Proposition 2.2.2.label Let $X$ and $Y$ be sets, then:
- (1)
For any $f: X \to Y$, the mapping $S \mapsto f^{-1}(S)$ is a total preimage function.
- (2)
For any total preimage function $P: 2^{Y} \to 2^{X}$, there exists a unique $f: X \to Y$ such that $P(S) = f^{-1}(S)$ for all $S \in 2^{Y}$.
Proof. (2): Let $x, y \in Y$ with $x \ne y$, then by (PF1) and (PF3),
\[P(\bracs{x}) \cap P(\bracs{y}) = P(\bracs{x}\cap \bracs{y}) = P(\emptyset) = \emptyset\]
By (T) and (PF2), $X = P(Y) = P\paren{\bigcup_{y \in Y}\bracs{y}}= \bigcup_{y \in Y}P(\bracs{y})$. Therefore $X = \bigsqcup_{y \in Y}P(\bracs{y})$.
Therefore for each $x \in X$, there exists a unique $f(x) \in Y$ such that $x \in P(f(x))$. The association $x \mapsto f(x)$ then is the unique desired function.$\square$
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