When is the image of a $\sigma$-algebra a $\sigma$-algebra? Let $(E,\mathcal{E})$ and $(F,\mathcal{F})$ be measurable spaces and $f:E \rightarrow F$ with $f$ $\mathcal{E}/\mathcal{F}$ measureable. When is $f(\mathcal{E})$ a $\sigma$-algebra? 
I am aware that the inverse image of a $\sigma$-algebra is a $\sigma$-algebra. It is obvious that $f$ must be surjective. Also since $f(\bigcup_{n \in \mathbb{N}}A_n)= \bigcup_{n \in \mathbb{N}} f(A_n)$ and $f(\emptyset ) = \emptyset $, $f(\mathcal{E})$ is $\sigma$-algebra iff for each $A \in \mathcal{E}$ there exists $A' \in \mathcal{E}$ with $f(A')=f(A)^C$ ($f(\mathcal{E})$ is closed under complements).
Is there anyway of simplifying this condtion?
 A: For reference, I'm proving the equivalence suggested in this comment.
Let $(E, \mathcal{E})$ be a measurable space, $F$ be a set and $f: E \to F$ be a function. Then the following two conditions are equivalent:


*

*$f(\mathcal{E})$ is a $\sigma$-algebra on $F$ and $f$ is $\mathcal{E}/f(\mathcal{E})$-measurable;

*$f$ is surjective and, for all $A \in \mathcal{E}, f^{-1}(f(A)) \in \mathcal{E}$.


First 1. $\implies$ 2.


*

*$f$ is surjective. Since $f(\mathcal{E})$ is a $\sigma$-algebra over $F$ it must contain $F$ and hence we can write $F = f(A)$ for some $A \subseteq E$.

*For all $A \in \mathcal{E},\ f^{-1}(f(A)) \in \mathcal{E}$. Take any $A \in \mathcal{E}$. Clearly, $f(A) \in f(\mathcal{E})$. Moreover, since $f(\mathcal{E})$ is a $\sigma$-algebra on $F$ and $f$ in $\mathcal{e}/f(\mathcal{E})$-measurable then $f^{-1}(f(A)) \in \mathcal{E}$.
Now, 2. $\implies$ 1.


*

*$f(\mathcal{E})$ is a $\sigma$-algebra on $F$.


*

*$F = f(E) \in \mathcal{E}$ since $f$ is surjective.

*For any $B \in f(\mathcal{E})$ we must show that $B^\complement \in f(\mathcal{E})$. Since $f$ is surjective, there is an $A^\complement \in E$ such that $f(A^\complement) = B^\complement$. By hypothesis we know that $f^{-1}(f(A^\complement)) \in \mathcal{E} \implies A^\complement \in \mathcal{E} \implies f(A^\complement) = B^\complement \in f(\mathcal{E})$.

*As before, for any $(B_n)_{n\in\mathbb{N}} \subset f(\mathcal{E})$, there exists a sequence $(A_n)_{n\in\mathbb{N}} \subset \mathcal{E}$ such that $f(A_n) = B_n$. Since $\mathcal{E}$ is itself a $\sigma$-algebra, it holds that $\bigcup_{n\in\mathbb{N}} A_n \in \mathcal{E}$ and thus
$$\bigcup_{n\in\mathbb{N}} B_n = \bigcup_{n\in\mathbb{N}} f(A_n) = f(\bigcup_{n\in\mathbb{N}} A_n) \in f(\mathcal{E}).$$


*$f$ is $\mathcal{E}/f(\mathcal{E})$ measurable. Take any $B \in f(\mathcal{E})$. By hypothesis we now that $B = f(A)$ for some $A \in \mathcal{E}$ and also $f^{-1}(f(A)) = f^{-1}(B) \in \mathcal{E}$, hence $f$ is $\mathcal{E}/f(\mathcal{E})$ measurable.
