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In Folland, there is a statement as follows:

Any mapping $f: X \to Y$ between two sets induces a mapping $f^{-1}: \mathcal{P}(Y) \to \mathcal{P}({X})$ (these are power sets) defined by $f^{-1}(E) = \{x \in X: f(x) \in E\}$ which preserves unions, intersections and complements.

Here is my question: why does it like this? I mean, $f$ is a mapping from $X \to Y$, why its inverse mapping induces a mapping between power sets?

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up vote 4 down vote accepted

An arbitrary mapping may not have an inverse. $f^{-1}$ is not the inverse mapping, it's preimage mapping. For any $A\subseteq Y$ you can find $f^{-1}[A]=\{x\in X\vert f(x)\in A\}\subseteq X$, and it is not hard to see that it satisfies the properties listed.

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OK, thanks. Then (I know I ask very basic questions), why we write the mapping as $X \to Y$ at the first place? If we write $f: X \to Y$ also induces $f: \mathcal{P}(X) \to \mathcal{P}(Y)$, is this statement true? – oeda Nov 7 '12 at 20:13
@John: because $f$ is from $X$ to $Y$. $f$ also induces the image mapping from $\mathcal P(X)$ to $\mathcal P (Y)$ (similarly defined), but it does not necessarily preserve intersections or complements (though it does preserve unions). – tomasz Nov 7 '12 at 20:18

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