# Composition of Functions

## Definition and Properties

Similarly to relations, we can compose two or more functions to create a new function. This operation is called the composition of functions.

Let $$g: A \to B$$ and $$f: B \to C$$ be two functions such that the range of $$g$$ equals the domain of $$f.$$ The composition of the functions $$f$$ and $$g,$$ denoted by $$f \circ g,$$ is another function defined as

$y = \left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right).$

Since functions are a special case of relations, they inherit all properties of composition of relations and have some additional properties. We list here some of them:

1. The composition of functions is associative. If $$h: A \to B,$$ $$g: B \to C$$ and $$f: C \to D,$$ then $$\left( {f \circ g} \right) \circ h = f \circ \left( {g \circ h} \right).$$
2. The composition of functions is not commutative. If $$g: A \to B$$ and $$f: B \to C,$$ then, as a rule, $$f \circ g \ne g \circ f.$$
3. Let $$g: A \to B$$ and $$f: B \to C$$ be injective functions. Then the composition of the functions $$f \circ g$$ is also injective.
4. Let $$g: A \to B$$ and $$f: B \to C$$ be surjective functions. Then the composition of the functions $$f \circ g$$ is also surjective.
5. It follows from the last two properties that if two functions $$g$$ and $$f$$ are bijective, then their composition $$f \circ g$$ is also bijective.

## Examples

### Example 1. Composition of Functions Defined on Finite Sets

Consider the sets $$A = \left\{ {1,2,3,4} \right\},$$ $$B = \left\{ {a,b,c,d} \right\}$$ and $$C = \left\{ \alpha, \beta, \gamma, \delta \right\}.$$ The functions $$g: A \to B$$ and $$f:B \to C$$ are defined as

$g = \left\{ {\left( {1,b} \right),\left( {2,b} \right),\left( {3,a} \right),\left( {4,d} \right)} \right\},\;\;f = \left\{ {\left( {a,\gamma } \right),\left( {b,\alpha } \right),\left( {c,\delta } \right),\left( {d,\alpha } \right)} \right\}.$

It is convenient to illustrate the mapping between the sets in an arrow diagram:

Given the mapping, we see that

$\left( {f \circ g} \right)\left( 1 \right) = f\left( {g\left( 1 \right)} \right) = f\left( b \right) = \alpha ;$
$\left( {f \circ g} \right)\left( 2 \right) = f\left( {g\left( 2 \right)} \right) = f\left( b \right) = \alpha ;$
$\left( {f \circ g} \right)\left( 3 \right) = f\left( {g\left( 3 \right)} \right) = f\left( a \right) = \gamma ;$
$\left( {f \circ g} \right)\left( 4 \right) = f\left( {g\left( 4 \right)} \right) = f\left( d \right) = \alpha.$

Hence, the composition of functions $$f \circ g$$ is given by

$f \circ g = \left\{ {\left( {1,\alpha } \right),\left( {2,\alpha } \right),\left( {3,\gamma } \right),\left( {4,\alpha } \right)} \right\}.$

This is represented in the following diagram:

### Example 2. Composition of Functions Defined on Infinite Sets

Let $$g: \mathbb{R} \to \mathbb{R}$$ and $$f: \mathbb{R} \to \mathbb{R}$$ be two functions defined as

$g\left( x \right) = {x^2} + 3x + 1,\;\;f\left( x \right) = \cos x.$

Determine the composite functions $$f \circ g,$$ $$g \circ f,$$ $$f \circ f,$$ $$g \circ g.$$

The first composite function $$\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right)$$ is formed when the inner function $${g\left( x \right)}$$ is substituted for $$x$$ in the outer function $${f\left( x \right)}.$$ This yields:

$\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right) = \cos \left( {g\left( x \right)} \right) = \cos \left( {{x^2} + 3x + 1} \right).$

Similarly we find the other composite functions:

$\left(g \circ f\right)\left(x\right) = g\left( {f\left( x \right)} \right) = {f^2}\left( x \right) + 3f\left( x \right) + 1 = {\cos ^2}x + 3\cos x + 1.$
$\left(f \circ f\right)\left(x\right) = f\left( {f\left( x \right)} \right) = \cos \left( {f\left( x \right)} \right) = \cos \left( {\cos x} \right).$
$\left(g \circ g\right)\left(x\right) = g\left( {g\left( x \right)} \right) = {g^2}\left( x \right) + 3g\left( x \right) + 1 = {\left( {{x^2} + 3x + 1} \right)^2} + 3\left( {{x^2} + 3x + 1} \right) + 1 = \left( {\color{magenta}{x^4} + \color{red}{9{x^2}} + \color{black}{1} + \color{green}{6{x^3}} + \color{red}{2{x^2}} + \color{blue}{6x}} \right) + \left( {\color{red}{3{x^2}} + \color{blue}{9x} + \color{black}{3}} \right) + \color{black}{1} = \color{magenta}{x^4} + \color{red}{9{x^2}} + \color{black}{1} + \color{green}{6{x^3}} + \color{red}{2{x^2}} + \color{blue}{6x} + \color{red}{3{x^2}} + \color{blue}{9x} + \color{black}{3} + \color{black}{1} = \color{magenta}{x^4} + \color{green}{6{x^3}} + \color{red}{14{x^2}} + \color{blue}{15x} + \color{black}{5}.$

## Compositions Involving Inverse Functions

Let $$f: A \to B$$ be a bijective function from domain $$A$$ to codomain $$B.$$ Then it has an inverse function $${f^{-1}}$$ that maps $$B$$ back to $$A.$$ Then

${f^{ - 1}} \circ f = {I_A}\;\text{ or }\;{f^{ - 1}}\left( {f\left( x \right)} \right) = x,$

where $${I_A}$$ is the identity function in the domain $$A$$ and $$x$$ is any element of $$A.$$

Similarly,

$f \circ {f^{ - 1}} = {I_B}\;\text{ or }\;f\left( {{f^{ - 1}}\left( y \right)} \right) = y,$

where $${I_B}$$ is the identity function in the codomain $$B$$ and $$y$$ is any element of $$B.$$