Compositions of Closure Operators are both Closure Operators iff Operators Commute

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

Let $f$ and $g$ be closure operators on $S$.

Then the following are equivalent:


 * $(1)\quad f \circ g$ and $g \circ f$ are both closure operators.
 * $(2)\quad f$ and $g$ commute (that is, $f \circ g = g \circ f$).

Inflationary
Follows from Composition of Inflationary Mappings is Inflationary.

Increasing
Follows from Composition of Increasing Mappings is Increasing.

Idempotent
Follows from Composition of Commuting Idempotent Mappings is Idempotent.

Thus if $f$ and $g$ commute, $f \circ g$ and $g \circ f$ are closure operators.

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

Then $f \circ g$ is idempotent, so:
 * $(f \circ g) \circ (f \circ g) = f \circ g$

By Composition of Mappings is Associative and the definition of composition:
 * $f(g(f(g(x)))) = f(g(x))$

Because an $\preceq$ is reflexive:
 * $f(g(f(g(x)))) \preceq f(g(x))$

Thus since $f$ is a closure operator:
 * $g(f(g(x))) \preceq f(g(x))$

Since $g$ is inflationary and $\preceq$ is antisymmetric:
 * $g(f(g(x))) = f(g(x))$

Since this holds for all $x \in S$:
 * $g \circ f \circ g = f \circ g$

A similar argument shows that:
 * $f \circ g \circ f = g \circ f$

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