# Is the pull back of a generated $\sigma$-algebra itself a generated $\sigma$-algebra?

Let $f : \Omega_{1} \rightarrow \Omega_{2}$ be a map from a measurable space $\Omega_{1}$ to another measurable spacce $(\Omega_{2},\mathcal{B})$, where $\mathcal{B}$ is the generated $\sigma$-algebra of some $\textbf{countable}$ collection of subsets of $\Omega_{2}$:

$$\mathcal{B}=\sigma(\mathcal{C}),$$ where $\mathcal{C}$ is countable.

Do we have that the pull-back $\sigma$-algebra on $\Omega_{1}$, $f^{-1}(\mathcal{B})$, also a generated-$\sigma$ algebra? I think that the answer should be yes and $f^{-1}(\mathcal{B})$ is generated by $f^{-1}(\mathcal{C})$ but I don't know how to prove this. Could anyone help on this? Thank you so much!

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Yes, and you don't need to assume that $\mathcal{C}$ is countable. Show that pullback preserves every operation. – Qiaochu Yuan Oct 12 '12 at 1:40
@QiaochuYuan Can you be more specific? I don't know how to show $f^{-1}(\mathcal{B}$ is the smallest $\sigma$-algebra that contains $f^{-1}(\mathcal{C})$ – user7762 Oct 12 '12 at 2:17

For any (not necessarily countable) collection ${\cal C}$ of subsets of $\Omega_2$ and any function $f:\Omega_1 \rightarrow \Omega_2$, we have: $$f^{-1}(\sigma({\cal C})) =\sigma(f^{-1}({\cal C})).$$
The "$\supseteq$" is clear as $f^{-1}({\cal C})\subseteq f^{-1}(\sigma({\cal C}))$ and $f^{-1}(\sigma({\cal C}))$ is a $\sigma$-algebra, hence including $\sigma(f^{-1}({\cal C}))$ (the minimal $\sigma$-algebra including collection $f^{-1}({\cal C})$).
The "$\subseteq$" is not immediately obvious. Let us denote by ${\cal G}$ the collection of all sets $G\subseteq \Omega_2$ such that $f^{-1}(G) \in \sigma(f^{-1}({\cal C}))$. We note that ${\cal G}$ is a $\sigma$-algebra and that ${\cal C}\subseteq {\cal G}$, so $\sigma({\cal C})\subseteq {\cal G}$ (again by minimality of generated $\sigma$-algebra). This means $f^{-1}(\sigma({\cal C})) \subseteq \sigma(f^{-1}({\cal C}))$.