Order of Two Operators Let's start by looking at the following example from ring theory:
Let $A$ and $B$ be commutative rings with identity and let $f:A\to B$ be a ring homomorphism. For every ideal $I$ of $A$, the extension $I^\mathrm e$ is defined to be the ideal of B which is generated by $f(I)$. For every ideal $J$ of $B$, the contraction $J^\mathrm c$ is defined to be the ideal $f^{-1}(J)$ of A. Then the following are true:


*

*$I\subseteq I^\mathrm {ec}$.

*$J\supseteq J^\mathrm {ce}$.

*$(I_1\cap I_2)^\mathrm e\subseteq I_1^\mathrm e\cap I_2^\mathrm e$.

*$(J_1+J_2)^\mathrm c\supseteq J_1^\mathrm c+J_2^\mathrm c$.


It seems that whenever we have two operations where one operation makes the input "larger" (such as extension) and the other one makes the input "smaller" (such as contraction), then for an input, if one applies the "large" operator and then the "smaller" operator, the result will generally be "larger" than the input, or the output when one applies the "smaller" operator and then the "larger" operator,.. On the other hand, if one applies the "smaller" operator and then the "larger" operator, the result will generally be "smaller" than the original input, or the output when one applies the "larger" operator and then the "smaller" operator,.
For example, if we start with an ideal $I$, taking the extension and then the contraction of the extension, the result will contain $I$ as a subset. Similarly, equation 3 above can be interpreted as "taking intersection(which makes the input smaller) then the extension (which makes then input larger), the result will be smaller than the result obtained from taking the extension first, then the intersection".
Similar ideas are found in other branches of maths. For example, let $A$ be a subset in a topological space, then $A\subseteq \text{the interior of the closure of }A$ and $A\supseteq \text{the closure of the interior of }A$.
My Question:
Is there any general formulated rule about the observation above? Very informally, I am thinking of something like $$A\subseteq \text{smaller operator }(\text{larger operator}(A)).$$
 A: As for (3) and (4), they follow from the general result true for any increasing (order-preserving) function $f$, namely $f(A \wedge B) \leq f(A) \wedge f(B)$ and
$f(A) \vee f(B) \leq f(A \vee B)$ (assuming the relevant meets and joins are defined). The first inequality becomes an equality when $f$ is a right adjoint that has a left adjoint. Similarly, the second inequality becomes an equality when $f$ is a left adjoint that has a right adjoint. What do I mean by adjoint? When for all $A, B$, $lA \leq B$ if and only if $A \leq rB$, we say that there is a Galois correspondence between the left adjoint $l$ and the right adjoint $r$ (where $l$ and $r$ are functions between partially ordered sets). The standard example is $fA \leq B$ if and only if $A \leq f^{-1}B$. By reversing orders in one of the posets, one can also define adjoints which have adjoints that have the same handed-ness (all of which are order-reversing), which yields very analogous results that are hard to keep straight, though.
As for (1) and (2), those are from a general result about Galois correspondences and from there being a Galois correspondence. If $r$ has a left adjoint $l$, one always has $lrA \leq A$ and $A \leq rlA$. Actually, you only need the correct half of a Galois condition to show either. E.g., $lA \leq B \rightarrow A \leq rB$ holding generally is sufficient to show that $A \leq rlA$ holds generally. And there's a somewhat converse result. If $r$ is increasing and $A \leq rlA$ holds generally, then $lA \leq B \rightarrow A \leq rB$ holds generally (the latter I call a $l$-Galois half-$r$ condition, because I'm weird enough to often like to name and consider separately the two directions of the equivalence that defines Galois correspondence).
