Types of Functions (One-One, Onto, Bijective)

MATFunctions

A function $f:A \to B$ is one-one (injective) if distinct inputs give distinct outputs, onto (surjective) if every element of the codomain $B$ is hit, and bijective if it is both. Bijection is the key idea because only a bijection has an inverse $f^{-1}$ .

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ISC problems ask you to test these properties for linear, quadratic and rational functions and then either find $f^{-1}$ or supply a domain/codomain restriction that forces bijectivity.

One-one (injective)

f(x_1) = f(x_2) \Rightarrow x_1 = x_2

Equivalently,

x_1 \ne x_2 \Rightarrow f(x_1) \ne f(x_2)

; this is the standard test to start an algebraic proof.

Onto (surjective)

\forall y \in B,\ \exists x \in A \ \text{such that}\ f(x) = y

B

is the codomain; onto means range of

f

equals

B

. Solve

y = f(x)

for

x

and check

x \in A

Bijective

f \ \text{is bijective} \iff f \ \text{is one-one and onto}

A bijection is invertible; its inverse

f^{-1}:B \to A

exists and is unique.

Linear example

f(x) = 3x - 7, \quad f^{-1}(y) = \dfrac{y+7}{3}

f:\mathbb{R}\to\mathbb{R}

; every non-constant linear map is a bijection on

\mathbb{R}

Finite-set shortcut

n(A) = n(B) < \infty \Rightarrow \big(f \text{ one-one} \iff f \text{ onto}\big)

For equal finite sets, proving either property gives the other, hence bijection; this fails on infinite sets.

Standard injectivity proof: assume $f(x_1)=f(x_2)$ , simplify algebraically, and conclude $x_1=x_2$ .
Standard surjectivity proof: take arbitrary $y \in B$ , solve $y=f(x)$ for $x$ , and verify the solved $x$ lies in the domain $A$ (this also hands you a candidate for $f^{-1}$ ).
$f(x)=x^2$ on $\mathbb{R}$ is neither one-one (since $f(-2)=f(2)$ ) nor onto (negatives are never outputs); restricting domain to $[0,\infty)$ and codomain to $[0,\infty)$ makes it bijective.
Every linear $f(x)=mx+c$ with $m \ne 0$ on $\mathbb{R}$ is a bijection, so it always has an inverse.
For a rational function like $f(x)=\dfrac{x-2}{x-3}$ from $\mathbb{R}\setminus\{3\}$ to $\mathbb{R}\setminus\{1\}$ , exclude the domain value making the denominator zero and the unattained output (the horizontal asymptote value).
The piecewise $\mathbb{N}\to\mathbb{N}$ map sending odd $n \mapsto n+1$ and even $n \mapsto n-1$ is a bijection — it just swaps consecutive pairs $1\leftrightarrow2,\ 3\leftrightarrow4,\dots$
Always state the codomain explicitly: the same rule can be onto for one codomain and not onto for another.

Where the marks go

Declaring $f$ onto without solving $y=f(x)$ and checking the pre-image actually lies in the domain.
Treating $f(x)=x^2$ on $\mathbb{R}$ as one-one; it is not, because $f(-a)=f(a)$ for $a \ne 0$ .
Forgetting to remove the excluded point from the domain/codomain of a rational function, which breaks the bijection claim.
Assuming one-one automatically implies onto on infinite sets — that shortcut only holds for equal FINITE sets.

How the board asks it

Derive / provelinear function injectivity and surjectivity proof
Show that the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=\dfrac{3x+4}{5}$ is one-one and onto, and hence a bijection.
Give reasons $f(x)=x^2$ on $\mathbb{R}$ fails one-one and onto
State whether the function $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^2$ is one-one and onto. Justify your answer with a suitable counter-example.
Numericalrestricting domain and codomain to force bijectivity
Let $f:[0,\infty)\to[0,\infty)$ be defined by $f(x)=x^2$ . Show that $f$ is a bijection and find $f^{-1}$ .
Derive / proverational function with excluded domain and codomain points
Prove that the function $f:\mathbb{R}\setminus\{3\}\to\mathbb{R}\setminus\{1\}$ defined by $f(x)=\dfrac{x-2}{x-3}$ is bijective, and obtain $f^{-1}$ .
Multiple choicecounting one-one functions between finite sets
If $A=\{1,2,3\}$ and $B=\{a,b,c,d\}$ , the number of one-one functions from $A$ to $B$ is (a) $12$ (b) $24$ (c) $64$ (d) $81$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.