Constant Mapping is Non-Commutative

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 $f \circ g \ne g \circ f$.

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:

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