Key takeaways
- (g ∘ f)(x) = g(f(x)): apply f first, then g, and the composition is only defined when f's outputs lie in g's domain.
- Composition is associative but not commutative: h ∘ (g ∘ f) = (h ∘ g) ∘ f, while g ∘ f and f ∘ g usually differ.
- A function has an inverse exactly when it is a bijection, and the inverse undoes it: f⁻¹(f(x)) = x and f(f⁻¹(y)) = y.
- To find an inverse, solve y = f(x) for x, then check both compositions give the identity.
- The inverse of a composition reverses the order: (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹.
Functions become much more useful once you can chain them together and, sometimes, run them backwards. Composition is the chaining; the inverse is the running backwards. Both are short definitions with a few traps, mostly to do with order.
Composition: one function after another
Given f: A → B and g: B → C, the composition g ∘ f: A → C is defined by
(g ∘ f)(x) = g(f(x)).
Read it "g after f" or "g of f of x". The function written on the right acts first. Applied Discrete Structures defines it with exactly these domains and codomains: the outputs of f must land in the set where g is defined, or the composition does not exist.
Example on a finite set. Let A = {1, 2, 3}, with f(1) = 2, f(2) = 3, f(3) = 3 and g(1) = 1, g(2) = 1, g(3) = 2.
| x | f(x) | g(f(x)) | g(x) | f(g(x)) |
|---|---|---|---|---|
| 1 | 2 | 1 | 1 | 2 |
| 2 | 3 | 2 | 1 | 2 |
| 3 | 3 | 2 | 2 | 3 |
Compare the third and fifth columns: g ∘ f and f ∘ g are different functions.
Example on the integers. With f(n) = n + 1 and g(n) = 2n:
- (g ∘ f)(n) = g(n + 1) = 2n + 2
- (f ∘ g)(n) = f(2n) = 2n + 1
Composition is not commutative. Order matters, and swapping it changes the answer.
What composition does preserve
Composition is associative: for any f, g and h with compatible domains, h ∘ (g ∘ f) = (h ∘ g) ∘ f, because both sides send x to h(g(f(x))). So a chain like h ∘ g ∘ f needs no brackets.
Applied Discrete Structures states two more facts, leaving their proofs as exercises; both are worth being able to reproduce:
- The composition of injections is an injection. If g(f(a)) = g(f(b)), then f(a) = f(b) because g is one-to-one, and a = b because f is one-to-one.
- The composition of surjections is a surjection. For any c ∈ C, pick b with g(b) = c, then a with f(a) = b; then g(f(a)) = c.
Together these show the composition of two bijections is a bijection. The definitions behind them are in injective, surjective and bijective functions.
The identity function
For any set A, the identity function i_A(x) = x leaves everything where it is. It plays the role in composition that 1 plays in multiplication: f ∘ i_A = f and i_B ∘ f = f for any f: A → B.
Inverses: running a function backwards
A function g: B → A is the inverse of f: A → B if
g(f(x)) = x for every x ∈ A, and f(g(y)) = y for every y ∈ B.
In other words, g ∘ f = i_A and f ∘ g = i_B. The inverse is written f⁻¹. Note that f⁻¹ means the inverse function, not 1/f.
When does an inverse exist? Exactly when f is a bijection.
- If f were not injective, two inputs a ≠ b would share an output y, and f⁻¹(y) would have to equal both.
- If f were not surjective, some y would have no input, and f⁻¹(y) would have nothing to be.
MathWorld defines a bijection as a transformation that is both one-to-one and onto, which is precisely the condition for "undo" to be a function.
Finding an inverse
On a finite set, reverse every arrow. Applied Discrete Structures uses this example: on A = {1, 2, 3}, f(1) = 2, f(2) = 3, f(3) = 1. Reversing the arrows gives f⁻¹(2) = 1, f⁻¹(3) = 2, f⁻¹(1) = 3.
With a formula, write y = f(x) and solve for x.
Example. f: ℝ → ℝ, f(x) = 5x − 3. From y = 5x − 3, x = (y + 3)/5, so f⁻¹(y) = (y + 3)/5. Check both directions:
- f⁻¹(f(x)) = ((5x − 3) + 3)/5 = x
- f(f⁻¹(y)) = 5 · (y + 3)/5 − 3 = y
The check matters because solving for x can hide a domain problem. The same formula on the integers, f: ℤ → ℤ, is injective but not surjective (y = 0 would need x = 3/5), so it has no inverse there.
Restricting the domain. g(x) = x² is not a bijection on ℝ, but restricted to [0, ∞) → [0, ∞) it is, with inverse √y. Likewise eˣ: ℝ → (0, ∞) has inverse ln y.
Inverses in modular arithmetic. On the remainders {0, 1, …, 6} under multiplication mod 7, the function x ↦ 3x is a bijection, and its inverse is x ↦ 5x, because 3 · 5 = 15 ≡ 1. Finding such inverses systematically is the subject of modular inverses.
The inverse of a composition
If f and g are both bijections, then
(g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹.
To undo "f, then g", undo g first, then undo f. The familiar analogy is getting dressed: socks then shoes, so to undress, shoes then socks. A one-line check: (f⁻¹ ∘ g⁻¹) ∘ (g ∘ f) = f⁻¹ ∘ (g⁻¹ ∘ g) ∘ f = f⁻¹ ∘ f = i, using associativity.
Four errors with composition and inverses
- Applying the left function first. In g ∘ f, f acts first.
- Assuming every function has an inverse. Check it is a bijection between the stated domain and codomain.
- Checking only one composition. On infinite sets especially, g ∘ f = i alone does not make g the inverse; you need f ∘ g = i as well.
- Writing (g ∘ f)⁻¹ = g⁻¹ ∘ f⁻¹. The order reverses.
Levin's Discrete Mathematics: An Open Introduction is a good companion for the underlying definitions of domain, codomain and range. Functions belong to the sets, functions and countability unit of Discrete Math: Study & Quiz, whose lessons include knowledge checks; it is listed with our education apps, and the discrete math hub links the neighbouring topics.
Good to know
Frequently asked questions
What does g ∘ f mean?
Is function composition commutative?
When does a function have an inverse?
What is the inverse of g ∘ f?
Sources
- Function Composition (Applied Discrete Structures) (opens in a new tab)discretemath.org — accessed
- Functions (Discrete Mathematics: An Open Introduction, 3rd edition, Oscar Levin) (opens in a new tab)Open Math Books — accessed
- Bijection (opens in a new tab)Wolfram MathWorld — accessed
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