Question

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

Short answer

Through the stepwise proof, it has been established that \(f^{-1}(Y \cap Z) = f^{-1}(Y) \cap f^{-1}(Z)\). This mathematical concept basically states that the pre-image of the intersection of two subsets equals to the intersection of the pre-image of these subsets under a function.

Step-by-step solution

Pre-Image of Intersection as Subsets of Intersection of Pre-Images

To begin with, it should be assessed that if an element is in \(f^{-1}(Y \cap Z)\), it is in \(f^{-1}(Y) \cap f^{-1}(Z)\). This implies that if a set X is such that \(f(X) \in Y \cap Z\), X will belong to both \(f^{-1}(Y)\) and \(f^{-1}(Z)\), hence belongs to the intersection.

Intersection of Pre-Images as Subset of Pre-Image of Intersection

Now, moving to the other side of proof, consider a set X is such that \(f(X) \in f^{-1}(Y) \cap f^{-1}(Z)\), then by definition of intersection, \(f(X) \in f^{-1}(Y)\) and \(f(X) \in f^{-1}(Z)\). As a result, \(f(X) \in Y \cap Z\), i.e., X is such that \(f(X) \in Y \cap Z\), so X belongs to the preimage \(f^{-1}(Y \cap Z)\). This completes the proof.

Finalizing Proof

Since each set is a subset of the other, they are equal. \(f^{-1}(Y \cap Z) = f^{-1}(Y) \cap f^{-1}(Z)\).