Let \(f: A \to B\) and let \(A_i \subset A\) and \(B_i \subset B\) for \(i = 0\) and \(i = 1\) . Show that \(f^{-1}\) preserves inclusions, unions, intersections, and differences of sets:
\(B_0 \subset B_1 \Rightarrow f^{-1}(B_0) \subset f^{-1}(B_1)\) .
\(f^{-1}(B_{0} \cup B_{1}) = f^{-1}(B_{0}) \cup f^{-1}(B_{1}).\)
\(f^{-1}(B_{0} \cap B_{1}) = f^{-1}(B_{0}) \cap f^{-1}(B_{1}).\)
\(f^{-1}(B_0 - B_1) = f^{-1}(B_0) - f^{-1}(B_1)\) .
Show that \(f\) preserves inclusions and unions only:
\(A_0 \subset A_1 \Rightarrow f(A_0) \subset f(A_1)\) .
\(f(A_{0} \cup A_{1}) = f(A_{0}) \cup f(A_{1}).\)
\(f(A_0 \cap A_1) \subset f(A_0) \cap f(A_1)\) ; give an example where equality fails.
\(f(A_0 - A_1) \supset f(A_0) - f(A_1)\) ; give an example where equality fails.
Solution
For preimages, all four assertions follow directly from the equivalence \[ x\in f^{-1}(S)\iff f(x)\in S. \] Thus:
If \(B_0\subset B_1\) and \(x\in f^{-1}(B_0)\), then \(f(x)\in B_1\), so \(x\in f^{-1}(B_1)\).
\[ x\in f^{-1}(B_0\cup B_1) \iff f(x)\in B_0\text{ or }f(x)\in B_1 \iff x\in f^{-1}(B_0)\cup f^{-1}(B_1). \]
The same argument with “and” gives \[ f^{-1}(B_0\cap B_1)=f^{-1}(B_0)\cap f^{-1}(B_1). \]
Likewise, \[ f^{-1}(B_0-B_1)=f^{-1}(B_0)-f^{-1}(B_1). \]
For images:
If \(A_0\subset A_1\), every value attained on \(A_0\) is attained on \(A_1\), so \(f(A_0)\subset f(A_1)\).
A value is attained on \(A_0\cup A_1\) exactly when it is attained on at least one of \(A_0,A_1\), hence \[ f(A_0\cup A_1)=f(A_0)\cup f(A_1). \]
If \(y=f(x)\) for some \(x\in A_0\cap A_1\), then \(y\in f(A_0)\cap f(A_1)\), so \[ f(A_0\cap A_1)\subset f(A_0)\cap f(A_1). \] Equality can fail: let \(f:\{0,1\}\to\{*\}\) be constant, \(A_0=\{0\}\), and \(A_1=\{1\}\). Then the left side is empty and the right side is \(\{*\}\). If \(f\) is injective, equality does hold, since equal image values then come from the same point.
If \(y\in f(A_0)-f(A_1)\), write \(y=f(x)\) with \(x\in A_0\). Since \(y\notin f(A_1)\), necessarily \(x\notin A_1\), so \(x\in A_0-A_1\). Thus \[ f(A_0-A_1)\supset f(A_0)-f(A_1). \] Equality can fail for the same constant map with \(A_0=\{0,1\}\), \(A_1=\{1\}\): the left side is \(\{*\}\), while the right side is empty. If \(f\) is injective, equality holds.