# Constant Mapping is Non-Commutative

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## Theorem

Let $S$ be a set whose cardinality is greater than one.

Let $f: S \to S$ and $g: S \to S$ be constant mappings on $S$.

Then:

- $f \circ g \ne g \circ f$

where $\circ$ denotes composition of mappings.

## Proof

First note that if $S$ is a singleton, then there exists only one constant mapping on $S$.

In such a circumstance, $f = g$ and so $f \circ g \ne g \circ f$.

$\Box$

So, let $\card S > 1$.

Then there exist at least $2$ distinct elements $a$ and $b$ of $S$.

Thus, let $f$ and $g$ be defined as:

- $\forall x \in S: \map f x = a$
- $\forall x \in S: \map g x = b$

for some $a, b \in S$ such that $a \ne b$.

We have that:

\(\ds \map {\paren {f \circ g} } x\) | \(=\) | \(\ds \map f {\map g x}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map f b\) | Definition of $\map g x$ | |||||||||||

\(\ds \) | \(=\) | \(\ds a\) | Definition of $\map f x$ | |||||||||||

\(\ds \map {\paren {g \circ f} } x\) | \(=\) | \(\ds \map g {\map f x}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map g a\) | Definition of $\map f x$ | |||||||||||

\(\ds \) | \(=\) | \(\ds b\) | Definition of $\map g x$ |

So $f \circ g \ne g \circ f$ as required.

$\blacksquare$

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $8$