Relations and Functions (Class 12 Maths): Types, Composition and Inverse Explained

Why this chapter is worth taking seriously

Relations and Functions usually opens the Class 12 maths year, and because the early questions look easy, many students give it a quick reading and move on. That is a mistake for two reasons. First, the proofs in this chapter are marked strictly — a correct idea written loosely does not score. Second, the vocabulary defined here (domain, range, one-one, onto, inverse) is used without further explanation in inverse trigonometric functions, in calculus and in linear algebra later in the year.

Relations: the three properties

A relation R on a set A is simply a subset of A × A. Everything asked about it reduces to testing three properties.

A relation that is all three is an equivalence relation.

A worked example

Let R on the set of integers be defined by (a, b) ∈ R if a − b is divisible by 3.

• Reflexive: a − a = 0, which is divisible by 3. True for every integer, so yes.
• Symmetric: if a − b = 3k then b − a = −3k, also divisible by 3. Yes.
• Transitive: if a − b = 3k and b − c = 3m, then a − c = 3(k + m). Yes.

So R is an equivalence relation. Note how each step is a general argument using arbitrary k and m — not a set of sample numbers.

Equivalence classes

An equivalence relation partitions the set into disjoint classes, where every element of a class is related to every other. In the example above there are exactly three classes, according to whether the remainder on division by 3 is 0, 1 or 2. Questions asking “how many equivalence classes” or “write the class containing 2” are testing precisely this partition.

Functions: one-one and onto

A function f : A → B assigns exactly one element of B to each element of A. Two properties are then examined:

• One-one (injective): different inputs give different outputs. The standard proof is to assume f(x₁) = f(x₂) and show algebraically that x₁ = x₂.
• Onto (surjective): every element of the codomain is achieved. The standard proof is to take an arbitrary y in B and construct an x in A with f(x) = y.
• Bijective: both. Only a bijective function has an inverse, which is why these two properties are always established before an inverse is found.

Codomain versus range is the distinction that decides most onto questions. The range is what the function actually produces; the codomain is what it was declared to map into. A function is onto exactly when these coincide. Many questions become straightforward once you write down both and compare them — and many marks are lost by students who never distinguish them.

Composition of functions

The order matters and is the most common source of error: g ∘ f is generally not equal to f ∘ g. Composition also requires that the range of f lies within the domain of g, a condition worth stating explicitly in a proof. Two useful results: the composition of two one-one functions is one-one, and the composition of two onto functions is onto.

Inverse functions

If f : A → B is bijective, its inverse f⁻¹ : B → A satisfies f⁻¹(f(x)) = x and f(f⁻¹(y)) = y. The method is mechanical:

1. Prove f is one-one.
2. Prove f is onto.
3. Set y = f(x) and solve for x in terms of y.
4. Write f⁻¹(y) with that expression, and state its domain.

Steps 1 and 2 are not optional. A question that asks you to find the inverse is implicitly asking you to justify that one exists, and answers that jump straight to the algebra lose those marks even when the final expression is correct.

Where marks are lost

• Proving a property with examples instead of a general argument.
• Testing symmetry or transitivity on only one pair and concluding it holds.
• Ignoring the codomain when deciding whether a function is onto.
• Reversing the order in composition.
• Finding an inverse without first establishing that the function is bijective.
• Writing the answer without stating the domain of the inverse.

How to revise

1. Write the three relation properties and the two function properties from memory, in full formal language.
2. Test five relations for all three properties, writing complete proofs and counterexamples.
3. Prove two functions bijective and find their inverses, showing all four steps.
4. Do three composition questions, checking the order deliberately each time.
5. Work through NCERT and previous years’ board questions, which follow these forms closely.

The chapter is short, but the writing discipline it teaches carries through the whole year — which is a good reason to do it carefully rather than quickly.

FAQs

What is the difference between codomain and range?

The codomain is the set the function is declared to map into. The range is the set of values it actually takes. The range is always a subset of the codomain, and the function is onto precisely when the two are equal.

How do I prove a relation is not transitive?

Produce a single counterexample: find elements a, b and c such that (a, b) and (b, c) are both in R while (a, c) is not. One such case is sufficient and is faster and safer than any general argument.

Why must a function be bijective to have an inverse?

It must be one-one so that each output comes from only one input, otherwise the inverse would not be well defined. It must be onto so that every element of the codomain has a preimage, otherwise the inverse would not be defined everywhere on that set.

Is g composed with f the same as f composed with g?

Generally no. (g ∘ f)(x) means apply f first and then g, while (f ∘ g)(x) reverses that order. They coincide only in special cases, and questions are often set specifically to test whether a student has noticed the difference.

Is this chapter difficult?

The ideas are not difficult, but the marking is strict about how proofs are written. Students who state the definition, test it properly and write in complete sentences generally find it one of the more secure chapters of the year.