Relations and Functions

Which pairs are related

Class 11 introduced sets, and with them the idea that elements of one set can be paired with elements of another. This chapter makes that pairing precise and studies exactly what patterns such a pairing can carry.

A relation $R$ on a set $A$ is any collection of ordered pairs $(a, b)$ with $a, b \in A$ — formally, any subset of $A \times A$. Nothing in that definition forces the pairs to follow a pattern; an arbitrary, scattered subset of $A \times A$ already counts as a relation.

Three patterns recur often enough to earn names: reflexive, symmetric, and transitive. A relation carrying all three at once is an equivalence relation. An equivalence relation does something a generic relation cannot: it splits $A$ into non-overlapping classes, each holding everything related to a common member. A narrower pattern matters just as much: the case where every element of one set is related to exactly one element of another. That pattern is a function, already familiar from Class 11, now placed inside the wider family of relations it belongs to.

This chapter studies both patterns in turn. First come the three properties and the equivalence relations they build; then a function’s own properties — one-one, onto, invertible — and how functions combine under composition.

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Reflexive, symmetric, transitive

A relation $R$ on $A$ can carry three properties, and none of the three implies another: reflexive, symmetric, and transitive.

$R$ is reflexive when $(a, a) \in R$ for every $a \in A$ — every element is related to itself. $R$ is symmetric when $(a, b) \in R$ always forces $(b, a) \in R$ — the relation never runs one way without also running the other. $R$ is transitive when $(a, b) \in R$ and $(b, c) \in R$ together force $(a, c) \in R$ — a chain of two related pairs always closes into a third.

Two extreme relations test the definitions cleanly. The empty relation, $R = \emptyset$, is symmetric and transitive vacuously. There are no pairs in $R$ at all, so neither condition has anything to violate. It is reflexive only when $A$ itself is empty; if $A$ has any element $a$, the pair $(a, a)$ is simply absent from $R$, and reflexivity fails outright.

The universal relation, $R = A \times A$, contains every possible ordered pair. It satisfies all three properties at once — reflexivity, symmetry, and transitivity impose no restriction it could ever fail to meet.

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Equivalence relations and equivalence classes

A relation that is reflexive, symmetric, and transitive all at once is called an equivalence relation. The three properties together do something none of them does alone: they let $A$ be organised into groups of mutually related elements.

For an equivalence relation $R$ on $A$ and any element $a \in A$, the equivalence class of $a$ is $[a] = {x \in A : (x, a) \in R}$ — everything in $A$ related to $a$. Because $R$ is reflexive, $a$ itself always belongs to $[a]$. No class is ever empty.

Two equivalence classes are either identical or completely disjoint; partial overlap is impossible. If $[a]$ and $[b]$ share even one element, symmetry and transitivity chain that shared element back to both $a$ and $b$. That forces $(a, b) \in R$ and, from there, every member of one class into the other. The classes therefore either coincide exactly or share nothing at all.

Every $a \in A$ belongs to its own class $[a]$, and classes never partially overlap. Together, the classes partition $A$ into non-overlapping groups whose union is all of $A$. The relation and the partition carry exactly the same information, read two different ways. Given the relation, the classes are built as above. Given the partition instead, the relation is recovered: declare two elements related exactly when they share a class.

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Two properties do not buy the third

Symmetric and transitive both talk about pairs already in $R$ closing up into further pairs, which invites treating reflexivity as something the other two properties deliver automatically. If every pair reverses and every chain closes, it can look as though every element must eventually pair with itself too.

That reasoning has a genuine gap: symmetric and transitive statements only ever talk about elements that already appear in some pair of $R$. Neither property says anything about an element $R$ never mentions at all — and an element outside every pair of $R$ is exactly where reflexivity can fail.

$R = {(1,1), (2,2), (1,2), (2,1)}$ on $A = {1, 2, 3}$ makes the gap concrete. $R$ is symmetric — the reverse of every pair present is present too. $R$ is transitive — $(1,2)$ and $(2,1)$ force $(1,1)$, and $(2,1)$ and $(1,2)$ force $(2,2)$, both already there; every other combination involves a diagonal pair and forces nothing new. Yet $R$ is not reflexive: $(3,3)$ is missing, because $3$ never appears in any pair of $R$ at all.

*Symmetric and transitive together force reflexivity only on the elements $R$ actually relates to something*. An element $R$ never mentions is invisible to both properties, and invisible to both is exactly how it dodges reflexivity.

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One-one, onto, both

A function $f: A \to B$ can carry two independent properties, one-one and onto, and a function can have either without the other, both together, or neither.

$f$ is one-one, or injective, when distinct inputs always produce distinct outputs: $f(x_1) = f(x_2)$ forces $x_1 = x_2$. Equivalently, no two different elements of $A$ ever share an image in $B$.

$f$ is onto, or surjective, when every element of $B$ is actually hit: for every $b \in B$, some $a \in A$ satisfies $f(a) = b$. Nothing in $B$ is left without a preimage.

A function that is both one-one and onto is called bijective. The two properties test different things. One-one asks whether outputs ever collide; onto asks whether the codomain is fully covered. Knowing one tells nothing about the other — each has to be checked on its own terms.

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Composing functions

Given two functions $f: A \to B$ and $g: B \to C$, their composition builds a new function that sends $x$ to $g(f(x))$. Apply $f$ first, then apply $g$ to whatever $f$ produced. The composition is written $g compose f$. It is defined precisely because $f$’s output lands in $B$, exactly where $g$ expects its input.

Composition is associative. Composing $f$, then $g$, then $h$ gives the same function no matter which two are grouped and applied first. $(h compose g) compose f$ and $h compose (g compose f)$ are the same function. Parentheses in a chain of compositions can therefore be dropped without ambiguity.

Composition is not generally commutative. $g(f(x))$ and $f(g(x))$ can disagree even when both are defined. One of the two can even fail to be defined, if the domains and codomains do not line up the other way round: $g$’s output may not land inside $f$’s domain at all. Order matters, and it has to be tracked explicitly rather than assumed.

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Invertible functions

A function $f: A \to B$ is invertible when some function $g: B \to A$ undoes it completely in both directions. $g(f(a)) = a$ for every $a \in A$, and $f(g(b)) = b$ for every $b \in B$. Undoing it in only one direction is not enough — both conditions have to hold.

*Such a $g$ exists if and only if $f$ is bijective*. One-one guarantees that $g$ can send each $b \in B$ back to a single well-defined $a$, without ambiguity. Onto guarantees that every $b \in B$ actually has some $a$ to send back to — so $g$ is defined on the whole of $B$. When it exists, the inverse $g$ is unique, and it is written $f^{-1}$.

Finding $f^{-1}$ in practice follows a fixed procedure. Write $y = f(x)$, solve that equation for $x$ in terms of $y$, then swap the names $x$ and $y$, so the result reads as a function of $x$ again. The solved expression is $f^{-1}(x)$.

Onto guarantees that nothing in $B$ is left without a preimage. That invites a shortcut: onto alone should be enough to build a reverse function, since every output already has something to send back to.

That shortcut ignores what a function actually requires: a single, unambiguous output for every input. Onto only guarantees that a preimage exists; it says nothing about whether that preimage is the only one.

$f: \mathbb{R} \to [0, \infty)$, $f(x) = x^2$ makes the gap concrete. $f$ is onto: every non-negative number is hit by some real $x$. But $f$ is not one-one — $f(-2) = f(2) = 4$, so $4$ has two preimages, not one. A candidate reverse rule sending $4$ back to $x$ has no way to choose between $-2$ and $2$. Both are equally valid preimages, and a function permits only one output per input. The candidate reverse is not a well-defined function at all.

Invertibility needs both properties together, not onto alone. One-one guarantees the reverse rule has no ambiguous choice to make; onto guarantees it has an output to assign for every element of $B$. Bijective is the name for carrying both at once, and nothing weaker suffices.

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Sameness, and reversibility

Two ideas run through this chapter, both concerned with turning an informal notion into something checkable: sameness and reversibility.

A relation that is reflexive, symmetric, and transitive at once — an equivalence relation — turns “related to” into a rigorous version of “the same as”. It partitions a set into non-overlapping classes, each holding everything that counts as the same under that relation, from congruent triangles to any other pattern satisfying the three properties.

A function that is bijective turns “maps to” into a rigorous version of “reversible”. One-one rules out ambiguity in the reverse direction; onto guarantees the reverse direction has something to send back everywhere; together they guarantee a unique inverse function exists.

Both ideas share the same discipline: check whether a relation or function actually has the property in question, rather than assuming it from the shape of the definition. Symmetric and transitive together do not guarantee reflexive; onto alone does not guarantee invertible. Composition extends the same checking discipline to combining functions: it is associative, but not generally commutative, and each property is verified in turn, never assumed.

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Practice set

Exercise 1.1 practises testing a given relation on a set for reflexivity, symmetry, and transitivity, checking each property directly against the definition. It also classifies each relation as an equivalence relation outright, or names the specific property that fails.

Exercise 1.2 practises testing whether a given function is one-one, onto, both, or neither. Several problems reuse the same formula across different domain and codomain choices, so the same rule can answer differently depending on where it is applied.

Exercise 1.3 practises composing two given functions in both orders, then checking whether the composite is bijective. Where it is, the exercise finds the inverse by the solve-and-swap method demonstrated earlier in the chapter.

The miscellaneous exercise mixes all three sections of the chapter — relations, functions, and composition — into a single problem set. No problem announces which technique applies, so identifying the right check is itself part of the exercise.