Functions · step 6 of 13

Images & preimages of unions and intersections

Preimages are beautifully behaved: f⁻¹[S ∪ T] = f⁻¹[S] ∪ f⁻¹[T], f⁻¹[S ∩ T] = f⁻¹[S] ∩ f⁻¹[T], and f⁻¹[Sᶜ] = (f⁻¹[S])ᶜ. Images are not: f[X ∪ Y] = f[X] ∪ f[Y] holds, but for intersections only f[X ∩ Y] ⊆ f[X] ∩ f[Y], with equality exactly when f is injective. Knowing which inclusion can be strict is the examinable content.

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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.

Prove preimage identities by unfoldinga ∈ f⁻¹[S ∪ T] iff f(a) ∈ S ∪ T iff f(a) ∈ S or f(a) ∈ T. The logic is immediate.
Prove the image inclusionIf b ∈ f[X ∩ Y] then b = f(a) for some a in both X and Y, so b lies in both images.
Build the counterexample for the reverseChoose f non-injective and X, Y disjoint sets whose images overlap. Then f[X ∩ Y] = ∅ but f[X] ∩ f[Y] ≠ ∅.
State the injectivity conditionEquality holds for all X, Y exactly when f is injective — a standard iff to prove.

Worked example

Show f[X ∩ Y] ⊊ f[X] ∩ f[Y] can happen, using f(x) = x².

Take X = {2} and Y = {−2}.
X ∩ Y = ∅, so f[X ∩ Y] = ∅.
f[X] = {4} and f[Y] = {4}.
So f[X] ∩ f[Y] = {4} ≠ ∅.

Answer. The inclusion is strict here: ∅ ⊊ {4}. Equality would require f to be injective, and x² is not.

Where marks get dropped

These are the specific errors that cost credit on images & preimages of unions and intersections questions — QED's rubric penalises each of them separately.

Assuming images behave like preimages. The intersection identity fails for images, and complements fail badly too.
Believing f[Xᶜ] = f[X]ᶜ. This is false in general even for injective f, since the image may not be all of the codomain.
Forgetting the direction of the inclusion. It is f[X ∩ Y] ⊆ f[X] ∩ f[Y], never the reverse.

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Images & preimages of unions and intersections — frequently asked questions

Why are preimages better behaved?

Because f⁻¹[S] is defined by a condition on f(a), and logical connectives commute with membership tests. Images require existential quantification over inputs, which does not commute with ∧.

When does f[X ∩ Y] = f[X] ∩ f[Y] hold for all X, Y?

Exactly when f is injective. Injectivity guarantees the two preimages of a common output coincide, so the intersections match.

Does f⁻¹[f[X]] = X?

Only ⊇ holds in general; equality needs injectivity. Dually f[f⁻¹[S]] ⊆ S with equality when f is surjective.

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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