Question

Consider \(f: A \rightarrow B\). Prove that \(f\) is injective if and only if \(X=f^{-1}(f(X))\) for all \(X \subseteq A .\) Prove that \(f\) is surjective if and only if \(f\left(f^{-1}(Y)\right)=Y\) for all \(Y \subseteq B\).

Short answer

The function \(f: A \rightarrow B\) is injective if and only if \(X=f^{-1}(f(X))\) for all \(X \subseteq A\) and the function is surjective if and only if \(f(f^{-1}(Y))=Y\) for all \(Y \subseteq B\). These relations were proved using the definitions of injective and surjective functions in a detailed step-by-step manner.

Step-by-step solution

Proving Injectiveness 'if' clause

Assume \(f: A \rightarrow B\) such that \(X=f^{-1}(f(X))\) for all \(X \subseteq A \). Take \(a_1, a_2 \in A\) such that \(f(a_1)=f(a_2)\). Now, we create a subset \(X=\{a_1, a_2\}\), then by our assumption, \(X=f^{-1}(f(X))\). Therefore, if \(f(a_1)=f(a_2)\) then \(a_1, a_2 \in f^{-1}(f(X)) = X\), therefore \(a_1 = a_2\). Thus, \(f\) is injective.

Proving Injectiveness 'only if' clause

Assume \(f: A \rightarrow B\) is injective. For \(X=f^{-1}(f(X))\) to be true for all \(X \subseteq A\), we need that all \(a \in f^{-1}(f(X))\) are also in \(X\), and vice-versa. Given \(a \in f^{-1}(f(X))\), then there is a \(x \in X\) with \(f(a) = f(x)\). But since \(f\) is injective, we have \(a = x\) and also \(a \in X\), proving the 'only if' clause.

Proving Surjectiveness 'if' clause

Assume \(f: A \rightarrow B\) such that \(f(f^{-1}(Y))=Y\) for all \(Y \subseteq B \). Now, take \(b \in B\), then the set \(Y=\{b\}\) is a subset of \(B\), and by our assumption, \(f(f^{-1}(Y)) = Y\), so \(b\) is in the range of \(f\), which means \(f\) is surjective.

Proving Surjectiveness 'only if' clause

Assume \(f: A \rightarrow B\) is surjective. For \(f(f^{-1}(Y))=Y\) to be true for all \(Y \subseteq B\), we need that all \(b \in f(f^{-1}(Y))\) are also in \(Y\), and vice-versa. Given \(b \in f(f^{-1}(Y))\), then there is an \(a \in A\) such that \(f(a) = b\). But since \(f\) is surjective, there is an \(a\) such that \(f(a) = b\) for every \(b \in B\), proving 'only if' clause.