Fixed Point of Composition of Inflationary Mappings

Theorem
Let $\left({S, \preceq}\right)$ be a ordered set.

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

Let $x \in S$.

Then:
 * $x$ is a fixed point of $f \circ g$


 * $x$ is a fixed point of both $f$ and $g$.
 * $x$ is a fixed point of both $f$ and $g$.

Necessary Condition
Follows from Fixed Point of Mappings is Fixed Point of Composition.

Sufficient Condition
Let $h = f \circ g$.

Let $x$ be a fixed point of $h$.

Then by the definition of composition:


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

Since $f$ is inflationary:


 * $x \preceq g \left({x}\right)$

Suppose for the sake of contradiction that $x \ne g \left({x}\right)$.

Then $x \prec g \left({x}\right)$.

Since $f$ is also inflationary:


 * $g \left({x}\right) \preceq f \left({g \left({x}\right)}\right)$

Thus by Extended Transitivity:


 * $x \prec f \left({g \left({x}\right)}\right)$

But this contradicts the assumption that $x$ is a fixed point of $f \circ g$.

Therefore, $x = g \left({x}\right)$, and $x$ is a fixed point of $g$.

Suppose for the sake of contradiction that $f \left({x}\right) \ne x$.

Then $x \prec f \left({x}\right)$.

As we have shown that $x = g \left({x}\right)$, it follows that:


 * $x \prec f \left({g \left({x}\right)}\right)$

But this contradicts assumption that $x$ is a fixed point of $f \circ g$.

Hence, $x$ is also a fixed point of $f$.