# 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