Problem 13.3 of Probability and Measure by Billingsley states:
$(\Omega, \mathcal{F})$ and $(\Omega', \mathcal{F}')$ are two measurable spaces. Suppose that $f: \Omega \rightarrow \mathbb{R}^1$ and $T:\Omega \rightarrow \Omega' $. Show that $f$ is measurable $T^{-1}\mathcal{F}':= \{ T^{-1}A': A' \in \mathcal{F}' \}$ if and only if there exists a map $\phi: \Omega' \rightarrow \mathbb{R}^1$ such that $\phi$ is measurable $\mathcal{F}'$ and $f= \phi T$.
Hint: First consider simple functions and then use Theorem 13.5.
where Theorem 13.5 states
If $f$ is real and measurable $\mathcal{F}$, there exists a sequence $\{f_n\}$ of simple functions, each measurable $\mathcal{F}$, such that $$0 \leq f_n(\omega)\uparrow f(\omega) \text{ if }f(\omega) \geq 0$$ and $$ 0 \geq f_n(\omega) \downarrow f(\omega) \text{ if }f(\omega) \leq 0.$$
I would like to consider a general case of Problem 13.3 where the codomain of $f$ is a general measurable space $(X, \mathcal{N})$ rather than $(\mathbb{R}^1, \mathcal{B}(\mathbb{R}^1))$, i.e. when it can be true that
$(\Omega, \mathcal{F})$ and $(\Omega', \mathcal{F}')$ are two measurable spaces. Suppose that $f: \Omega \rightarrow X$ and $T:\Omega \rightarrow \Omega' $. $f$ is measurable $T^{-1}\mathcal{F}'/\mathcal{N}$ if and only if there exists a map $\phi: \Omega' \rightarrow X$ such that $\phi$ is measurable $\mathcal{F}'/\mathcal{N}$ and $f= \phi T$.
noticing that Theorem 13.5 does not apply here.
Thanks and regards!