$\mathcal A=\{E\subseteq Y:\; f^{-1}(E)\in \sigma(f^{-1}(\mathcal C)\}$ is a $\sigma-algebra$ in $Y$ and $C \subset \mathcal A$ Let $f: X \to Y$ be a function, $\mathcal C$ a family of subsets  of $Y$, prove that
$\mathcal A=\{E\subseteq Y:\; f^{-1}(E)\in \sigma(f^{-1}[\mathcal C])\}$ is a $\sigma$-algebra in $Y$ and $\mathcal C \subset \mathcal A$
my attempt:
if $C \in \mathcal C$, then  $ f^{-1}(C)\in f^{-1}[\mathcal{C}]\subseteq\sigma(f^{-1}[\mathcal{C}])$ so $C\in\mathcal{A}$
but I have not been able to prove that $\mathcal A=\{E\subseteq Y:\; f^{-1}(E)\in \sigma(f^{-1}[\mathcal C])\}$ is a $\sigma$-algebra
 A: So a $\sigma$-algebra is by definition closed under countable set theoretic operations. It is enough to check that $\mathcal A$ is closed under countable unions and complement.
Let $\{U_{i}\}$ be a countable family contained in $\mathcal A$. Then $f^{-1}(U_{i})\in\sigma(f^{-1[\mathcal C]})$ for every $i$. Hence $\bigcup_{i}f^{-1}(U_{i})\in\sigma(f^{-1}[\mathcal C])$, since $\sigma(f^{-1[\mathcal C]})$ is closed under countable unions. We then have $f^{-1}(\bigcup_{i}U_{i})\in\sigma(f^{-1}[\mathcal C])$, which is equivalent to the statement $\bigcup_{i}U_{i}\in\mathcal A$.
We check that if $A\in\mathcal A$, then $A^\complement\in\mathcal A$. Statement $A\in\mathcal A$ implies $f^{-1}(A)\in\sigma(f^{-1}[\mathcal C])$ once again using the definition of $\sigma$-algebra we have that $\sigma(f^{-1}[\mathcal C])$ is closed under taking complements, hence we obtain $Y\setminus f^{-1}(A)= f^{-1}(Y\setminus A)\in\sigma(f^{-1}[\mathcal C])$ which proves that $A\complement\in\mathcal A$.
A: Note $\sigma(f^{-1}(\mathcal{C}))$ is a $\sigma$ algebra in X.

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*$f^{-1}(Y)= X \in \sigma(f^{-1}(\mathcal{C}))$. Thus $X\in \mathcal{A}$


*Let $W \in \mathcal{A}$ then $f^{-1}(Y \setminus W)= X\setminus f^{-1}(W) \in \sigma(f^{-1}(\mathcal{C}))$ Thus $Y \setminus W \in \mathcal{A}$


*Let $W_i \in \mathcal{A}$ then $f^{-1}(\bigcup_{i=1}^{\infty} W_i)= \bigcup_{i=1}^{\infty} f^{-1} W_i \in \sigma(f^{-1}(\mathcal{C}))$ Thus $\bigcup_{i=1}^{\infty} W_i\in \mathcal{A} $
