Fixed Point of Composition of Inflationary Mappings

Theorem
Let $\struct {S, \preceq}$ be an 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 definition of composition:


 * $\map f {\map g x} = x$

Since $f$ is inflationary:


 * $x \preceq \map g x$

$x \ne \map g x$.

Then $x \prec \map g x$.

Since $f$ is also inflationary:


 * $\map g x \preceq \map f {\map g x}$

Thus by Extended Transitivity:


 * $x \prec \map f {\map g x}$

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

Therefore, $x = \map g x$, and $x$ is a fixed point of $g$.

$\map f x \ne x$.

Then $x \prec \map f x$.

As we have shown that $x = \map g x$, it follows that:


 * $x \prec \map f {\map g x}$

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

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