Composition of Inflationary and Idempotent Mappings

Theorem
Let $(S, \preceq)$ 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, $f \circ g = g \circ f$).
 * $(3)\quad \operatorname{img}(f \circ g) = \operatorname{img}(g \circ f)$

where $\circ$ represents composition and $\operatorname{img}$ 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 $(f \circ g) \circ (f \circ g) = f \circ g$

By Composition of Mappings is Associative and the definition of composition:


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

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


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

Since $f$ is inflationary:


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

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


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

Since $g$ is inflationary:


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

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


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

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:


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

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.