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Let $f:X\rightarrow Y$ and let $\mathcal{B}$ be an algebra over $Y$. I would like to show that $\lbrace f^{-1}[B]\,|\,B\in \mathcal{B}\rbrace$ is an algebra over $X$.

Here is my attempt:\\\

I let $D=\lbrace f^{-1}[B]\,|\,B\in \mathcal{B}\rbrace$. I know I have to show $M,N\in D \implies M\cup N\in D$ and if $\in D$ , then $M^c\in D.$
This is where I'm stuck and I'll need a little help. I don't know what $D$ looks like.

Added: From @Jonas's comments, I make some additions.
$M\in D\implies$ $M=f^{-1}[B]$ for some $B\in \mathcal{B}$.
$N\in D\implies$ $N=f^{-1}[A ]$ for some $A\in \mathcal{B}$.
$M \cup N = f^{-1}[B]\cup f^{-1}[A]=f^{-1}[B\cup A]$, $B\cup A\in \mathcal{B}$. So $M \cup N \in D.$

Left to show that $M^c\in D$.
$M^c=\left[f^{-1}[B]\right]^c=f^{-1}[B^c]$. So $M^c\in D$ since $B\in \mathcal{B} \implies B^c\in \mathcal{B}.$

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Well, you know that inverse images commute with the common set operations right? – Jonas Teuwen Oct 1 '11 at 20:09
Yes, I know that...the inverse image of the union is the union of the inverse image...etc...I guess my question really is what it means to say $M\in D$? – Jack Oct 1 '11 at 20:12
$M \in D$ means that $M = f^{-1}[B]$ for some $B \in \mathcal B$. – Jonas Teuwen Oct 1 '11 at 20:15
Okay. So now the same works for $M^c$. – Jonas Teuwen Oct 1 '11 at 20:41
@Jonas: Is what I did for the complement okay? – Jack Oct 1 '11 at 20:48
up vote 1 down vote accepted

To add an answer (which has been given in the comments):

Note that $f^{-1}$ commutes with unions, intersections and complements. Then your result follows immediately after you noticed that $M \in D$ means that $M = f^{-1}[B]$ for some $B \in \mathcal B$.

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