Skip to content
Free public beta until December 4, 2026: every course is open, no card needed. After that, the first chapter of each course stays free and full access is a one-time $39 ($19 for up to 100 founding members chosen for helpful feedback). Your progress is kept. How the beta works

Recipe

Test whether a function is onto

Use when: you're given f: A \to B and asked whether it's onto (surjective), meaning every element of the codomain actually gets produced

Goal: either a construction showing an input exists for every element of the codomain, or one specific codomain element no input reaches

Taught in Onto functions

Quick recipe

  1. Take an arbitrary element of the codomain and try to solve f(a) = b for a
  2. Check whether the constructed input always lands back in the domain
  3. If onto fails, exhibit one codomain element no input reaches
  4. State the final verdict with its justification
  1. 1

    Set the function's rule equal to an arbitrary output b, then attempt to isolate the input a in terms of b.

    Example

    For f(x) = 3x - 7: set 3x - 7 = y for an arbitrary real y, then solve: 3x = y+7, so x = \dfrac{y+7}{3}.

    Why: constructing an explicit input for an arbitrary output is what "every element of B is reached" actually means, not just checking a handful of values.

  2. 2

    Confirm whether the expression from the previous step always produces a valid member of the domain, for every b in the codomain.

    Example

    \dfrac{y+7}{3} is a real number for every real y, so it always lies in the domain \mathbb{R}: f is onto. By contrast, solving 2n = 3 for an integer n gives n = 1.5, not an integer.

    • the constructed input is always in the domain, for every b — the function is onto; write up the construction as the proof.
    • the constructed input sometimes falls outside the domain, such as not being an integer — look for one specific b that breaks it.

    Why: a formula for the input only proves onto if it genuinely stays inside the domain for every codomain element, not just for some of them.

  3. 3

    Pick a candidate output the constructed formula could not produce as a valid domain element, then confirm directly that no input works.

    Example

    For f(n) = 2n on the integers: 2n = 3 has no integer solution, since n = 1.5 is not an integer, so 3 is never reached.

    Why: one confirmed unreachable element disproves "every b is hit" completely, the same way one counterexample disproves any "for all" claim.

  4. 4

    Write "onto," backed by the completed construction, or "not onto," backed by the specific unreached element.

    Why: onto is a claim about the entire codomain, so the verdict needs either a general construction or a specific counterexample behind it.

Common mistakes

  • Checking that a few codomain elements are reached and concluding onto without a general construction.
  • Forgetting to confirm the constructed input actually lies in the stated domain, not just that the algebra worked.
  • Picking a b value that happens to be reachable when hunting for a counterexample, then wrongly concluding the function is onto.
  • Confusing onto with one-to-one, since one asks whether every output is reached and the other asks whether outputs ever repeat.