Let $(\Omega, \mathcal F)$ be a measurable space and $A \in \sigma(f_k : k \in \mathbb N)$ for some measurable functions $f_k:\Omega \rightarrow E $ into another measurable space $(E, \mathcal E)$.
Now consider $g:\Omega \rightarrow \Omega$ measurable.
Do we have $g^{-1}(A) \in \sigma(f_k\circ g: k \in \mathbb N)$?
My attempt: I suspect that the statement holds, since we can verify the statement directly for the sets of the form $f_k^{-1}(B)$ such that $B\in \mathcal E$, the sets which generate $\sigma(f_k : k \in \mathbb N)$. So the question arises whether it is enough to prove the result only for the generator?
Help appreciated.
It is quite easy to verify that $\{A\in \sigma \{f_1,f_2,\cdots\}:g^{-1}(A) \in \sigma \{f_1\circ g,f_2\circ g,\cdots\}\}$ is a sigma algebra. Since it contains the generating sets it conatins the entire sigma algebra $\sigma \{f_1,f_2,\cdots\}$ and this is what we want to porve.