Section 1.4: Composition of Functions

Solution
For every \(a \in X\), \((a, a) \in R\), since \(f(a) = f(a)\), showing that \(R\) is reflexive.
Similarly, \((a, b) \in R \Rightarrow f(a) = f(b) \Rightarrow f(b) = f(a) \Rightarrow (b, a) \in R\).
Therefore, \(R\) is symmetric.
Further, \((a, b) \in R\) and \((b, c) \in R \Rightarrow f(a) = f(b)\) and \(f(b) = f(c) \Rightarrow f(a) = f(c) \Rightarrow (a, c) \in R\), which implies that \(R\) is transitive.
Hence, \(R\) is an equivalence relation.