have chosen. Now let $A = C = \Sigma^*$, and $B = \{\langle w, H(w)\rangle \mid w \in \Sigma^*\}$. Let $f = \mathrm{id}$ and $g = \pi_1$, i.e. $g(\langle w,u\rangle) = w$. Now $A,B,C$ and $f,g$ meet the criteria of the claim, and the only map $h$ that works is $h(w) = \langle w, H(w)\rangle$, which is essentially as hard as $H$.