Functions · step 2 of 13

Injective, surjective & bijective

f is injective if distinct inputs give distinct outputs, surjective if every element of the codomain is hit, and bijective if both. The proof templates are fixed: for injectivity assume f(x) = f(y) and derive x = y; for surjectivity take an arbitrary b in the codomain and construct an x mapping to it. Both properties depend on the declared domain and codomain, not just the formula.

Practise injective, surjective & bijective free → All Functions subtopics

✓ Unlimited questions · marked criterion by criterion · no card needed

Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Injectivity — assume equal outputsStart "suppose f(x) = f(y)", manipulate, and conclude x = y. To refute, give two distinct inputs with the same output.
Surjectivity — solve for the inputTake arbitrary b in the codomain, solve f(x) = b for x, and check x lies in the domain. To refute, name a b with no solution.
Watch the domain restrictionx² is not injective on ℝ but is on [0,∞). Always check the stated domain before deciding.
Bijective means bothState both proofs. For finite sets of equal size, either property implies the other — but that shortcut needs the sizes to be equal and finite.

Worked example

Is f : ℝ → ℝ, f(x) = 2x + 3 bijective?

Injective: suppose 2x + 3 = 2y + 3. Then 2x = 2y, so x = y ✓.
Surjective: take arbitrary b ∈ ℝ and solve 2x + 3 = b.
x = (b − 3)/2, which is a real number for every real b ✓.
Both hold.

Answer. Yes, bijective, with inverse f⁻¹(y) = (y − 3)/2.

Where marks get dropped

These are the specific errors that cost credit on injective, surjective & bijective questions — QED's rubric penalises each of them separately.

Refuting injectivity with two inputs that are actually equal. The witnesses must be distinct.
Claiming surjectivity without checking the solution lies in the domain. For f : ℕ → ℕ, f(n) = 2n, solving 2n = 3 gives n = 1.5 ∉ ℕ.
Assuming injective implies surjective. That holds for functions between finite sets of the same size, not in general — f : ℕ → ℕ, f(n) = n + 1 is injective and not surjective.

Practise this until it is automatic

Unlimited fresh questions

QED generates new injective, surjective & bijective problems on demand at warm-up, exam and challenge level, so you can drill this one skill until it stops costing you marks.

Marked like an examiner

Every answer is scored against a point-by-point rubric with partial credit, so you see exactly which step of the method broke down — not just a tick or a cross.

Answer in real notation

A one-tap symbol palette, a visual equation editor and a truth-table builder — or photograph your handwritten working and QED converts it to LaTeX.

Saved to your library

Every question you generate is kept and re-takeable as a timed exam, and your Functions mastery is tracked so you know when this is exam-ready.

Injective, surjective & bijective — frequently asked questions

How do I refute injectivity fastest?

Find a symmetry. For even functions f(−x) = f(x), so any non-zero x gives an instant counterexample.

Does a bijection require equal cardinalities?

Yes, and it is the definition of equal cardinality for infinite sets. A bijection exists between ℕ and ℤ despite ℤ appearing larger.

What if the codomain is the image?

Then f is surjective by construction. This is why questions always state the codomain explicitly — otherwise surjectivity would be vacuous.

The rest of Functions

Injective, surjective, bijective, composition, inverse. Each subtopic below has its own method, worked example and mark-losing traps.

Ready to make injective, surjective & bijective exam-proof?

Generate your first questions free — no card, no setup, no personal data stored. Practise until the method is second nature.

Start practising free →