Let $P,Q$ be two posets. Recall that a function $f:P\to Q$ is order preserving iff $x\le y$ implies $f(x)\le f(y)$ . This is equivalent to saying that the preimage of any down set under $f$ is a down set.

Definition A function $f:P\to Q$ between ordered sets $P,Q$ is said to be residuated if the preimage of any principal down set is a principal down set. In other words, for any $y\in Q$ , there is some $x\in P$ such that $$f^{-1}(\down{y}) = \down{x},$$ where $f^{-1}(A)$ for any $A\subseteq Q$ is the set $\lbrace a\in X\mid f(a)\in A\rbrace$ , and $\down b = \lbrace c \mid c\le b\rbrace$ .

Let us make some observations on a residuated function $f:P\to Q$ :

1. $f$ is order preserving.
   Proof. Suppose $a\le b$ in $P$ . Then there is some $c\in P$ such that $\down c = f^{-1}(\down f(b))$ . Since $f(b) \in \down f(b)$ , $b \in \down c$ , or $b\le c$ , which means $a\le c$ , or $a\in \down c$ . But this implies $a\in f^{-1}(\down f(b))$ , so that $f(a) \in \down f(b)$ , or $f(a)\le f(b)$ .
2. The $x$ in the definition above is unique.
   Proof. If $\down x = \down z$ , then $x\in \down z$ , or $x\le z$ . Similarly, $z\le x$ , so that $x=z$ .
3. $g:Q\to P$ given by $g(y)=x$ is a well-defined function, with the following properties:
   1. $g$ is order preserving,
   2. $1_P \le g\circ f$ ,
   3. $f\circ g\le 1_Q$ .
   Proof. $g$ is order preserving: if $r\le s$ in $Q$ , then $\down r \subseteq \down s$ , so that $\down u = f^{-1}(\down r)\subseteq f^{-1}(\down s) = \down v$ , where $u=g(r)$ and $v=g(s)$ . But $\down u\subseteq \down v$ implies $u\le v$ , or $g(r)\le g(s)$ .

   $1_P \le g\circ f$ : let $x\in P$ , $y=f(x)$ , and $z=g(y)$ . Then $x\in f^{-1}(y) \subseteq f^{-1}(\down y) = \down z$ , which implies $x\le z=g(f(x))$ as desired.

   $f\circ g\le 1_Q$ : pick $y\in Q$ and let $x=g(y)$ . Then $f^{-1}(\down y)=\down x$ , so that $f(x)\in f(\down x) = f(f^{-1}( \down y))=\down y$ , or $f(x)\le y$ , or $f(g(y))=f(x) \le y$ as a result.

Actually, the third observation above characterizes $f$ being residuated:

Definition. Given a residuated function $f:P\to Q$ , the unique function $g:Q\to P$ defined above is called the residual of $f$ , and is denoted by $f^+$ .

For example, given any function $f: A\to B$ , the induced function $f: P(A)\to P(B)$ (by abuse of notation, we use the same notation as original function $f$ ), given by $f(S)=\lbrace f(a)\mid a\in S\rbrace$ is residuated. Its residual is the function $f^{-1}: P(B)\to P(A)$ , given by $f^{-1}(T)=\lbrace a \in A\mid f(a)\in T\rbrace$ .

Here are some properties of residuated functions and their residuals:

• A bijective residuated function is an order isomorphism, and conversely. Furthermore, the residual is residuated, and is its inverse.
• If $f: P\to Q$ is residuated, then $f\circ f^+ \circ f = f$ and $f^+ \circ f \circ f^+ = f^+$ .
• If $f: P\to Q$ and $g:Q\to R$ are residuated, so is $g\circ f$ and $(g\circ f)^+=f^+\circ g^+$ .
• If $f: P\to Q$ is residuated, then $f^+\circ f: P \to P$ is a closure map on $P$ . Conversely, any closure function can be decomposed as the functional composition of a residuated function and its residual.

Remark. Residuated functions and their residuals are closely related to Galois connections. If $f:P\to Q$ is residuated, then $(f,f^+)$ forms a Galois connection between $P$ and $Q$ . On the other hand, if $(f,g)$ is a Galois connection between $P$ and $Q$ , then $f: P\to Q$ is residuated, and $g: Q\to P$ is $f^+$ . Note that PM defines a Galois connection as a pair of order-preserving maps, where as in Blyth, they are order reversing.

Bibliography

1

T.S. Blyth, Lattices and Ordered Algebraic Structures, Springer, New York (2005).

2

G. Grätzer, General Lattice Theory, 2nd Edition, Birkhäuser (1998)