Onto (Surjective) Functions and Bijections

4. Onto (Surjective) Functions and Bijections

While injectivity is about no repeats, surjectivity is about nothing left out — every target in $B$ must receive at least one arrow.

Method — Proving Surjectivity

Given $y \in B$, solve $f(x) = y$ for $x$ and check the solution genuinely lies in the domain $A$.

Illustration. Let $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 5x - 7$. For any $y \in \mathbb{R}$, solving $5x - 7 = y$ gives $x = \dfrac{y+7}{5}$, which is a real number for every real $y$. So $f$ is onto.

Method — Disproving Surjectivity

Exhibit one element of the codomain with no preimage in the domain.

Illustration. Let $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$. Since $x^2 \geq 0$ for every real $x$, no negative real number (e.g. $y = -1$) is ever attained. So $f$ is not onto $\mathbb{R}$ (though it is onto if the codomain is narrowed to $[0, \infty)$).

Bijective Functions

A bijection pairs up every element of $A$ with exactly one element of $B$ and leaves nothing in $B$ unused — this is precisely the condition under which $f$ has a genuine inverse function (Section 6).

Illustration. $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 5x - 7$ is one-one (Section 3) and onto (above), hence bijective. …