ThmDex – An index of mathematical definitions, results, and conjectures.
Result R4537 on D528: Map image
Inclusion of inverse image does not imply inclusion of image
Formulation 0
Let $f : \{ 0, 1 \} \to \{ a, b \}$ be a D18: Map such that
(i) \begin{equation} f(0) = a \end{equation}
(ii) \begin{equation} f(1) = b \end{equation}
(iii) \begin{equation} A : = \{ 0, 1 \} \end{equation}
(iv) \begin{equation} B : = \{ b \} \end{equation}
Then
(1) $f$ is a D468: Bijective map
(2) \begin{equation} f^{-1}(B) \subseteq A \end{equation}
(3) \begin{equation} f(A) \not\subseteq B \end{equation}
Proofs
Proof 0
Let $f : \{ 0, 1 \} \to \{ a, b \}$ be a D18: Map such that
(i) \begin{equation} f(0) = a \end{equation}
(ii) \begin{equation} f(1) = b \end{equation}
(iii) \begin{equation} A : = \{ 0, 1 \} \end{equation}
(iv) \begin{equation} B : = \{ b \} \end{equation}
Bijectivity is clear from the definition. Further, we have \begin{equation} f^{-1} B = f^{-1} \{ b \} = \{ 1 \} \subseteq \{ 0, 1 \} = A \end{equation} and \begin{equation} f A = f \{ 0, 1 \} = \{ a, b \} \supset \{ b \} = B \end{equation} $\square$