Proof that the preimage of generated $\sigma$-algebra is the same as the generated $\sigma$-algebra of preimage. This question has also been asked here, but the answer there didn't help me.
I am trying to prove that, given some measurable space $(X, \Sigma)$, if $G$ is a collection of subset of $X$ such that $\sigma(G) = \Sigma$, then
$$f^{-1}(\sigma(G)) = \sigma(f^{-1}(G)).$$
So far, I have been able to show that $\sigma(f^{-1}(G)) \subseteq f^{-1}(\sigma(G))$, but I am having trouble with the opposite inclusion. I have shown that $f^{-1}(G)$ is a $\sigma$-algebra, from which it follows that $f^{-1}(G) \subseteq \sigma(f^{-1}(G))$, but that's about all I have been able to show. How do I proceed from there?
The answer in the other question suggests the following approach:
Prove that


*

*$f^{-1}(G) \in \sigma(f^{-1}(G))$,

*if $f^{-1}(A_i) \in \sigma(f^{-1}(G))$ for $i \in \mathbb{N}$ (i.e. a countable collection), then $f^{-1}(\bigcup_i A_i) \in \sigma(f^{-1}(G))$, and

*if $f^{-1}(A) \in \sigma(f^{-1}(G))$ then $f^{-1}(Y \setminus A) \in \sigma(f^{-1}(G))$.


However, I am not sure how to actually prove this, and even then, I'm not sure how it follows from this that $f^{-1}(\sigma(G)) \subseteq \sigma(f^{-1}(G))$.
 A: After working on it some more, I have come up with a proof. It follows the ideas of the question I linked, and uses the properties of preimage that saz listed in the comments. I also found out that the proof does not actually need the premise that $\sigma(\mathcal{G}) = \Sigma$, and it is in fact not even necessary that $X$ is a measurable space.
The proof is as follows. We wish to prove that $f^{-1}(\sigma(\mathcal{G})) \subseteq \sigma(f^{-1}(\mathcal{G}))$.
  First we define
  $$D = \{G \subseteq Y \mid f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))\}.$$
  Observe that what we want to show will follow if we prove that
  $D$ is a $\sigma$-algebra and that $\mathcal{G} \subseteq D$.
  This is because it will imply that $\sigma(\mathcal{G}) \subseteq D$,
  which, by the definition of $D$, will imply that if $G \in \sigma(\mathcal{G})$,
  then $f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))$.
  This further implies that
  $$f^{-1}(\sigma(\mathcal{G})) = \{f^{-1}(G) \mid G \in \sigma(\mathcal{G})\} \subseteq \sigma(f^{-1}(\mathcal{G})).$$
  Hence, we now show that $D$ is a $\sigma$-algebra and that $\mathcal{G} \subseteq D$.
  First observe that $f^{-1}(\mathcal{G}) \subseteq \sigma(f^{-1}(\mathcal{G}))$.
  This implies that if $G \in \mathcal{G}$ then $f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))$.
  By the definition of $D$, this further implies that $\mathcal{G} \subseteq D$.
  To show that $D$ is a $\sigma$-algebra,
  we verify properties of a $\sigma$-algebra.


*

*Since $\emptyset = f^{-1}(\emptyset)$ and
                         $\emptyset \in \sigma(f^{-1}(\mathcal{G}))$,
                         we have $\emptyset \in D$.

*Assume $G \in D$. Then $f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))$.
                         Since $\sigma(f^{-1}(\mathcal{G}))$ is closed under complement,
                         we get
                         $$X \setminus f^{-1}(G) = f^{-1}(Y) \setminus f^{-1}(G) = f^{-1}(Y \setminus G) = f^{-1}(G^c) \in \sigma(f^{-1}(\mathcal{G})),$$
                         which implies that $G^c \in D$.

*Assume $G_1,G_2,\dots \in D$. Then for all $G_i$
                         we have $f^{-1}(G_1) \in \sigma(f^{-1}(\mathcal{G}))$.
                         Since $\sigma(f^{-1}(\mathcal{G}))$ is closed under
                         countable union, we get
                         $$\bigcup_i f^{-1}(G_i) = f^{-1}(\bigcup_i G_i) \in \sigma(f^{-1}(\mathcal{G})),$$
                         which implies that $\bigcup_i G_i \in D$.
A: This is based on the following properties of $f^{-1}$:
$$
f^{-1}(\bigcup_{i=1}^{\infty}G_i)=\bigcup_{i=1}^{\infty}f^{-1}(G_i)\tag1
$$
$$
f^{-1}(\bigcap_{i=1}^{\infty}G_i)=\bigcap_{i=1}^{\infty}f^{-1}(G_i)\tag2
$$
$$
f^{-1}(G_i^c)=f^{-1}(G_i)^c\tag3
$$
Suppose $A\in \sigma(G)$ and $\:A=\bigcup_{i=1}^{\infty}G_i,\:G_i\in G$, i.e. $A$ is formed by the countable union of sets in $G$. Then 
$$
f^{-1}(A)=f^{-1}(\bigcup_{i=1}^{\infty}G_i)=\bigcup_{i=1}^{\infty}f^{-1}(G_i)\in\sigma(f^{-1}(G))
$$
For $A\in \sigma(G)$ and $\:A=\bigcap_{i=1}^{\infty}G_i,\:G_i\in G$, i.e. $A$ is formed by the countable intersection of sets in $G$. Then 
$$
f^{-1}(A)=f^{-1}(\bigcap_{i=1}^{\infty}G_i)=\bigcap_{i=1}^{\infty}f^{-1}(G_i)\in\sigma(f^{-1}(G))
$$
For $A\in \sigma(G)$ and $\:A=G_i^c,\:G_i\in G$, i.e. $A$ is formed by the complement of set in $G$. Then 
$$
f^{-1}(A)=f^{-1}(G_i^c)=f^{-1}(G_i)^c\in \sigma(f^{-1}(G))
$$
$G_i\in \sigma(G)$ is again the countable union, intersection and complement of sets and so $f^{-1}(G_i)\in \sigma(f^{-1}(G))$. 
So we have
$$f^{-1}(\sigma(G)) \subseteq \sigma(f^{-1}(G))$$
A: mrp and hermes answers are great! I just want to add a solution which does not hinge on the construction of our $\sigma$-algebra as combinations of set operations but rather the definiton of our the sigma algebra of a set $G$ being the intersection of all sigma algebras containing $G$, or the smallest sigma algebra containing $G$.
$f^{-1}(\sigma(G))\subseteq \sigma (f^{-1}(G))$: Lets prove the contrapositive that $\neg (f^{-1}(\sigma(G))\supset \sigma (f^{-1}(G)))$. Assume $f^{-1}(\sigma(G))\supset \sigma (f^{-1}(G))$ and let $f^{-1}(\sigma(G))\setminus \sigma (f^{-1}(G))=\{f^{-1}(x_1),f^{-1}(x_2),...,f^{-1}(x_n),...\}$ for some $x_i \in \sigma(G)$. It follows that $f^{-1}(\sigma(G))\setminus \{f^{-1}(x_1),f^{-1}(x_2),...,f^{-1}(x_n),...\}=f^{-1}(\sigma(G)\setminus\{x_1,x_2,...,x_n,...\})$ this will ofcourse be the same as $\sigma(f^{-1}(G))$. Now as $\sigma(G)\setminus\{x_1,x_2,...,x_n,...\}$ is smaller than $\sigma(G)$ we must either (1) have that it does not contain $G$ or (2) its not a $\sigma$-algebra, the equality $f^{-1}(\sigma(G)\setminus\{x_1,x_2,...,x_n,...\})=\sigma (f^{-1}(S))$ rules out (2) so it must be that $\sigma(G)\setminus\{x_1,x_2,...,x_n,...\}$ is no $\sigma$-algebra, this means that there are sets $A_n$ in $\sigma(G)\setminus\{x_1,x_2,...,x_n,...\}$ such that $\bigcup_n A_n\notin \sigma(G)\setminus\{x_1,x_2,...,x_n,...\}$(or some $A\setminus B$) this means that $f^{-1}(\bigcup_n A_n)=\bigcup_n f^{-1}(A_n)\notin f^{-1}(\sigma(G)\setminus \{ x_1, x_2,...\})$ but $f^{-1}(A_n)$ must be in $\sigma(f^{-1}(G))$ as we otherwise would have removed it so $\bigcup_n f^{-1}(A_n)\in \sigma(f^{-1}(G))=f^{-1}(\sigma(G)\setminus \{ x_1, x_2,...\}$ a contradiction so our assumption must be wrong. This proves the inclusion
$f^{-1}(\sigma(G))\supseteq \sigma (f^{-1}(G))$: here it is enough to remember that the preimage of a $\sigma$-ring being a $\sigma$-ring so $f^{-1}(\sigma (G))$ is a $\sigma$-ring and as $G\subseteq\sigma (G)$ we also have $f^{-1}(G)\subseteq f^{-1}(\sigma(G))$ so $f^{-1}(\sigma(G))$ is a $\sigma$-ring containing $f^{-1}(G)$ from which it follows that $f^{-1}(\sigma(G))\supseteq \sigma(f^{-1}(G))$ as $\sigma(f^{-1}(G))$ is the smallest $\sigma$-ring containing $f^{-1}(G)$ $\square$
I am still fairly new to this area(the course started just over a week ago) so my argument may be messy, sorry about that. But i do think this is a nice alternative.
