Functions · step 1 of 13

Domain, codomain, image & preimage

A function f : A → B carries three pieces of data: the domain A, the codomain B, and the rule. The image f[A] is what is actually hit, and it is generally a proper subset of the codomain — which is exactly why surjectivity is a real condition. The preimage f⁻¹[S] = {a ∈ A : f(a) ∈ S} always makes sense, even when f has no inverse function.

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Method: how to approach it

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

Read off domain and codomain from the declarationThey are part of the function, not derived from the formula. f : ℝ → ℝ and f : ℝ → [0,∞) with f(x) = x² are different functions.
Compute the image by sweeping the domainf[S] = {f(a) : a ∈ S}. For finite sets, apply f to each element and collect the distinct outputs.
Compute the preimage by solvingf⁻¹[T] collects every input landing in T. Solve f(a) ∈ T; the answer may be empty, a single point, or many points.
Keep preimage notation separate from inversef⁻¹[T] is a set and always defined; f⁻¹(b) as a single element requires f to be bijective.

Worked example

For f : ℝ → ℝ, f(x) = x², find the image of ℝ, f[{−1, 0, 2}] and f⁻¹[{4, 9}].

Image: squares of reals are exactly the non-negative reals, so f[ℝ] = [0, ∞).
f[{−1,0,2}] = {1, 0, 4} = {0, 1, 4}.
f⁻¹[{4,9}] solves x² = 4 or x² = 9.
x = ±2 or x = ±3.

Answer. f[ℝ] = [0,∞) — a proper subset of the codomain ℝ; f[{−1,0,2}] = {0,1,4}; f⁻¹[{4,9}] = {−3,−2,2,3}.

Where marks get dropped

These are the specific errors that cost credit on domain, codomain, image & preimage questions — QED's rubric penalises each of them separately.

Equating image and codomain. They coincide exactly when f is surjective, and the whole point of the distinction is that they usually do not.
Forgetting negative preimages. x² = 4 has two solutions, and dropping −2 is the most common slip.
Writing f⁻¹ for the preimage of a single element as though it were a number. f⁻¹(4) as a set is {−2, 2}.

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Domain, codomain, image & preimage — frequently asked questions

Can a preimage be empty?

Yes. f⁻¹[{−1}] = ∅ for f(x) = x², because nothing squares to a negative number. Empty preimages are exactly the elements of the codomain outside the image.

Does f⁻¹ exist for non-invertible f?

The preimage operator on SETS always exists. The inverse FUNCTION exists only when f is bijective, and the shared notation is unfortunate.

Why does the codomain matter?

Because surjectivity is defined relative to it. Restricting the codomain to the image makes any function surjective, which is why the codomain must be declared.

The rest of Functions

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

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