Functions · step 3 of 13

Composition g∘f and its properties

(g∘f)(x) = g(f(x)) — f runs first, despite g being written first. Composition is associative but not commutative, and it interacts predictably with the other properties: if f and g are both injective so is g∘f, and likewise for surjective. The converses are one-sided and are the standard exam trap: g∘f injective forces f injective, but says nothing about g.

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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.

Check the types matchg∘f needs the codomain of f to be contained in the domain of g. Otherwise the composition is undefined.
Substitute inner into outerWrite g(f(x)) and simplify. Doing it the other way computes f∘g, a different function.
Transfer properties forwardsInjective ∘ injective = injective; surjective ∘ surjective = surjective. Prove by chaining the definitions.
Transfer backwards carefullyIf g∘f is injective then f is injective. If g∘f is surjective then g is surjective. The other halves can fail.

Worked example

For f(x) = x + 1 and g(x) = x², compute g∘f and f∘g, and evaluate both at x = 2.

(g∘f)(x) = g(f(x)) = g(x + 1) = (x + 1)².
(f∘g)(x) = f(g(x)) = f(x²) = x² + 1.
At x = 2: (g∘f)(2) = 3² = 9.
(f∘g)(2) = 4 + 1 = 5.

Answer. g∘f = (x+1)² and f∘g = x² + 1 — different functions, giving 9 and 5 at x = 2.

Where marks get dropped

These are the specific errors that cost credit on composition g∘f and its properties questions — QED's rubric penalises each of them separately.

Composing in the reading order. g∘f applies f first, and reversing this changes the answer in almost every case.
Assuming g∘f injective makes g injective. Take f : {1} → {1,2} and g : {1,2} → {1} — the composite is injective while g is not.
Ignoring domain compatibility. Composing √x after x − 5 needs x ≥ 5, so the domain shrinks.

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Composition g∘f and its properties — frequently asked questions

Is composition associative?

Yes: h∘(g∘f) = (h∘g)∘f, since both send x to h(g(f(x))). This is why iterated compositions need no brackets.

When is g∘f = f∘g?

Only in special cases — for instance when both are powers of the same function, or when one is the identity. Commutativity is the exception.

What does g∘f = id tell me?

That f is injective and g is surjective. Both directions (g∘f = id and f∘g = id) together mean f and g are mutually inverse bijections.

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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