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Section 5.2 Image and Preimage
Definition 5.5 .
Let \(P \subseteq A\text{.}\) The image of \(P\) under \(f\text{,}\) denoted \(f(P)\text{,}\) is the set defined by
\begin{equation*}
f (P) = {b\in B | b = f (p) for some p\in P}.
\end{equation*}
The range of \(f\) (also called the image of \(f\) ) is the set \(f(a)\text{.}\)
Definition 5.6 .
Let \(Q \subseteq B\text{.}\) The inverse image of \(Q\) under \(f\text{,}\) denoted \(f\inv (Q)\text{,}\) is the set defined by
\begin{equation*}
f\inv(Q) = {a\in A | f(a)\in Q}.
\end{equation*}
Theorem 5.7 .
Let \(A\) and \(B\) be sets, let \(C, D \subseteq A\) and \(S,\sqcap \subseteq B\) be subsets, and let \(f : A \to B\) be a function. Let \(I\) and \(K\) be non-empty sets, let \({Ui}i\in I\) be a family of subsets of \(A\) indexed by \(I\text{,}\) and let \({Vk}k\in K\) be a family of subsets of \(B\) indexed by \(K\text{.}\)
\(f ( \es ) = \es \) and \(f\inv ( \es ) = \es \text{.}\)
\(f\inv (B) = A\text{.}\)
\(f (C) \subseteq S\) if and only if \(C \subseteq f\inv (S)\text{.}\)
If \(C \subseteq D\text{,}\) then \(f (C) \subseteq f (D)\text{.}\)
If \(S \subseteq T\text{,}\) then \(f\inv (S) \subseteq f\inv (T )\text{.}\)
\(f (\bigcup i\in I Ui) = \bigcup i\in I f (Ui)\text{.}\)
\(f (⋂i\in I Ui) \subseteq ⋂i\in I f (Ui)\text{.}\)
\(f\inv (\bigcup k\in K Vk) = \bigcup k\in K f\inv (Vk)\text{.}\)
\(f\inv (⋂k\in K Vk) = ⋂k\in K f\inv (Vk)\text{.}\)
Exploration 5.8 .
Let \(A\) and \(B\) be sets, let \(P, Q \subseteq A\) be subsets and let \(f : A \to B\) be a function.
Prove that \(f (P) - f (Q) \subseteq f (P - Q)\text{.}\)
Is it necessarily the case that \(f (P - Q) \subseteq f (P) - f (Q)\text{?}\) Give a proof or a counterexample.
