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I stumbled upon this problem:

Suppose $X$ and $Y$ are two sets, $f: X \to Y$ is a function and $A \subseteq 2^Y$. Show that $$m(f^{-1}(A)) = f^{-1}(m(A))$$ Where $m$ denotes the monotone class generated by $A$.

For me, $f^{-1}(A)$ does not quite make sense, since $A$ is a set of subsets of $Y$, wheras usually the argument of $f^{-1}$ is just a subset of $Y$. Can anyone shed light on this?

Found in Exercises in Analysis Part I.

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    Presumably $f^{-1}(A)$ is the set $\{f^{-1}(a):a\in A\}$ from the context it is used in, but I have no expertise here. – 2017-01-21

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It is a slight abuse of notation. In this case $$ f^{-1}(A)=\bigcup_{S\in A}\{f^{-1}(S)\} $$

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    Shouldn't it be $\bigcup_{S \in A} \{f^{-1}(S)\}$? We want to end up with a set of sets. – 2017-01-22
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    yes, you are right. I edited it – 2017-01-22