$\mathcal{B}$ an algebra $\implies f^{-1}(\mathcal{B}) $ an algebra Im trying to prove the following statement:
Let $f: \Omega \to E$.
$\mathcal{B}$ an algebra on $E$ $\implies f^{-1}(\mathcal{B}) $ an algebra on $\Omega$.
To show that $f^{-1}(\mathcal{B})$ is an algebra, following has to hold:

*

*$\Omega \in f^{-1}(\mathcal{B})$

*Stable under complementation

*Stable under finite union

My question:
In the first point, since $\mathcal{B}$ is an algebra on $E$, it follows that $E \in \mathcal{B}$. But how do I even know if $f^{-1}(E)$ is defined? Don't we need the restriction that the function needs to be bijective in order to gurantee that $f^{-1}$ is possible?
EDIT (after suggestion by Henno Brandsma)

*

*$\Omega \in f^{-1}(\mathcal{B})$ since $f^{-1}[E] = \Omega$.And $E \in \mathcal{B}$

*Then regarding the complement: I take any $A \in \mathcal{B}$ so the complement would be $A^c = E \setminus A$. Note that $A^c \in \mathcal{B}$. So $f^{-1}[E\setminus A]= \Omega \setminus f^{-1}[A]$ and $f^{-1}[A]$ is defined since we know $f^{-1}[E] = \Omega$ and $A \subset E$. So we have the complement of $f^{-1}[A]$ as well.

*This one follows from the fact that $f^{-1}[\bigcup_i A_i] = \bigcup_i f^{-1}[A_i]$.


Use that

*

*$f^{-1}[E]=\Omega$

*$f^{-1}[E\setminus A] = \Omega\setminus f^{-1}[A]$ for all $A \subseteq E$,

*$f^{-1}[\bigcup_i A_i] = \bigcup_i f^{-1}[A_i]$
Note that $f^{-1}[A]$ is always defined, it's just the set $$\{x \in \Omega\mid f(x) \in A\}$$
