Q.Let $f: X \to Y$ be a function. Define a relation $R$ in $X$ given by $R = \{(a, b): f(a) = f(b)\}$. Examine whether $R$ is an equivalence relation or not.

Concept: Equivalence Relation Proof — We check reflexivity, symmetry, and transitivity.

1. Reflexive: For any $a \in X$, $f(a) = f(a)$, so $(a, a) \in R$.
2. Symmetric: If $(a, b) \in R$, then $f(a) = f(b)$, so $f(b) = f(a)$, giving $(b, a) \in R$. …

The core idea here is simple: we are grouping elements of $X$ based on whether they map to the same output in $Y$ under $f$. This is exactly the "kernel" of the function — the relation that identifies elements that are indistinguishable from the function's perspective. Proving it's an equivalence relation means checking three properties, each of which flows naturally from the fact that equality itself is an equivalence relation.

Let's walk through it step by step.

1. Reflexive: For any $a \in X$, we need $(a, a) \in R$.

   Since $f(a) = f(a)$ always (a value is equal to itself), the condition $f(a) = f(b)$ holds when $b = a$. So $(a, a) \in R$ for every $a \in X$.

   Reasoning: This works because equality is reflexive — nothing deeper.

2. Symmetric: If $(a, b) \in R$, then $f(a) = f(b)$. But equality is symmetric: if $f(a) = f(b)$, then $f(b) = f(a)$. So $(b, a) \in R$ as well.

   Reasoning: The condition is an equation; swapping the two sides doesn't change the truth.

3. Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $f(a) = f(b)$ and $f(b) = f(c)$. By the transitivity of equality, $f(a) = f(c)$. Hence $(a, c) \in R$.

   Reasoning: Equality chains together — if two values both equal the same third value, they equal each other. …

Method: Checking Equivalence for a Relation Built From a Function

This method applies whenever a relation is defined as "$aRb$ when $f(a) = f(b)$" (the kernel of a function). Because the condition is an equality of outputs, all three equivalence properties are borrowed directly from the properties of equality itself.

Steps

Step 1: Identify the condition as an equality statement.

Rewrite the relation so the defining test is visibly "two things are equal": $aRb \iff f(a) = f(b)$. Equality is itself reflexive, symmetric, and transitive, and that is the engine of the whole proof.

Step 2: Reflexivity — equality of a value with itself.

For any $a$, $f(a) = f(a)$ is automatic, so $(a,a) \in R$.

Step 3: Symmetry — swap the two sides of the equation.

If $f(a) = f(b)$ then $f(b) = f(a)$, so $(b,a) \in R$. …

Common Mistakes

Mistake 1: Thinking the relation fails when $f$ is not one-one.

Why it's wrong: the relation $R = \{(a,b): f(a)=f(b)\}$ is an equivalence relation for every function $f$; injectivity of $f$ is irrelevant. Correct approach: the properties come from equality of outputs, which always behaves reflexively, symmetrically, and transitively.

Mistake 2: Concluding "not an equivalence relation" because many elements share an image. …