let $(X,\mathfrak{F},\mu)$ be a measure space and $(Y,\mathfrak{G})$ a mesurable space, $t:X \rightarrow Y$ a measurable function and $\sigma(t)$ the $\sigma$-algebra generated by $t$. Is it true that for each $\sigma(t)$-measurable function $f:X\rightarrow \mathbb{R}$ it exists a measurable function $\phi:Y \rightarrow \mathbb{R}$ such that $f(x)=\phi(t(x))$ $\mu-a.e.$ ??
look: taking $Y=\{y\}$ the $\sigma$-algebra generated by $t$ is $\{X,\varnothing\}$ and the only $\{X,\varnothing\}$-measurable function is constant, so (unfortunately) this does not work as a counterexample.
The answer seems to be yes.
Note first that $\sigma(t)=\{t^{-1}(A):A\in\mathfrak{G}\}$. So if $f=\chi_E$ then $$f=\chi_{t^{-1}(A)}=\chi_A\circ T.$$
So it's true if $f$ is simple. Now if $f\ge0$ there exists a sequence $(f_n)$ of simple functions increasing to $f$. If $f_n=\phi_n\circ t$ then $(\phi_n)$ is increasing. So $\phi_n\to\phi$ pointwise, hence $f=\phi\circ t$.