Types of functions: one-one, onto, bijective

A function $f : A \to B$ associates to each $a \in A$ a unique $b = f(a) \in B$. That definition is silent about whether different inputs may share an output or whether every output of $B$ is actually hit. Class XII makes these questions central. A function is one-one (injective) if no two inputs collide; it is onto (surjective) if every output is achieved; it is bijective if both. Only bijections admit inverses, so this classification is the gatekeeper for the rest of the chapter.

Definitions

Let $f : A \to B$ be a function.

• $f$ is one-one (injective) if for all $x_1, x_2 \in A$, $f(x_1) = f(x_2) \implies x_1 = x_2.$
• $f$ is onto (surjective) if for every $y \in B$ there exists $x \in A$ with $f(x) = y$. Equivalently, the range of $f$ equals $B$.
• $f$ is bijective (one-one and onto) if both hold.

The contrapositive of injectivity is often more usable: $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$.

Tests in practice

Algebraic test for one-one. Set $f(x_1) = f(x_2)$ and try to deduce $x_1 = x_2$. If you can, $f$ is injective. If you find $x_1 \neq x_2$ producing the same output, $f$ is not injective.

Algebraic test for onto. Pick an arbitrary $y \in B$, solve $f(x) = y$ for $x$, and check $x \in A$.

Graphical test. A function $f : \mathbb{R} \to \mathbb{R}$ is injective iff every horizontal line meets the graph in at most one point. It is surjective iff every horizontal line meets the graph in at least one point. It is bijective iff every horizontal line meets the graph exactly once.

Calculus test. A differentiable function with $f'(x) > 0$ (or $< 0$) everywhere on an interval is one-one on that interval, because it is strictly monotonic. We will exploit this constantly in inverse trigonometric functions and in calculus.

Detailed example

Consider $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 2x + 3$.

One-one? If $2x_1 + 3 = 2x_2 + 3$ then $2x_1 = 2x_2$ so $x_1 = x_2$. Yes.

Onto? Take any $y \in \mathbb{R}$, set $x = (y - 3)/2 \in \mathbb{R}$ and verify $f(x) = y$. Yes.

Now consider $g : \mathbb{R} \to \mathbb{R}$, $g(x) = x^2$.

One-one? $g(1) = g(-1) = 1$, so no.

Onto? Take $y = -1$. There is no real $x$ with $x^2 = -1$. So no.

Restricting codomain and domain can change everything. The map $g : [0, \infty) \to [0, \infty)$, $g(x) = x^2$, is bijective.

A worked theorem

Theorem. A function $f : A \to B$ between finite sets of the same cardinality is one-one iff it is onto.

Proof. Suppose $|A| = |B| = n$.

($\Rightarrow$) If $f$ is one-one, the image $f(A)$ contains $n$ distinct elements. Since $f(A) \subseteq B$ and $|B| = n$, we must have $f(A) = B$. So $f$ is onto.

($\Leftarrow$) If $f$ is onto, every element of $B$ has a preimage. If two distinct elements $x_1, x_2$ of $A$ shared an image, then $f(A)$ would have at most $n - 1$ elements, contradicting surjectivity. So $f$ is one-one. $\blacksquare$

The theorem fails for infinite sets: $f : \mathbb{N} \to \mathbb{N}$, $f(n) = 2n$ is injective but not surjective.

Worked examples

Example 1. Is $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^3$, bijective?

One-one: If $x_1^3 = x_2^3$ then $x_1 = x_2$ (cube root is single-valued on $\mathbb{R}$). Onto: every real $y$ has cube root $y^{1/3}$. So yes.

Example 2. Is $f : \mathbb{N} \to \mathbb{N}$, $f(n) = n^2 + 1$, one-one? Onto?

One-one: $n_1^2 + 1 = n_2^2 + 1 \Rightarrow n_1 = n_2$ (since $\mathbb{N}$ is positive). Yes. Onto: $f(\mathbb{N}) = \{2, 5, 10, 17, \dots\}$, missing $1, 3, 4, 6, \dots$. So no.

Example 3. $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = \tfrac{x}{1 + |x|}$. Show $f$ is one-one and find its range.

If $x \ge 0$, $f(x) = x/(1 + x) \in [0, 1)$. If $x < 0$, $f(x) = x/(1 - x) \in (-1, 0)$. Strictly increasing on each piece and continuous, with matching value $0$ at $x = 0$. Range is $(-1, 1)$. One-one because strictly increasing.

Example 4. Define $f : \{1, 2, 3\} \to \{4, 5, 6\}$, $f(1) = 4, f(2) = 5, f(3) = 6$. Is $f$ bijective?

Yes , both sets have size $3$ and $f$ is clearly one-one and onto.

Example 5. $f : \mathbb{R} \setminus \{2\} \to \mathbb{R} \setminus \{1\}$, $f(x) = \tfrac{x - 1}{x - 2}$. Show $f$ is bijective.

One-one: $(x_1 - 1)(x_2 - 2) = (x_2 - 1)(x_1 - 2)$ expands to $-x_2 = -x_1$ (algebra). Yes.

Onto: Given $y \neq 1$, solve $y = (x - 1)/(x - 2)$. Cross-multiply: $y(x - 2) = x - 1$, so $x(y - 1) = 2y - 1$, $x = (2y - 1)/(y - 1)$. Defined since $y \neq 1$. So $f$ is onto.

Example 6. $f : \mathbb{R} \to \mathbb{R}$, $f(x) = \sin x$. One-one? Onto?

Not one-one: $f(0) = f(\pi) = 0$. Not onto: range is $[-1, 1]$, not all of $\mathbb{R}$.

Restricting domain to $[-\pi/2, \pi/2]$ and codomain to $[-1, 1]$ makes $f$ a bijection. This restriction is the foundation of $\arcsin$ in the next chapter.

Try it yourself

1. Is $f : \mathbb{Z} \to \mathbb{Z}$, $f(n) = n + 7$, bijective?
2. Is $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^2 + 1$, one-one?
3. Is $f : [0, \pi] \to [-1, 1]$, $f(x) = \cos x$, bijective?
4. Show $f : \mathbb{N} \to \mathbb{N}$, $f(n) = 2n - 1$, is one-one but not onto.
5. Count the one-one functions from $\{1, 2, 3\}$ to $\{a, b, c, d\}$.
6. Count the onto functions from $\{1, 2, 3, 4\}$ to $\{a, b\}$. (Hint: $2^4 - 2 = 14$.)
7. Show that $f : \mathbb{R} \to \mathbb{R}$, $f(x) = ax + b$ ($a \neq 0$), is bijective.
8. Is $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = e^x$ one-one? Onto?
9. Find the range of $f : \mathbb{R} \to \mathbb{R}$, $f(x) = \tfrac{1}{1 + x^2}$. Is $f$ onto?
10. Show $f : \mathbb{Z} \to \mathbb{Z}$, $f(n) = n^3$, is one-one but not onto.
11. Prove: if $f : A \to B$ and $g : B \to C$ are both injective, then $g \circ f$ is injective.
12. Prove: if $g \circ f$ is one-one, must $f$ be one-one? Must $g$?
13. Construct a function $\mathbb{N} \to \mathbb{N}$ which is onto but not one-one.
14. The greatest integer function $\lfloor x \rfloor : \mathbb{R} \to \mathbb{Z}$. Is it one-one? Onto?

Pitfalls / Tricks

• A function from $\mathbb{R}$ to $\mathbb{R}$ that is monotonic on intervals is not necessarily one-one globally; check the entire domain.
• Surjectivity depends crucially on the codomain. Changing the codomain to the range always makes a function onto.
• For finite sets of the same size, one-one and onto are equivalent. For infinite sets they are not , every Hilbert hotel argument depends on this gap.
• The number of one-one functions from $|A| = m$ to $|B| = n$ ($m \le n$) is $\tfrac{n!}{(n - m)!}$. The number of onto functions from a set of size $m$ to a set of size $n$ ($m \ge n$) is $\sum_{k = 0}^{n} (-1)^k \binom{n}{k}(n - k)^m$ , useful in JEE counting problems.

The next subtopic builds on bijections to define composition and inverses.