Let \(f: A \to B\) . Let \(A_0 \subset A\) and \(B_0 \subset B\) .
Show that \(A_0 \subset f^{-1}(f(A_0))\) and that equality holds if \(f\) is injective.
Show that \(f(f^{-1}(B_0)) \subset B_0\) and that equality holds if \(f\) is surjective.
Solution
If \(a\in A_0\), then \(f(a)\in f(A_0)\), hence \[ a\in f^{-1}(f(A_0)). \] Thus \[ A_0\subset f^{-1}(f(A_0)). \] If \(f\) is injective and \(a\in f^{-1}(f(A_0))\), then \(f(a)=f(a_0)\) for some \(a_0\in A_0\). Injectivity gives \(a=a_0\in A_0\), so equality holds.
If \(b\in f(f^{-1}(B_0))\), then \(b=f(a)\) for some \(a\) with \(f(a)\in B_0\). Hence \(b\in B_0\), so \[ f(f^{-1}(B_0))\subset B_0. \] If \(f\) is surjective and \(b\in B_0\), choose \(a\in A\) with \(f(a)=b\). Then \(a\in f^{-1}(B_0)\), hence \(b\in f(f^{-1}(B_0))\). Thus equality holds.