Fixed Point of Mappings is Fixed Point of Composition

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Theorem

Let $S$ be a set.

Let $f, g: S \to S$ be mappings.

Let $x \in S$ be a fixed point of both $f$ and $g$.


Then $x$ is also a fixed point of $f \circ g$, the composition of $f$ and $g$.


General Result

Let $S$ be a set.

Let $n \in \N$ be a strictly positive integer.

Let $\N_n$ be the initial segment of $n$ in $\N$.

That is, let $\N_n = \left\{{0, 1, \dots, n-1}\right\}$.

For each $i \in \N_n$, let $f_i: S \to S$ be a mapping.

Let $x \in S$ be a fixed point of $f_i$ for each $i \in \N_n$.

Let $g = f_0 \circ f_1 \circ \dots \circ f_{n-1}$ be the composition of all the $f_i$s.


Then $x$ is a fixed point of $g$.


Proof

Since $x$ is a fixed point of $g$:

$g \left({x}\right) = x$

Thus:

$f \left({g \left({x}\right)}\right) = f \left({x}\right)$

Since $x$ is a fixed point of $f$:

$f \left({x}\right) = x$

It follows that:

$\left({f \circ g}\right) \left({x}\right) = f \left({g \left({x}\right)}\right) = x$

Thus $x$ is a fixed point of $f \circ g$.

$\blacksquare$