Fixed Point of Composition of Inflationary Mappings

Theorem
Let $(S, \preceq)$ be a ordered set.

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

Let $x \in S$.

Then $x$ is a fixed point of $f \circ g$ iff it is a fixed point of $f$ and of $g$.

Reverse Implication
Follows from Fixed Point of Mappings is Fixed Point of Composition.

Forward Implication
Let $h = f \circ g$.

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

Then by the definition of composition:


 * $f(g(x)) = x$

Since $f$ is inflationary:


 * $x \preceq g(x)$

Suppose for the sake of contradiction that $x \ne g(x)$.

Then $x \prec g(x)$.

Since $f$ is also inflationary:


 * $g(x) \preceq f(g(x))$

Thus by Extended Transitivity:


 * $x \prec f(g(x))$

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

Thus $x = g(x)$.

Suppose for the sake of contradiction that $f(x) \ne x$.

Then $x \prec f(x)$.

As we have shown that $x = g(x)$:


 * $x \prec f(g(x))$

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