Question

Given \(f: A \rightarrow B\) and subsets \(Y, Z \subseteq B,\) prove \(f^{-1}(Y \cup Z)=f^{-1}(Y) \cup f^{-1}(Z)\).

Short answer

Proven that \(f^{-1}(Y \cup Z) = f^{-1}(Y) \cup f^{-1}(Z)\) for a function \(f: A \rightarrow B\) and subsets \(Y, Z \subseteq B\).

Step-by-step solution

Understanding the Definitions

Define the function \(f: A \rightarrow B\), and subsets \(Y, Z \subseteq B\). The preimage of a set, denoted \(f^{-1}(Y)\), includes all the elements in the domain A that get mapped to the elements in Y under the function f.

Proving the Left Subset Relation

To show \(f^{-1}(Y \cup Z) = f^{-1}(Y) \cup f^{-1}(Z)\), first prove the left subset relation \(f^{-1}(Y \cup Z) \subseteq f^{-1}(Y) \cup f^{-1}(Z)\). Assume \(x \in f^{-1}(Y \cup Z)\). This means \(f(x) \in Y \cup Z\), which implies \(f(x) \in Y\) or \(f(x) \in Z\). Thus, \(x \in f^{-1}(Y)\) or \(x \in f^{-1}(Z)\) which means \(x \in f^{-1}(Y) \cup f^{-1}(Z)\). We have shown if \(x \in f^{-1}(Y \cup Z)\), then \(x \in f^{-1}(Y) \cup f^{-1}(Z)\), hence \(f^{-1}(Y \cup Z) \subseteq f^{-1}(Y) \cup f^{-1}(Z)\).

Proving the Right Subset Relation

Now prove the right subset relation \(f^{-1}(Y \cup Z) \supseteq f^{-1}(Y) \cup f^{-1}(Z)\). Assume \(x \in f^{-1}(Y) \cup f^{-1}(Z)\). This means \(x \in f^{-1}(Y)\) or \(x \in f^{-1}(Z)\), which implies \(f(x) \in Y\) or \(f(x) \in Z\). Therefore, \(f(x) \in Y \cup Z\), which implies \(x \in f^{-1}(Y \cup Z)\). If \(x \in f^{-1}(Y) \cup f^{-1}(Z)\), then \(x \in f^{-1}(Y \cup Z)\), hence \(f^{-1}(Y \cup Z) \supseteq f^{-1}(Y) \cup f^{-1}(Z)\).

Equality Concluding

Since \(f^{-1}(Y \cup Z) \subseteq f^{-1}(Y) \cup f^{-1}(Z)\) and \(f^{-1}(Y \cup Z) \supseteq f^{-1}(Y) \cup f^{-1}(Z)\), by definition of set equality, it follows that \(f^{-1}(Y \cup Z) = f^{-1}(Y) \cup f^{-1}(Z)\).