Definition:Composition of Mappings

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Definition

Let $S_1$, $S_2$ and $S_3$ be sets.

Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that the domain of $f_2$ is the same set as the codomain of $f_1$.

Definition 1

The composite mapping $f_2 \circ f_1$ is defined as:

$\forall x \in S_1: \map {\paren {f_2 \circ f_1} } x := \map {f_2} {\map {f_1} x}$


Definition 2

The composite of $f_1$ and $f_2$ is defined and denoted as:

$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$


Definition 3

The composite of $f_1$ and $f_2$ is defined and denoted as:

$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z}$


CompositeMapping.png

General Definition

Let $f_1: S_1 \to S_2, f_2: S_2 \to S_3, \ldots, f_n: S_n \to S_{n + 1}$ be mappings such that the domain of $f_k$ is the same set as the codomain of $f_{k - 1}$.


Then the composite of $f_1, f_2, \ldots, f_n$ is defined and denoted as:

\(\displaystyle \forall x \in S_1: \map {\paren {f_n \circ \cdots \circ f_2 \circ f_1} } x\) \(:=\) \(\displaystyle \begin {cases} \map {f_1} x & : n = 1 \\ \map {f_n} {\map {\paren {f_{n - 1} \circ \cdots \circ f_2 \circ f_1} } x} : & n > 1 \end {cases}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {f_n} {\dotsm \map {f_2} {\map {f_1} x} \dotsm}\)


Commutative Diagram

The concept of composition of mappings can be illustrated by means of a commutative diagram.

This diagram illustrates the specific example of $f_2 \circ f_1$:

$\begin{xy}\[email protected]+1em{ S_1 \ar[r]^*+{f_1} \[email protected]{-->}[rd]_*[l]+{f_2 \mathop \circ f_1} & S_2 \ar[d]^*+{f_2} \\ & S_3 }\end{xy}$


Composition as a Binary Operation

Let $\sqbrk {S \to S}$ be the set of all mappings from a set $S$ to itself.

Then the concept of composite mapping defines a binary operation on $\sqbrk {S \to S}$:

$\forall f, g \in \sqbrk {S \to S}: g \circ f = \set {\tuple {s, t}: s \in S, \tuple {f \paren s, t} \in g} \in \sqbrk {S \to S}$


Thus, for every pair $\tuple {f, g}$ of mappings in $\sqbrk {S \to S}$, the composition $g \circ f$ is another element of $\sqbrk {S \to S}$.


Warning

Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that the domain of $f_2$ is the same set as the codomain of $f_1$.


If $\Dom {f_2} \ne \Cdm {f_1}$, then the composite mapping $f_2 \circ f_1$ is not defined.


This definition is directly analogous to that of composition of relations owing to the fact that a mapping is a special kind of relation.


Also known as

In the context of analysis, this is often found referred to as a function of a function, which (according to some sources) makes set theorists wince, as it is technically defined as a function on the codomain of a function.

Some sources call $f_2 \circ f_1$ the resultant of $f_1$ and $f_2$ or the product of $f_1$ and $f_2$.


Some authors write $f_2 \circ f_1$ as $f_2 f_1$.

Others, particularly in books having ties with computer science, write $f_1; f_2$ or $f_1 f_2$ (note the reversal of order), which is read as (apply) $f_1$, then $f_2$.


Examples

Compositions of $x^2$ with $2 x + 1$

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = x^2$

Let $g: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map g x = 2 x + 1$


Then the compositions of $f$ with $g$ are:

$f \circ g: \R \to \R$:

$\forall x \in \R: \map {\paren {f \circ g} } x = \paren {2 x + 1}^2$

$g \circ f: \R \to \R$:

$\forall x \in \R: \map {\paren {g \circ f} } x = 2 x^2 + 1$


Compositions of $\sin x$ with $2 x + 1$

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = \sin x$

Let $g: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map g x = 2 x + 1$


Then the compositions of $f$ with $g$ are:

$f \circ g: \R \to \R$:

$\forall x \in \R: \map {\paren {f \circ g} } x = \map \sin {2 x + 1}$

$g \circ f: \R \to \R$:

$\forall x \in \R: \map {\paren {g \circ f} } x = 2 \sin x + 1$


Arbitrary Finite Sets

Let:

\(\displaystyle A\) \(=\) \(\displaystyle \set {1, 2, 3}\)
\(\displaystyle B\) \(=\) \(\displaystyle \set {a, b}\)
\(\displaystyle C\) \(=\) \(\displaystyle \set {u, v, w}\)


Let $\theta: A \to B$ and $\phi: B \to C$ be defined in two-row notation as:

\(\displaystyle \theta\) \(=\) \(\displaystyle \binom {1 \ 2 \ 3} {a \ b \ a}\)
\(\displaystyle \phi\) \(=\) \(\displaystyle \binom {a \ b} {w \ v}\)


Then:

$\phi \circ \theta = \dbinom {1 \ 2 \ 3} {w \ v \ w}$


Arbitrary Finite Set with Itself

Let $X = Y = \set {a, b}$.


Consider the mappings from $X$ to $Y$:

\((1):\quad\) \(\displaystyle \map {f_1} a\) \(=\) \(\displaystyle a\)
\(\displaystyle \map {f_1} b\) \(=\) \(\displaystyle b\)


\((2):\quad\) \(\displaystyle \map {f_2} a\) \(=\) \(\displaystyle a\)
\(\displaystyle \map {f_2} b\) \(=\) \(\displaystyle a\)


\((3):\quad\) \(\displaystyle \map {f_3} a\) \(=\) \(\displaystyle b\)
\(\displaystyle \map {f_3} b\) \(=\) \(\displaystyle b\)


\((4):\quad\) \(\displaystyle \map {f_4} a\) \(=\) \(\displaystyle b\)
\(\displaystyle \map {f_4} b\) \(=\) \(\displaystyle a\)


The Cayley table illustrating the compositions of these $4$ mappings is as follows:

$\begin{array}{c|cccc} \circ & f_1 & f_2 & f_3 & f_4 \\ \hline f_1 & f_1 & f_2 & f_3 & f_4 \\ f_2 & f_2 & f_2 & f_2 & f_2 \\ f_3 & f_3 & f_3 & f_3 & f_3 \\ f_4 & f_4 & f_3 & f_2 & f_1 \\ \end{array}$

We have that $f_1$ is the identity mapping and is also the identity element in the algebraic structure under discussion


Also see

  • Results about composite mappings can be found here.


Sources