Composition of Inflationary and Idempotent Mappings

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

Let $f$ and $g$ be inflationary and idempotent mappings on $S$.

Then the following are equivalent:


 * $(1): \quad f \circ g$ and $g \circ f$ are both idempotent
 * $(2): \quad f$ and $g$ commute (that is, $f \circ g = g \circ f$)
 * $(3): \quad \operatorname{Im} \left({f \circ g}\right) = \operatorname{Im} \left({g \circ f}\right)$

where:
 * $\circ$ represents composition
 * $\operatorname{Im}$ represents the image of a mapping.

$(2)$ implies $(1)$
Follows from Composition of Commuting Idempotent Mappings is Idempotent.

$(1)$ implies $(2)$
Suppose that $f \circ g$ and $g \circ f$ are idempotent.

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

By Composition of Mappings is Associative and the definition of composition, we have for each $x \in S$:


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

Because $\preceq$ is an ordering and hence reflexive:


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

Since $f$ is inflationary:


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

Thus since $\preceq$ is an ordering and hence transitive:


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

Since $g$ is inflationary:


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

Thus since $\preceq$ is an ordering and hence antisymmetric:


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

Since this holds for all $x \in S$, Equality of Mappings shows that:


 * $g \circ f \circ g = f \circ g$

The same argument, with the roles of $f$ and $g$ reversed, shows that:


 * $f \circ g \circ f = g \circ f$

Combining everything, we obtain:


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

Thus $f \circ g = g \circ f$, so $f$ and $g$ commute.

$(2)$ implies $(3)$
Equals substitute for equals.

$(3)$ implies $(1)$
Follows from Composition of Idempotent Mappings.