Types of Functions

The two questions that classify every function

From Class XI you know functions and their graphs — identity, constant, polynomial, rational, modulus, signum, greatest integer. This section asks two sharper questions about any function $f : X \to Y$:

Question 1 — do distinct inputs stay distinct? (Is $f$ one-one?)

Question 2 — does every element of $Y$ get hit? (Is $f$ onto?)

Definition (one-one / injective). $f : X \to Y$ is one-one if the images of distinct elements are distinct — equivalently, for every $x_1, x_2 \in X$, $$f(x_1) = f(x_2) \implies x_1 = x_2$$ A function that is not one-one is called many-one.

Definition (onto / surjective). $f : X \to Y$ is onto if every $y \in Y$ is the image of some $x \in X$: for every $y \in Y$ there exists $x \in X$ with $f(x) = y$. Equivalently: Range of $f$ = $Y$ (the whole co-domain).

Definition (bijective). $f$ is bijective if it is both one-one and onto — a perfect pairing between $X$ and $Y$.

The proof templates (used in every board answer)

To prove one-one: start from $f(x_1) = f(x_2)$ and derive $x_1 = x_2$ by algebra. $$f(x) = 2x: \quad f(x_1) = f(x_2) \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2$$

To disprove one-one: exhibit two distinct inputs with the same image. $$f(x) = x^2 \text{ on } \mathbb{R}: \quad f(-1) = 1 = f(1), \text{ and } -1 \neq 1$$

To prove onto: take an arbitrary $y$ in the co-domain and solve $f(x) = y$ for $x$, checking the solution lies in the domain. $$f(x) = 2x \text{ on } \mathbb{R}: \quad \text{given } y, \text{ take } x = \frac{y}{2} \in \mathbb{R}, \ f\left(\frac{y}{2}\right) = y$$

To disprove onto: exhibit one element of the co-domain with no preimage. $$f(x) = 2x \text{ on } \mathbb{N}: \quad 1 \in \mathbb{N} \text{ but } 2x = 1 \text{ has no solution in } \mathbb{N}$$

Notice how the same formula $f(x) = 2x$ is bijective on $\mathbb{R}$ but only one-one on $\mathbb{N}$ — the domain and co-domain are part of the function. Changing them changes the classification; this is the single most tested subtlety of the section.

The standard gallery (memorise these verdicts)

The Finite-Set Phenomenon and Sharper Examples

One-one $\iff$ onto on a finite set

For maps of a finite set to itself the two properties collapse into one:

Fact. For a finite set $X$, a function $f : X \to X$ is one-one if and only if it is onto.

Why (for $X = \{1,2,3\}$): if $f$ is one-one, its three images are distinct elements of a three-element set — so they fill it, and $f$ is onto. Conversely if $f$ is onto but not one-one, two inputs share an image, leaving at most two distinct images to cover three elements — impossible.

This is a characteristic difference between finite and infinite sets: on $\mathbb{N}$, the map $f(x) = 2x$ is one-one but not onto, and the map $f(1) = f(2) = 1$, $f(x) = x - 1$ for $x > 2$ is onto but not one-one. Neither behaviour can occur on a finite set mapping to itself.

Two functions worth studying closely

The parity swap. $f : \mathbb{N} \to \mathbb{N}$ defined by $$f(x) = \begin{cases} x + 1, & \text{if } x \text{ is odd} \\ x - 1, & \text{if } x \text{ is even} \end{cases}$$ is bijective: it pairs $1 \leftrightarrow 2$, $3 \leftrightarrow 4$, $5 \leftrightarrow 6, \ldots$ Each number is both the image of exactly one number and the preimage of exactly one — a perfect shuffle of $\mathbb{N}$.

The floor-halving map. $f : \mathbb{N} \to \mathbb{N}$ with $f(n) = \frac{n+1}{2}$ for odd $n$ and $\frac{n}{2}$ for even $n$ sends $1 \mapsto 1, 2 \mapsto 1, 3 \mapsto 2, 4 \mapsto 2, \ldots$ It is onto (every $m$ is $f(2m)$) but not one-one ($f(1) = f(2) = 1$) — each output is hit exactly twice.

The rational-function workhorse

$f : \mathbb{R} - \{3\} \to \mathbb{R} - \{1\}$, $\ f(x) = \dfrac{x - 2}{x - 3}$ is one-one and onto:

One-one: $\dfrac{x_1 - 2}{x_1 - 3} = \dfrac{x_2 - 2}{x_2 - 3}$; cross-multiplying, $(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$, which expands to $-3x_1 - 2x_2 = -3x_2 - 2x_1$, i.e. $x_1 = x_2$.

Onto: given $y \neq 1$, solve $y = \dfrac{x - 2}{x - 3}$: $\ yx - 3y = x - 2$, so $x(y - 1) = 3y - 2$ and $x = \dfrac{3y - 2}{y - 1}$, which is defined (since $y \neq 1$) and never equals 3 (that would need $3y - 2 = 3y - 3$, impossible). So every target is hit.

The carefully punctured domain and co-domain are what make this work — another reminder that they are part of the function's identity.

Common mistakes to avoid

Mistake 1 — proving one-one by example. Checking $f(1) \neq f(2)$ proves nothing; the algebra must run for arbitrary $x_1, x_2$. Conversely a single collision disproves it.

Mistake 2 — confusing range with co-domain. Onto means range = co-domain. The function $x^2 : \mathbb{R} \to \mathbb{R}$ has range $[0, \infty) \neq \mathbb{R}$, so it is not onto — even though its range is a perfectly nice set.

Mistake 3 — ignoring the domain when solving $f(x) = y$. For $f(x) = 2x$ on $\mathbb{N}$, the equation $2x = 1$ has the solution $x = \frac{1}{2}$ — but $\frac{1}{2} \notin \mathbb{N}$, so 1 is not hit. The solution must live in the stated domain.

Mistake 4 — 'increasing means onto'. $f(x) = x^3$ is strictly increasing on $\mathbb{Z}$, hence one-one, but not onto $\mathbb{Z}$: the value 2 is no integer's cube. Monotonicity gives injectivity, never surjectivity.

Solved Examples

Example 1 — Roll numbers

Let $A$ be the set of all 50 students of Class X in a school and $f : A \to \mathbb{N}$ assign to each student their roll number. Show that $f$ is one-one but not onto.

Step 1 — one-one: no two different students share a roll number, so distinct students have distinct images.

Step 2 — not onto: with roll numbers running 1 to 50, the number $51 \in \mathbb{N}$ is nobody's roll number — an element of the co-domain with no preimage.

Answer: one-one but not onto. Any injective map of a 50-element set into the infinite set $\mathbb{N}$ must leave almost everything unhit.

Example 2 — Doubling on $\mathbb{N}$

Show that $f : \mathbb{N} \to \mathbb{N}$, $f(x) = 2x$, is one-one but not onto.

Step 1 — one-one: $f(x_1) = f(x_2) \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2$.

Step 2 — not onto: for $1 \in \mathbb{N}$, the equation $2x = 1$ demands $x = \frac{1}{2}$, which is not a natural number.

Answer: one-one, not onto — the image is only the even numbers.

Example 3 — Doubling on $\mathbb{R}$

Prove that $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 2x$, is one-one and onto.

Step 1 — one-one: $2x_1 = 2x_2 \Rightarrow x_1 = x_2$.

Step 2 — onto: given any $y \in \mathbb{R}$, the real number $\frac{y}{2}$ satisfies $f\left(\frac{y}{2}\right) = 2 \cdot \frac{y}{2} = y$.

Answer: bijective. Same formula as Example 2, different domain and co-domain, different verdict — the classification belongs to the whole package $(f, X, Y)$, not to the formula alone.

Example 4 — Onto but not one-one

Show that $f : \mathbb{N} \to \mathbb{N}$, given by $f(1) = f(2) = 1$ and $f(x) = x - 1$ for every $x > 2$, is onto but not one-one.

Step 1 — not one-one: $f(1) = f(2) = 1$ with $1 \neq 2$.

Step 2 — onto: given $y \in \mathbb{N}$ with $y \neq 1$, choose $x = y + 1 > 2$; then $f(y + 1) = y$. And $1 = f(1)$ is also hit.

Answer: onto but not one-one — the map shifts everything down by one and dumps both 1 and 2 onto 1.

Example 5 — The parabola fails both tests

Show that $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$, is neither one-one nor onto.

Step 1 — not one-one: $f(-1) = 1 = f(1)$.

Step 2 — not onto: $-2$ in the co-domain is no real number's square, since squares are nonnegative.

Answer: neither. (On $\mathbb{N} \to \mathbb{N}$ the same formula is one-one — there are no negatives to collide — but still not onto, since 2 is not a perfect square.)

Example 6 — The parity swap is a bijection

Show that $f : \mathbb{N} \to \mathbb{N}$ given by $f(x) = x + 1$ if $x$ is odd and $f(x) = x - 1$ if $x$ is even, is both one-one and onto.

Step 1 — collisions across parities are impossible: if $x_1$ odd and $x_2$ even had $f(x_1) = f(x_2)$, then $x_1 + 1 = x_2 - 1$, i.e. $x_2 - x_1 = 2$ — but then $x_1$ and $x_2$ would have the same parity, a contradiction.

Step 2 — collisions within a parity force equality: both odd gives $x_1 + 1 = x_2 + 1 \Rightarrow x_1 = x_2$; both even similarly. So $f$ is one-one.

Step 3 — onto: any odd number $2r + 1$ is the image of the even number $2r + 2$, and any even number $2r$ is the image of the odd number $2r - 1$.

Answer: bijective — $f$ swaps each odd number with its even successor.

Example 7 — The finite-set collapse

Show that an onto function $f : \{1, 2, 3\} \to \{1, 2, 3\}$ is always one-one, and that a one-one function $f : \{1, 2, 3\} \to \{1, 2, 3\}$ must be onto.

Step 1 — onto $\Rightarrow$ one-one (by contradiction): if $f$ were not one-one, two elements would share an image, so the range would contain at most two of the three co-domain elements — contradicting onto.

Step 2 — one-one $\Rightarrow$ onto: three distinct images inside a three-element co-domain must exhaust it.

Answer: on a finite set mapping to itself, one-one and onto are equivalent — and this equivalence fails on infinite sets (Examples 2 and 4 are the counterexamples).