Question regarding measurable map and doubt clearing Suppose a map $T:\Omega_1\rightarrow\Omega_2$ and $\sigma$-fields $\mathcal{F}_1$ and $\mathcal{F}_2$.
Define $T^{-1}\mathcal{F}_2 = \{T^{-1}A_2 | A_2\in \mathcal{F}_2\}$ and $T\mathcal{F}_1 = \{A_2 | T^{-1}A_2\in \mathcal{F}_1\}$
Now it is easy to show that $T^{-1}\mathcal{F}_2$ and $T\mathcal{F}_1$ are a $\sigma$-fields. But now my aim is show $T$ is $\mathcal{F}_1/\mathcal{F}_2$ measurable if and only if $T^{-1}\mathcal{F}_2\subset \mathcal{F}_1$ and to $\mathcal{F}_2\subset T\mathcal{F}_1$.
My approach
(Only if part) Given $T$ is measurable $\mathcal{F}_1/\mathcal{F}_2$ that imply $T^{-1}A_2 \in \mathcal{F}_1$ for each $A_2\in \mathcal{F}_2$
Therefore $T^{-1}\mathcal{A}_2 \subset \mathcal{F}_1$ for each $\mathcal{A}_2\subset \mathcal{F}_2$ for a arbitrary collection of set $\mathcal{A}_2$ but I cannot show that $\sigma\{T^{-1}\mathcal{A}_2\}$ = $T^{-1}\mathcal{F}_2$.
then we can conclude that $T^{-1}\mathcal{F}_2\subset \mathcal{F}_1$.
and from definition of measurable map, $A_2\in \mathcal{F}_2\Rightarrow T^{-1}A_2\in \mathcal{F}_1$
Therefore  $A_2\in T\mathcal{F}_1$ then $\mathcal{F}_2\subset T\mathcal{F}_1$.(is this right conclusion?)
(if part) Given that $T^{-1}\mathcal{F}_2\subset \mathcal{F}_1$ and $\mathcal{F}_2\subset T\mathcal{F}_1$.
$T^{-1}\mathcal{F}_2\subset \mathcal{F}_1 \iff A_2\in\mathcal{F}_2\Rightarrow T^{-1}A_2\in\mathcal{F}_1$.............(1)
and
$\mathcal{F}_2\subset T\mathcal{F}_1 \iff T^{-1}A_2\in\mathcal{F}_1\Rightarrow A_2\in \mathcal{F}_2$..................(2)
Therefore from (1) and (2) we can say that $T$ is measurable. (Is my logic correct?)
New short approach appreciable. Thanks in advance
 A: You want to prove that:
$T$ is $\mathcal{F}_1/\mathcal{F}_2$ measurable if and only if $T^{-1}\mathcal{F}_2\subset \mathcal{F}_1$ if and only if $\mathcal{F}_2\subset T\mathcal{F}_1$.
Proof. $T$ is $\mathcal{F}_1/\mathcal{F}_2$ measurable if and only if, for each $A_2\in \mathcal{F}_2$, $T^{-1}A_2 \in \mathcal{F}_1$ , that means,
$$T^{-1}\mathcal{F}_2 = \{T^{-1}A_2 | A_2\in \mathcal{F}_2\}\subset \mathcal{F}_1$$
So we have proved the first equivalence.
Now let us prove that  $T^{-1}\mathcal{F}_2\subset \mathcal{F}_1$ if and only if $\mathcal{F}_2\subset T\mathcal{F}_1$. We have
\begin{align*}
T^{-1}\mathcal{F}_2\subset \mathcal{F}_1 & \Leftrightarrow 
\{T^{-1}A_2 | A_2\in \mathcal{F}_2\} \subset \mathcal{F}_1 \\
& \Leftrightarrow 
\textrm{for all } A_2\in \mathcal{F}_2, T^{-1}A_2 \in  \mathcal{F}_1\\
&\Leftrightarrow 
\mathcal{F}_2 \subset \{A_2 | T^{-1}A_2\in \mathcal{F}_1\} \\
& \Leftrightarrow 
\mathcal{F}_2\subset T\mathcal{F}_1
\end{align*}
Remark: We have proved the $T$ is $\mathcal{F}_1/\mathcal{F}_2$ measurable if and only if $T^{-1}\mathcal{F}_2 \subset \mathcal{F}_1$.
So $T^{-1}\mathcal{F}_2$ is the smallest $\sigma$-algebra that we can take in $\Omega_1$ that make $T$ is $\mathcal{F}_1/\mathcal{F}_2$ measurable.
If we take $\mathcal{F}_1$ to be smaller that $T^{-1}\mathcal{F}_2$, then $T^{-1}\mathcal{F}_2\subset \mathcal{F}_1$ no longer holds and $T$ will not be  $\mathcal{F}_1/\mathcal{F}_2$ measurable.
