Composition of Mappings/Examples

Examples of Compositions of Mappings

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:

\(\ds A\)

\(=\)

\(\ds \set {1, 2, 3}\)

\(\ds B\)

\(=\)

\(\ds \set {a, b}\)

\(\ds C\)

\(=\)

\(\ds \set {u, v, w}\)

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

\(\ds \theta\)

\(=\)

\(\ds \binom {1 \ 2 \ 3} {a \ b \ a}\)

\(\ds \phi\)

\(=\)

\(\ds \binom {a \ b} {w \ v}\)

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 $Y$:

\(\text {(1)}: \quad\)

\(\ds \map {f_1} a\)

\(=\)

\(\ds a\)

\(\ds \map {f_1} b\)

\(=\)

\(\ds b\)

\(\text {(2)}: \quad\)

\(\ds \map {f_2} a\)

\(=\)

\(\ds a\)

\(\ds \map {f_2} b\)

\(=\)

\(\ds a\)

\(\text {(3)}: \quad\)

\(\ds \map {f_3} a\)

\(=\)

\(\ds b\)

\(\ds \map {f_3} b\)

\(=\)

\(\ds b\)

\(\text {(4)}: \quad\)

\(\ds \map {f_4} a\)

\(=\)

\(\ds b\)

\(\ds \map {f_4} b\)

\(=\)

\(\ds a\)

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.