Composition of Mappings is not Commutative

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Theorem

The composition of mappings is not in general a commutative binary operation:

$f_2 \circ f_1 \ne f_1 \circ f_2$


Proof

Proof by Counterexample

Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings.

First note that unless $S_1 = S_3$ then $f_2 \circ f_1$ is not even defined.

So in that case $f_2 \circ f_1$ is definitely not the same thing as $f_1 \circ f_2$.


So, let us suppose $S_1 = S_3$ and so we define $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_1$.

If $S_1 \ne S_2$ then:

$f_2 \circ f_1: S_1 \to S_1$
$f_1 \circ f_2: S_2 \to S_2$

and so by Equality of Mappings they are unequal because their domains and codomains are different.


Finally, suppose $S_1 = S_2$, and consider the following.

$S_1 = S_2 = \set {a, b}$
$f_1 = \set {\tuple {a, a}, \tuple {b, a} }$
$f_2 = \set {\tuple {a, b}, \tuple {b, b} }$

It is straightforward to check that $f_1$ and $f_2$ are mappings, and that:

$f_1 \circ f_2 = \set {\tuple {a, b}, \tuple {b, b} }$
$f_2 \circ f_1 = \set {\tuple {a, a}, \tuple {b, a} }$

Thus, even in this limitingly simple case, we see that:

$f_2 \circ f_1 \ne f_1 \circ f_2$

$\blacksquare$


Sources