Composition of Mappings/Examples/Arbitrary Finite Set with Itself

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Example of Compositions of Mappings

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

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

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

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

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

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

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