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:

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

$\blacksquare$


Examples

Sum of Squares not Square of Sum

Let $f: \R \times \R \to \R$ be the real-valued function defined as:

$\forall \tuple {x, y} \in \R \times \R: \map f {x, y} = x^2 + y^2$

Let $g: \R \times \R \to \R$ be the real-valued function defined as:

$\forall \tuple {x, y} \in \R \times \R: \map g {x, y} = x + y$

Let $h: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map g x = x^2$


Then we have that:

$\map h {\map g {x, y} } = \paren {x + y}^2$

while:

$\map g {\map h x, \map h y} = x^2 + y^2 = \map f {x, y}$


Hence the diagram:

$\begin{xy} \[email protected]+2mu@+1em{ \R \times \R \ar[r]^*{g} \ar@{-->}[rd]_*{f} & \R \ar[d]^*{h} \\ & \R }\end{xy}$

is not a commutative diagram.


Sources

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