Relations and Functions — Types, Composition, Inverse

The Foundation of All Higher Maths

Relations and functions form the language that the rest of mathematics is built on. A relation connects elements of two sets. A function is a special relation where every input has exactly one output. Understanding the types of relations (reflexive, symmetric, transitive) and functions (one-one, onto, bijective) is essential for Class 11-12 and competitive exams.

This chapter appears in CBSE Class 12 boards (4-6 marks) and JEE Main (1 question on average). The questions are mostly conceptual — checking properties, finding compositions, verifying invertibility.

graph TD
    A[Relation on Set A] --> B{Check properties}
    B --> C{Reflexive?}
    C -->|aRa for all a| D[Yes]
    C -->|Not for some a| E[No]
    B --> F{Symmetric?}
    F -->|aRb implies bRa| G[Yes]
    F -->|Not always| H[No]
    B --> I{Transitive?}
    I -->|aRb and bRc implies aRc| J[Yes]
    I -->|Fails somewhere| K[No]
    D --> L{All three?}
    G --> L
    J --> L
    L -->|Yes| M[Equivalence Relation]
    L -->|No| N[Not equivalence]

Key Terms & Definitions

Relation — A subset of the Cartesian product $A \times B$ . If $(a, b) \in R$ , we write $aRb$ .

Function — A relation from $A$ to $B$ where every element of $A$ has exactly one image in $B$ .

Domain — The set of all valid inputs.

Codomain — The set into which the function maps (declared set).

Range — The actual set of outputs (subset of codomain).

Types of Relations

Property	Condition	Example on `{1,2,3}`
Reflexive	$(a,a) \in R$ for all $a$	`{(1,1),(2,2),(3,3),...}`
Symmetric	$(a,b) \in R \Rightarrow (b,a) \in R$	If `(1,2)` in R, then `(2,1)` must be too
Transitive	$(a,b), (b,c) \in R \Rightarrow (a,c) \in R$	`(1,2)` and `(2,3)` imply `(1,3)`
Equivalence	All three above	Partitions the set into classes

The empty relation on a non-empty set is symmetric and transitive (vacuously true — there are no pairs to violate the conditions) but NOT reflexive (since $(a,a)$ is not in the empty relation). This catches many students off guard.

Types of Functions

Type	Condition	Also Called
One-one (Injective)	$f(a) = f(b) \Rightarrow a = b$	No two inputs share an output
Onto (Surjective)	Range = Codomain	Every element of codomain is hit
Bijective	One-one AND onto	Invertible

Composition: $(f \circ g)(x) = f(g(x))$ . Apply $g$ first, then $f$ . Note: $f \circ g \neq g \circ f$ in general.

Inverse: $f^{-1}$ exists only if $f$ is bijective. $f^{-1}(y) = x$ where $f(x) = y$ .

Solved Examples — Easy to Hard

Example 1 (Easy — CBSE)

Is the relation $R = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ on $A = \{1,2,3\}$ an equivalence relation?

Reflexive: $(1,1), (2,2), (3,3) \in R$ . Yes.

Symmetric: $(1,2) \in R$ and $(2,1) \in R$ . Check all pairs — yes.

Transitive: $(1,2)$ and $(2,1) \in R$ , so $(1,1)$ must be in $R$ — it is. All chains check out. Yes.

All three hold. $R$ is an equivalence relation.

Example 2 (Medium — CBSE Board)

Show that $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = 2x + 3$ is bijective. Find $f^{-1}$ .

One-one: $f(a) = f(b) \Rightarrow 2a+3 = 2b+3 \Rightarrow a = b$ . Yes.

Onto: For any $y \in \mathbb{R}$ , take $x = \frac{y-3}{2}$ . Then $f(x) = y$ . Every real is hit.

So $f$ is bijective. $f^{-1}(y) = \frac{y-3}{2}$ .

Example 3 (Medium — JEE Main)

If $f(x) = x^2$ and $g(x) = \sqrt{x}$ , find $f \circ g$ and $g \circ f$ . Are they equal?

$(f \circ g)(x) = f(\sqrt{x}) = (\sqrt{x})^2 = x$ (for $x \geq 0$ )

$(g \circ f)(x) = g(x^2) = \sqrt{x^2} = |x|$

No, they’re not equal. $(g \circ f)(x) = |x|$ , which equals $x$ only when $x \geq 0$ .

Example 4 (Hard — JEE Main)

Let $f: \mathbb{N} \to \mathbb{N}$ be defined by $f(n) = \begin{cases} \frac{n+1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases}$ . Is $f$ one-one? Is $f$ onto?

One-one: $f(1) = 1$ and $f(2) = 1$ . Two different inputs give the same output. Not one-one.

Onto: For any $m \in \mathbb{N}$ , we can find $n = 2m$ (even) such that $f(2m) = m$ . So every natural number is in the range. Onto.

Exam-Specific Tips

CBSE Board: 4-6 marks. Typical questions: verify equivalence relation, check one-one/onto for a given function, find the inverse of a bijection, find $f \circ g$ . Write every step — examiners look for explicit verification of each property.

JEE Main: Questions often involve piecewise functions where you must determine injectivity/surjectivity. Also tested: number of onto functions from a set with $m$ elements to a set with $n$ elements (using inclusion-exclusion).

Common Mistakes to Avoid

Mistake 1 — Confusing codomain with range. The codomain is the declared target set. The range is the actual set of outputs. A function is onto only when range = codomain.

Mistake 2 — Assuming $f \circ g = g \circ f$ . Composition is generally not commutative. Always apply in the correct order.

Mistake 3 — Declaring inverse without checking bijectivity. $f^{-1}$ exists only if $f$ is both one-one and onto. If either fails, the inverse doesn’t exist as a function.

Mistake 4 — Checking transitivity incompletely. You must check ALL chains $(a,b)$ and $(b,c)$ in $R$ , not just a few. Missing one counterexample means incorrectly declaring transitivity.

Mistake 5 — Confusing relation with function. A relation can map one input to multiple outputs. A function cannot. Always verify the “unique output” condition.

Practice Questions

Q1. Is $R = \{(a,b) : a \leq b\}$ on $\mathbb{R}$ an equivalence relation?

Reflexive: $a \leq a$ , yes. Symmetric: $1 \leq 2$ but $2 \not\leq 1$ , no. Not an equivalence relation (fails symmetry).

Q2. If $f(x) = 3x - 4$ and $g(x) = \frac{x+4}{3}$ , show $f \circ g = g \circ f = I$ (identity).

$(f \circ g)(x) = 3 \cdot \frac{x+4}{3} - 4 = x$ . $(g \circ f)(x) = \frac{(3x-4)+4}{3} = x$ . Both give identity.

Q3. Show that $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^3$ is bijective.

One-one: $a^3 = b^3 \Rightarrow a = b$ . Onto: for any $y$ , $x = y^{1/3}$ exists in $\mathbb{R}$ . Bijective.

Q4. Let $R$ on {1,2,3} be {(1,2), (2,3)}. Is it transitive?

$(1,2)$ and $(2,3)$ are in $R$ , but $(1,3)$ is not. Not transitive.

Q5. Is $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$ one-one? onto?

Not one-one: $f(2) = f(-2) = 4$ . Not onto: no $x$ gives $f(x) = -1$ . Neither.

Q6. Find $f^{-1}$ for $f(x) = \frac{4x+3}{6x-4}$ where $x \neq 2/3$ .

Let $y = \frac{4x+3}{6x-4}$ . Then $6xy - 4y = 4x + 3$ , so $x(6y-4) = 4y+3$ , giving $x = \frac{4y+3}{6y-4}$ . So $f^{-1}(x) = \frac{4x+3}{6x-4}$ . Remarkably, $f = f^{-1}$ (self-inverse).

Q7. How many functions from {1,2,3} to {a,b} are onto?

Total functions $= 2^3 = 8$ . Functions that miss $a$ : all map to $b$ , that’s 1. Functions that miss $b$ : all map to $a$ , that’s 1. By inclusion-exclusion: onto $= 8 - 1 - 1 = 6$ .

Q8. If $f(x) = x + 1$ and $g(x) = 2x - 1$ , find $(f \circ g)^{-1}$ .

$(f \circ g)(x) = 2x - 1 + 1 = 2x$ . $(f \circ g)^{-1}(x) = x/2$ .

FAQs

What is the difference between a relation and a function?

A relation is any set of ordered pairs. A function is a special relation where each input has exactly one output. Every function is a relation, but not every relation is a function.

What is an equivalence class?

If $R$ is an equivalence relation on $A$ , the equivalence class of element $a$ is the set of all elements related to $a$ : $[a] = \{x \in A : xRa\}$ . These classes partition $A$ into non-overlapping groups.

Can a function from a finite set be one-one but not onto?

Yes, if the codomain has more elements than the domain. From {1,2} to {a,b,c}, $f(1)=a, f(2)=b$ is one-one but not onto (c is not hit).

What is the composition of three functions?

$(f \circ g \circ h)(x) = f(g(h(x)))$ . Apply the innermost function first. Composition is associative: $f \circ (g \circ h) = (f \circ g) \circ h$ .

Advanced Concepts

Number of functions between finite sets

If $|A| = m$ and $|B| = n$ :

Type	Count
Total functions	$n^m$
One-one (injective)	$\frac{n!}{(n-m)!}$ (only if $m \leq n$ )
Onto (surjective)	$\sum_{k=0}^{n} (-1)^k \binom{n}{k}(n-k)^m$ (inclusion-exclusion)
Bijective	$n!$ (only if $m = n$ )

JEE Main frequently asks: “Find the number of onto functions from a set of 4 elements to a set of 2 elements.” Total = $2^4 = 16$ . Non-onto (missing at least one element): $\binom{2}{1} \times 1^4 - \binom{2}{2} \times 0^4 = 2$ . Onto = $16 - 2 = 14$ .

Equivalence classes and partitions

An equivalence relation on set $A$ partitions $A$ into disjoint equivalence classes. The number of distinct equivalence classes equals the number of blocks in the partition.

Example: On $\{1, 2, 3, 4, 5, 6\}$ , the relation “same remainder when divided by 3” gives three equivalence classes: $\{3, 6\}$ , $\{1, 4\}$ , $\{2, 5\}$ .

Binary operations (CBSE add-on)

A binary operation $*$ on set $A$ is a function from $A \times A$ to $A$ . Properties to check: closure, commutativity ( $a * b = b * a$ ), associativity ( $a * (b * c) = (a * b) * c$ ), identity element, inverse element.

Additional Practice Questions

Q9. How many equivalence relations are possible on $\{1, 2, 3\}$ ?

Each equivalence relation corresponds to a partition of the set. Partitions of $\{1,2,3\}$ : $\{\{1\},\{2\},\{3\}\}$ , $\{\{1,2\},\{3\}\}$ , $\{\{1,3\},\{2\}\}$ , $\{\{2,3\},\{1\}\}$ , $\{\{1,2,3\}\}$ . Total: 5 equivalence relations. (This is the Bell number $B_3 = 5$ .)

Q10. Is $*$ defined by $a * b = a + b - ab$ on $\mathbb{R}$ commutative? Associative? Find identity.

Commutative: $a * b = a + b - ab = b + a - ba = b * a$ . Yes. Identity: $a * e = a \implies a + e - ae = a \implies e(1-a) = 0$ . For all $a \neq 1$ , $e = 0$ . Check: $a * 0 = a + 0 - 0 = a$ . Identity element = 0. Associative: $(a * b) * c = (a+b-ab) + c - (a+b-ab)c = a+b+c-ab-ac-bc+abc$ . $a * (b * c) = a + (b+c-bc) - a(b+c-bc) = a+b+c-bc-ab-ac+abc$ . Equal. Yes, associative.

Why is bijectivity needed for inverse?

One-one ensures each output has at most one pre-image (so $f^{-1}$ gives a unique value). Onto ensures each element of the codomain has at least one pre-image (so $f^{-1}$ is defined everywhere on the codomain).