Questions: Function Composition and Functional Structure

Question 1

Given f: A → B defined by f(a) = a + 1 and g: B → C defined by g(b) = b², what is (g ∘ f)(3)?

• A. 10 (computing g(3) = 9, then f(9) = 10)
• B. 16 (computing f(3) = 4, then g(4) = 16)
• C. 7 (adding f(3) + g(3) = 4 + 9)
• D. 12 (multiplying f(3) × g(3) = 4 × 3)

Answer: (B)

The notation g ∘ f means 'apply f first, then g.' So (g ∘ f)(3) = g(f(3)) = g(4) = 16. Option A reverses the order — applying g first, then f — which is a common mistake. The symbol g ∘ f is read 'g after f,' and the right-to-left application order follows mathematical convention where the function closest to the argument is applied first.

Question 2

Suppose g ∘ f is an injective (one-to-one) function. What can we conclude?

• A. Both f and g must be injective
• B. f must be injective, but g need not be
• C. g must be injective, but f need not be
• D. Neither f nor g is required to be injective

Answer: (B)

If g ∘ f is injective and f(a₁) = f(a₂), then g(f(a₁)) = g(f(a₂)), so (g ∘ f)(a₁) = (g ∘ f)(a₂), which forces a₁ = a₂. So f must be injective. But g need not be: g could be non-injective on its full codomain, as long as it is injective on the image of f. The dual fact is: if g ∘ f is surjective, then g must be surjective (though f need not be).

Question 3

The composition of two bijections is always a bijection.

• T. True
• F. False

Answer: True

If f: A → B and g: B → C are both bijections (injective and surjective), then g ∘ f is injective (composition of injections is injective) and surjective (composition of surjections is surjective), hence a bijection. This also means g ∘ f has a two-sided inverse, namely f⁻¹ ∘ g⁻¹: C → A. This fact underlies why bijections form a group under composition.

Question 4

Function composition is commutative: for any functions f and g with compatible domains, f ∘ g typically equals g ∘ f.

• T. True
• F. False

Answer: False

Composition is generally not commutative. For example, if f(x) = x + 1 and g(x) = x², then (f ∘ g)(x) = x² + 1 but (g ∘ f)(x) = (x + 1)² = x² + 2x + 1 — clearly different. Furthermore, f ∘ g and g ∘ f may not even both be defined if the domains and codomains don't align. Composition IS associative (h ∘ (g ∘ f) = (h ∘ g) ∘ f), but associativity and commutativity are different properties.

Question 5

Why does the notation g ∘ f indicate that f is applied first, and why does this right-to-left order matter?

Model answer: The notation comes from how functions are written in mathematics: to apply g after f to an element a, you write g(f(a)), where the innermost function (f) acts first. The symbol g ∘ f mirrors this: f is on the right, closest to the input, so it is applied first. The order matters because composition is not commutative — f ∘ g and g ∘ f generally produce different functions, and may not even both be defined if the codomain of one doesn't match the domain of the other.

This right-to-left convention can feel counterintuitive but is consistent throughout mathematics. The codomain of f must equal the domain of g for g ∘ f to be defined — this domain-matching requirement enforces the order and prevents ambiguity. In category theory, this convention is standardized and extended to arbitrary morphisms.