Introductory notes on real-enriched categories

Dexue Zhang School of Mathematics, Sichuan University, Chengdu, China<EMAIL_ADDRESS>
The author gratefully acknowledges the support of the National Natural Science Foundation of China (No. 12371463).

[2020]18B35, 18C15, 18C20, 18D15, 18D20

Real-enriched categories are   categories with real numbers as enrichment. Precisely, a real-enriched category is a category enriched over the commutative and unital quantale composed of the unit interval and a continuous t-norm. These notes present a brief introduction to real-enriched categories, focusing on the presheaf monad  and its submonads in the category of such categories.

§ INTRODUCTION

Real-enriched categories are a very special kind of enriched categories, namely, those categories of which the enrichment are real numbers. Precisely, a real-enriched category is a category enriched over the unital quantale composed of the the interval  $[0,1]$ and  a continuous t-norm. To require   enrichment be real numbers limits drastically  the scope of the theory, but the benefit is also apparent.
First, real numbers posses much richer structures than a general quantale or more generally, a symmetric and monoidal closed category,   one expects  more can be said about such categories because of the richer structures of the enrichment. Second, the study of such categories is closely related to many-valued logic. In 1973 Lawvere [78] pointed out a deep connection between enriched categories and generalized pure logic.  For real-enriched categories, this “generalized pure logic” is the Basic Logic, called the BL-logic, developed by  Hájek [40]. As shown in Hájek [40] and  Cignoli,   Esteva,   Godo and Torrens [20],   BL-logic is a logic of continuous t-norms. In particular, real-enriched categories may be viewed as many-valued ordered sets, a viewpoint emphasized in these notes.

Though being a narrow class, real-enriched categories are sufficiently general to provide a common framework for ordered sets and quasi-metric spaces,  they are natural objects for Quantitative Domain Theory, see  e.g.  America and Rutten [5], Antoniuk and Waszkiewicz [6], Bonsangue,   van Breugel and Rutten [17], Flagg [27], Flagg and Kopperman [28], Gutierres and Hofmann [38], Hofmann and Waskiewicz [52, 53], Kostanek and Waszkiewicz [62], Künzi and Schellekens [63], and etc.

There are many texts on enriched categories, e.g. Eilenberg and Kelly [24],  Lawvere [78], the book of Kelly [58], Chapter 6 in the Handbook of Borceux [19], and the  works of Stubbe [105, 106, 107] on quantaloid-enriched categories. The aim of these notes is to explain basic ideas  about real-enriched categories,  focusing on the presheaf monad and its submonads. Special attention is paid to  how the logic structure of the truth-value set affects mathematical results, see e.g. Proposition <ref>, Theorem <ref>, Theorem <ref>. Many topics are not covered here.   MacNeille completion of real-enriched categories, Formal Concept Analysis  of fuzzy contexts  (see Bělohlávek [10]) are not touched;  topology of real-enriched categories, except the notion of open ball topology, is not touched either, interested readers are referred to Goubault-Larrecq [36], Hofmann [44],   Lowen [85, 86], and the book [50] edited by Hofmann, Seal and Tholen.

These notes are arranged into four parts. The first part consists of Section <ref> and Section <ref>. This part recalls, for convenience of the reader,  some basic ideas in theories of ordered sets and continuous t-norms. Many notions for real-enriched categories  find their roots in the theory of ordered sets; the interval $[0,1]$ together with a continuous t-norm is our truth-value set. The second part consists of Section <ref> – Section <ref>. This part deals with basic structures of real-enriched categories, including   (adjoint) functors, (adjoint) distributors, Yoneda lemma, weighted limits and weighted colimits.
The third part consists of Section <ref> – Section <ref>, focusing on special kinds of colimits and limits.
The last 5 sections constitute the fourth part,  dealing with the presheaf monad in the category of real-enriched categories,  its submonads and algebras.

§ GALOIS CONNECTION AND DOMAINS

This   section recalls some basic ideas in order theory, and fixes some notations. The materials can be  found in many places, e.g. Davey and Priestley [22], Gierz,   Hofmann,   Keimel et al [31], and  Goubault-Larrecq [36].

Let $X$ be a set. An order (or a preorder) on $X$ is a binary relation $\leq$ on $X$ such that, for all $x, y, z \in X$,

* $x \leq x$,

* $x\leq y$ and $y \leq z$ imply $x \leq z$.

These conditions are referred to, respectively, as reflexivity and transitivity. A set $X$ equipped with an order relation $\leq$ is called an ordered set (or a preordered set).
Usually we  say simply “$X$ is an ordered set”. When it is necessary to specify the order relation we write $(X,\leq)$. If $x\leq y$ and $y\leq x$, we say that $x$ and $y$ are equivalent and write $x\cong y$.

A partial order on a set $X$ is an order that satisfies additionally the axiom of antisymmetry:

(iii) $x \cong y$ implies $x = y$.

* The identity relation  on a set $X$ is a partial order, called the discrete order on $X$.

* For each set $X$, the inclusion relation $\subseteq$ is a partial order on the powerset $2^X\coloneqq\{A\mid A\subseteq X\}$ of $X$.

* The smaller than or equal to relation $\leq$ is a partial order on the set $\mathbb{R}$ of real numbers. But, the strict smaller than relation $<$ on $\mathbb{R}$ is not an order since it is not reflexive.

* The divisibility relation $\sqsubseteq$,  defined by  $m\sqsubseteq n$ if $m=kn$ for some natural number $k$, is a partial order on the set $\mathbb{N}$ of natural numbers.

If $\leq$ is an order on a set $X$, then the opposite relation $\leq^{\rm op}$, given by $x\leq^{\rm op}y$ if $y\leq x$, is also an order on $X$, called the opposite of $\leq$.

Suppose $X$ is an ordered set, $A$ is a subset and $x$ is an element of $X$.
We say that $x$  is  an upper bound of   subset $A$ if $a\leq x$ for all $a\in A$. We say that $x$ is a join, or  a supremum, of $A$ if $x$ is the least upper bound of $A$; that means, $x$ is an upper bound of $A$ and $x\leq y$ whenever  $y$ is an upper bound of $A$. It is clear that any two joins of a subset $A$ are equivalent, hence every subset of an ordered set has  at most one join, up to isomorphism. So, we will speak of the join of a subset.  Dually, $x$ is a lower bound of $A$ if $x\leq a$ for all $a\in A$. The  greatest lower bound of $A$, if exists, is called the meet, or the infimum, of $A$.

Suppose $X$ is an ordered set and $A\subseteq X$. We say that $A$ is

* a lower set if $x\in A$ and $y\leq x$ imply $y\in A$;
* an upper set if $x\in A$ and $x\leq y$ imply $y\in A$.
For a subset $A$ of an ordered set $X$, let
\[\ua A=\{x\in X\mid \exists\thinspace a\in A, a\leq x\}\]
\[\da A=\{x\in X\mid \exists\thinspace a\in A, x\leq a\}.\]
Then $\ua A$ and $\da A$ are, respectively, the smallest upper set and the smallest lower set containing $A$. So,   $\da A$ is the lower set and $\ua A$ the upper set generated by $A$, respectively. When $A$ is a singleton set $\{a\}$, we write $\ua a$ and $\da a$. A lower set of the form $\da a$ is called a principal lower set. Likewise, an upper set of the form $\ua a$ is called a principal upper set.

For each topological space $(X,\tau)$, define a relation  $\leq_\tau$ on the set $X$ by
\[x\leq_\tau y\iff x\in\overline{\{y\}}.\]
Then $\leq_\tau$ is an order on $X$, called the specialization order of the space $(X,\tau)$. The specialization  order $\leq_\tau$ is antisymmetric if and only if $(X,\tau)$ is   $T_0$; $\leq_\tau$ is the discrete order  if and only if $(X,\tau)$ is   $T_1$. Every open set is an upper set  and every closed set is a lower set with respect to the specialization order. For each $x\in X$, the principal lower set $\da x$ is the closure of $\{x\}$, i.e. $\da x=\overline{\{x\}}$.

Suppose $X$ is an ordered set and $A\subseteq X$. Let
\[\ub A=\{x\in X\mid \forall a\in A, a\leq x\}\]
\[\lb A=\{x\in X\mid \forall a\in A, x\leq a\}.\]
In other words, $\ub A $ and $\lb A $ are the set of all upper bounds and the set of all lower bounds of $A$, respectively. Then, $\ub A$ is an upper set and $\lb A$ is a lower set;    $A$ has a join  if and  only if $\ub A$ is a principal upper set;   $A$ has a meet if and  only if $\lb A$ is a principal lower set.

Suppose $X$ is an ordered set.

* A subset $D$  of  $X$ is   directed if $D$ is not empty and for all $x,y\in D$, there is some $z\in D$ such that $x\leq z$ and $y\leq z$. A  directed lower set of $X$ is called an ideal of $X$.
* A subset $F$ of $X$ is  filtered if $F$ is not empty and for all $x,y\in F$, there is some $z\in F$ such that $z\leq x$ and $z\leq y$. A filtered  upper set of $X$ is called a filter of $X$.

In particular, an ordered set $X$ is directed if it is a directed subset of itself. This means that $X$ is not empty and every pair of elements in $X$ has an upper bound.

Suppose $X$ is an ordered set.

* For each $x\in X$, the principal lower set $\da x$ is   an ideal.
* For each $x\in X$, the principal upper set $\ua x$ is   a filter.
* Let $X$ be a topological space. Then for each $x\in X$, the family $\CN_x=\{U\subseteq X\mid x\in U^\circ\}$ of all neighborhoods of $x$ is a filter of  $(2^X,\subseteq)$.
* For each set $X$, the family of finite subsets of $X$ is an ideal of $(2^X,\subseteq)$.
Suppose $X$ is an ordered set. We say that $X$ is

* bounded if it has a least and a greatest element.

* a join semilattice if it is antisymmetric and every pair of elements $x,y$ in $X$ has a join, denoted by $x\vee y$.
* a meet semilattice if it is antisymmetric and every pair of elements $x,y$ in $X$ has a meet, denoted by $x\wedge y$.
* a  lattice if it is at the same time a join semilattice and a meet semilattice.

If $X$ is a lattice, then so is the opposite $X^{\rm op}$ of $X$. The ordered set $(\mathbb{R},\leq)$ is a lattice, but not bounded.

For each ordered set $X$, the following   are equivalent:

* Every subset of $X$, including the empty one, has a join.
* Every subset of $X$, including the empty one,  has a meet.

$(1)\Rightarrow(2)$ We show that each subset $A$ of $X$  has a meet. For this consider the set $\lb A$ of lower bounds of $A$. By assumption,  $\lb A$ has a join, say $x$. If we could show that $x$ is a lower bound of $A$, then it would be the greatest lower bound, hence the meet of $A$. Since every element $a\in A$ is an upper bound of $\lb A$,  it follows that $A$ is a set of   upper bounds of $\lb A$. Since $x$, being the join of $\lb A$, is the least element of $\ub (\lb A)$, then $x\leq a$ for all $a\in A$, hence $x$ is a lower bound of $A$.

$(2)\Rightarrow(1)$ Similar.

Suppose $X$ is an ordered set. We say that $X$ is

* complete if it satisfies the equivalent conditions in Lemma <ref>.
* a complete lattice if it is complete and antisymmetric.

The empty subset of an ordered set $X$  has a join (meet, respectively)  if and only if $X$ has a least (greatest, respectively) element. So, every complete ordered set is bounded.

For each set $X$, $(2^X,\subseteq)$ is a complete lattice. For any $a\leq b$ in $\mathbb{R}$, the partially ordered set $([a,b],\leq)$ is a complete lattice. For each ordered set $X$, the set $\CP X$ of all lower sets of $X$ ordered by inclusion   is a complete lattice. In particular, if $(X,\leq)$ is discrete then $(\CP X,\subseteq)$ coincides with the complete lattice $(2^X,\subseteq)$.

Suppose $X,Y$ are ordered sets and  $f\colon X\lra Y$ is a map. We say that

* $f$  preserves order  if $x_1\leq x_2 \implies f(x_1)\leq f(x_2).$
* $f$  reflects order  if $f(x_1)\leq f(x_2) \implies x_1\leq x_2 $.
* $f$  preserves joins  (meets, resp.) if for each subset $A$ of $X$ and each join (meet, resp.) $a$ of $A$, $f(a)$ is a join (meet, resp.) of $\{f(x)\mid x\in A\}$.

The identity map on any ordered set is order-preserving; the composite of  order-preserving maps is order-preserving. So with ordered sets and order-preserving maps we have a category Ord.

It is clear that an order isomorphism (i.e. an isomorphism in the category Ord) preserves both joins and meets.
A map $f\colon X\lra Y$ is called an order-reversing isomorphism if $f\colon X^{\rm op}\lra Y$ is an order isomorphism, or equivalently, $f\colon X\lra Y^{\rm op}$ is an order isomorphism. Every order-reversing isomorphism transforms joins  to meets  and  meets to joins.

Suppose $\{X_i\}_{i\in J}$ is a family of ordered sets. Define a binary relation on the product set $\prod_{i\in J}X_i$ by
\[(x_i)_i\leq (y_i)_i ~\quad~ \text{if  for each $i$, $x_i\leq  y_i$ in $X_i$}.\]
Then $\leq$ is an order on $\prod_iX_i$ and for each $j$, the projection $p_j\colon  \prod_iX_i \lra X_j $ preserves order.

Suppose $X$ is an ordered set.

* The diagonal map

\[\Delta\colon X\lra X\times X, \quad \Delta(x)=(x,x)\]
preserves and reflects order.
* The map
\[\sy\colon X\lra\CP X, \quad x \mapsto\thinspace\da x\]
preserves and reflects order, where $\CP X$ is the set of all lower sets of $X$ ordered by  inclusion. This is a special case of   Yoneda embedding in category theory.

* The map
\[\syd\colon X\lra\CPd X, \quad x \mapsto\thinspace\ua x\]
preserves and reflects order, where $\CPd X$ is the set of all upper sets of $X$ ordered by converse inclusion. This is a special case of  coYoneda embedding in category theory.

5pt Galois connection

Suppose $X,Y$ are ordered sets; $f\colon X\lra Y$ and $g\colon Y\lra X$ are a pair of maps. We say that $f$ is left adjoint to $g$, or $g$ is right adjoint to $f$, and write $f\dashv g$,  if for all $x\in X$ and $y\in Y$ it holds that
\[f(x)\leq y\iff x\leq g(y).\]
In this case,  the pair $(f,g)$  is called a Galois connection, or an adjunction, with $f$ being the left adjoint and $g$ the right adjoint.

If $f\colon X\lra Y$ is an order isomorphism, then $f$ is both  left and right adjoint to its inverse $f^{-1}$.

If  both $f_1\colon X\lra Y$ and $f_2\colon X\lra Y$ are left adjoint to $g\colon Y\lra X$, then $f_1(x)\cong f_2(x)$ for all $x\in X$.

Therefore, the left adjoints of a map, if exist, are essentially  unique; likewise for right adjoints.  So, we shall speak of the left adjoint and the right adjoint.

Left adjoints  and  right adjoints  preserve  order.

Suppose   $f\colon X\lra Y$ is left adjoint to $g\colon Y\lra X$. We show that both $f$ and $g$ preserve order.   Assume that $x_1\leq x_2$. Since $f(x_2)\leq f(x_2)$, then $x_2\leq gf(x_2)$, so $x_1\leq gf(x_2)$, hence $f(x_1)\leq f(x_2)$. This shows that $f$ preserves order. Likewise, $g$ preserves order.

Each composite of left adjoints is a left adjoint.  Dually, each composite of  right  adjoints is a  right  adjoint.

Suppose $X$ is a topological space, $\CO(X)$ is the set of open sets and $\mathcal{C}(X)$ is the set of closed sets. Then the interior operator $( -)^\circ\colon 2^X \lra  \CO(X) $ is right adjoint to the inclusion   $  \CO(X) \lra 2^X ;$  the closure operator $(-)^-\colon 2^X \lra  \mathcal{C}(X) $ is  left adjoint to the inclusion  $  \mathcal{C}(X) \lra 2^X .$

Suppose $X$ is an ordered set; $\CP X$ and $\CPd X$ are the ordered sets of lower sets and upper sets of $X$, respectively, as in Example <ref>.  For all $A\in\CP X$ and $B\in\CPd X$, it holds that
\[\ub A\supseteq B\iff \forall x\in A, \forall y\in B,~ x\leq y\iff A\subseteq \lb B,\]
so $\ub\colon \CP X\lra\CPd X$ is left adjoint to $\lb\colon \CPd X\lra \CP X.$

Suppose $X,Y$ are sets and $f\colon X\lra Y$ is a map. Let
\[f_\exists\colon2^X \lra 2^Y\]
be the map that sends each subset $A$ of $X$ to its image $ \{f(a)\mid a\in A\}$; let
\[f^{-1}\colon2^Y \lra 2^X\]
be the map that sends each subset $B$ of $Y$ to its preimage $\{x\in X\mid f(x)\in B\}$.
Then $f_\exists$ is left adjoint to $f^{-1}$. The map $f^{-1}$ also has a right adjoint, given by
\[f_\forall \colon 2^X \lra  2^Y ,\quad f_\forall (A)=\{y\in Y\mid f^{-1}(y)\subseteq A\}. \]
Thus, we have a string  of adjunctions:
\[f_\exists\dashv f^{-1}\dashv f_\forall\colon 2^X\lra 2^Y .\]

Suppose $X,Y$ are ordered sets and $f\colon X\lra Y$  preserves order. For each lower set $B\subseteq Y$, the inverse image $f^{-1}(B)$ is a lower set of $X$, so we have an order-preserving map
\[f^{-1}\colon \CP Y  \lra \CP X .\]
For each lower set $A\subseteq X$,  let
\[f_\exists(A)=\{y\in Y\mid y\leq f(a) ~\text{for some}~ a\in A\}.\]

\[f_\exists\colon \CP X  \lra  \CP Y \]
is order-preserving and  it is left adjoint to $f^{-1}$. The map $f^{-1}$ also has a right adjoint
\[f_\forall \colon \CP X  \lra  \CP Y \]
given by
\[f_\forall (A)=\{y\in Y\mid f^{-1}(\da y)\subseteq A\}.\]
Thus, we  have a string of adjunctions:
\[f_\exists\dashv f^{-1}\dashv f_\forall\colon \CP X\lra\CP Y .\]

If both $X$ and $Y$ are discrete ordered sets, then this string of adjunctions reduces to that in the above example.

Suppose $X,Y$ are ordered sets; $f\colon X\lra Y$ and $g\colon Y\lra X$  preserve order. The following   are equivalent:

* $f\dashv g$.

* $\id_X\leq gf$ and $fg\leq\id_Y$.

* For each $x\in X$, $f(x)$ is a least element in $g^{-1}(\ua x)$.

* For each $y\in Y$, $g(y)$ is a greatest element in $f^{-1}(\da y)$.

For each $x\in X$, since $f(x)\leq f(x)$, then  $x\leq gf(x)$, so  $\id_X\leq gf$.  For each $y\in Y$, since $g(y)\leq g(y)$, then  $fg(y)\leq y$, so $fg\leq\id_Y$.

$(2)\Rightarrow(3)$ First, since $x\leq gf(x)$, then $f(x)\in g^{-1}(\ua x)$. Second, for each $y\in g^{-1}(\ua x)$, since $x\leq g(y)$, then  $f(x) \leq fg(y) \leq y$.  So  $f(x)$ is the least element in $g^{-1}(\ua x)$.

$(3)\Rightarrow(1)$ Suppose   $f(x)\leq y$. Since $f(x)\in g^{-1}(\ua x)$, then  $x\leq gf(x) \leq g(y)$. Conversely, suppose   $x\leq g(y)$. Since $fg(y)$ is the least element in $g^{-1}(\ua g(y))$ and $y\in g^{-1}(\ua g(y))$, then $f(x)\leq fg(y)\leq y$.   Therefore, $f$ is left adjoint to $g$.

That $(2)\Rightarrow(4) \Rightarrow(1)$ can be proved in a similar way.

Suppose   $X, Y$ are partially ordered sets;  $f\colon X\lra Y$ is left adjoint to $g\colon Y\lra X$. Then $f=fgf$ and $g=gfg$, hence both $gf\colon X\lra X$ and $fg\colon Y\lra Y$ are idempotent.

Suppose  $X, Y$ are partially ordered sets,   $f\colon X\lra Y$ is left adjoint to $g\colon Y\lra X$. The following conditions are equivalent:

* $g$ is surjective.

* $f(x)=\min g^{-1}(x)$ for all $x\in X$.

* $gf=1_X$.

* $f$ is injective.
Likewise, the following conditions are equivalent:

* $g$ is injective.

* $g(y)=\max f^{-1}(y)$ for all $y\in Y$.

* $fg=1_Y$.

* $f$ is surjective.

Every left adjoint  $f\colon  X\lra Y$  preserves joins. Conversely, if $X$ is complete and $f\colon X\lra Y$ preserves joins, then $f$ is a left adjoint.

Suppose   $f\colon  X\lra Y$ has a right adjoint, say $g\colon Y\lra X$. We show that for any subset $A$ of $X$, $f(\bv A)$ is a join of $\{f(a)\mid a\in A\}$ whenever $\bv A$ exists. Since $f$ preserves order, then $f(\bv A)$ is an upper bound of $\{f(a)\mid a\in A\}$. If $y$ is an upper bound of $\{f(a)\mid a\in A\}$, then $a\leq g(y)$ for all $a\in A$, so  $\bv A\leq g(y)$, hence $f(\bv A)\leq y$.

Conversely, suppose $X$ is complete and $f\colon X\lra Y$ preserves joins.  Since $f$ preserves joins, for each $y\in Y$  the join of   $f^{-1}(\da y)$ is also a member of $f^{-1}(\da y)$. Assigning to  $y$ the join of $f^{-1}(\da y)$ yields a right adjoint of $f$.

Dually, every right adjoint preserves meets. Conversely, if $f\colon X\lra Y$ preserves meets and $X$ is complete then $f$ is a right adjoint.

Galois connections    are of fundamental importance in order theory. Many concepts in order theory are postulated, or can be characterized, in terms of Galois connections.

For each ordered set $X$, the following statements are equivalent:

* $X$ is   complete.
* The Yoneda embedding $\sy\colon X\lra\CP X$ has a left adjoint.
* The coYoneda embedding $\syd\colon X\lra\CPd X$ has a right adjoint.

$(1)\Rightarrow(2)$ If  $s\colon\CP X\lra X$ is left adjoint to $\sy$, then for all $x\in X$ and $A\in\CP X$,
\[A\subseteq\thinspace\da x\iff s(A)\leq x,\]
which says that $x$ is an upper bound of $A$ if and only if $x\geq s(A)$.  So  $s(A)$ is a least upper bound of $A$.

$(2)\Rightarrow(1)$ Suppose that $X$ is complete. Let $s\colon\CP X\lra X$ be a map such that $s(A)$ is a join of $A$ for each $A\in\CP X$. Then $s$ is a left adjoint of $\sy$.

The equivalence $(1)\Leftrightarrow(3)$ is verified in a similar way.

Suppose $X$ is a partially ordered set.

* $X$ is a meet semilattice if and only if the diagonal map $\Delta\colon X\lra X\times X$ has a right adjoint.
* $X$ is a join semilattice if and only if the diagonal map $\Delta\colon X\lra X\times X$ has a left adjoint.
* $X$ is a lattice if and only if the  diagonal map $\Delta\colon X\lra X\times X$ has, at the same time, a left and a right adjoint; that is, there is a string of adjunctions: $\vee\dashv\Delta\dashv\wedge\colon X\times X\lra X.$

Quantales provide another kind of interesting examples.  Here we only include the definition of commutative and unital  quantales, standard reference for  quantales is Rosenthal [96]. A commutative and unital quantale
\[{\sf Q}=({\sf Q},\with,k)\]

is a commutative monoid with $k$ being the unit, such that the underlying set ${\sf Q}$ is a complete lattice  and for each $x\in{\sf Q}$, the map $x\with-\colon{\sf Q}\lra{\sf Q}$ has a right adjoint $x\ra-\colon{\sf Q}\lra{\sf Q}$. This is equivalent to that   $\with$ distributes over  joins.  In particular, a frame is a complete lattice $L$ such that $(L, \wedge,1)$ is a quantale; or equivalently,  for each $a\in L$ the meet operator $a\wedge-\colon L\lra L$ distributes over joins.

5pt Completely distributive lattices

Suppose $X$ is a partially ordered set. By Proposition <ref>, $X$ is a complete lattice if and only if the Yoneda embedding $\sy\colon X\lra\CP X$ has a left adjoint $\sup\colon \CP X\lra X$, which sends each subset $A$ of $X$ to its join $\bv A$.

A complete lattice $X$ is  completely distributive if the left adjoint of the Yoneda embedding $\sy\colon X\lra\CP X$ also has a left adjoint; that means, there is a string of adjunctions
\[\Downarrow\thinspace\dashv\sup\dashv\sy\colon X\lra\CP X.\]

Lattices satisfying the requirement in the above definition are also said to be constructively completely distributive, see e.g. Wood [117], because, compared with the complete distributivity law (see Theorem <ref>), it does not resort to the Axiom of Choice.

To characterize completely distributive lattices, we need the notion of totally below relation. Suppose $x,y$ are elements of a complete lattice $X$. We say that $x$ is totally below $y$, in symbols $x\lhd y$, if for each subset $A$ of $X$,
$$ y\leq\sup A \implies   x\leq a~\text{for some}~  a\in A.$$

The following  facts are easy to check:
* If $x\lhd y$ then $x\leq y$.
* $x\lhd y$ if and only if   $x\in \bigcap\{A\in \CP X\mid y\leq \sup A\}$.
* If $a\leq x\lhd y\leq b$  then $a\lhd b$.
* If $x\lhd \bv_{i\in I}y_i$  then $x\lhd y_i$  for some $i\in I$.

For each complete lattice $X$, the following   are equivalent:

* $X$ is completely distributive.

* For each family $\{A_i\}_{i\in I}$ of lower sets of $X$,
\[\sup\bigcap_{i\in I}A_i=\bw_{i\in I}\sup A_i.\]

* Every element $x$ of $X$ is the join of the elements   totally below it, i.e. $x=\bv\{z\in X\mid z\lhd x\}$.

* $X$ satisfies the complete distributivity law: for any family $\{x_{i,j}\mid i\in I, j\in J_i\}$ of elements of $X$,
\begin{equation*}  \label{CD law}\bw_{i\in I}\bv_{j\in J_i}x_{i,j} = \bv_{f\in\prod_{i\in I}J_i}\bw_{i\in I}x_{i,f(i)}.\end{equation*}

$(1)\Rightarrow(2)$ This follows from that $\sup\colon \CP X\lra X$ is a right adjoint.

$(2)\Rightarrow(3)$ This follows from that an element $z$  is totally below $x$ if and only if
\[z\in \bigcap\{A\in \CP X\mid x\leq \sup A\}.\]

$(3)\Rightarrow(4)$ Let
\[x=\bw_{i\in I}\bv_{j\in J_i}x_{i,j}\]
and suppose $z\lhd x$. By definition of the totally below relation, for each $i\in I$, there is some element in $J_i$, say $g(i)$, such that $z\leq x_{i,g(i)}$. Thus
\[z\leq \bv_{f\in\prod_{i\in I}J_i}\bw_{i\in I}x_{i,f(i)}\]
and consequently,
\[\bw_{i\in I}\bv_{j\in J_i}x_{i,j} \leq \bv_{f\in\prod_{i\in I}J_i}\bw_{i\in I}x_{i,f(i)}\]
by arbitrariness of $z$. The converse inequality is obvious.

$(4)\Rightarrow(1)$ For each $x\in X$, let
\[\Da x=\bigcap\{A\in\CP X\mid x\leq\sup A\}.\]
We show that for all $x\in X$ and $A\in\CP X$,
\[\Da x\subseteq A\iff x\leq\sup A,\]
which entails that
\[\Downarrow\colon X\lra \CP X\]
is left adjoint to
\[\sup\colon \CP X\lra X.\]
To this end, we check that   each $x\in X$  is the join of $\Da x$, i.e. $\sup\Da x=x$.

Index the family $\{A\in\CP X\mid x\leq\sup A\}$ as $\{A_i\}_{i\in I}$; for each $i$, index the elements of $A_i$ as $\{x_{i,j}\mid j\in J_i\}$. Since each $A_i$ is a lower set, it follows that for each $f\in\prod_{i\in I}J_i$,
\[\bw_{i\in I}x_{i,f(i)}\in\bigcap_{i\in I}A_i.\]
Then,  from the complete distributivity law it follows that $x$ is the join of $\Da x$ which is, by definition, the intersection $\bigcap_{i\in I}A_i$.

The totally below relation in a completely distributive lattice $X$ is interpolative in the sense that if $x\lhd y$ then there is some $z\in X$ such that $x\lhd z\lhd y$.

This follows immediately from the fact that the map $\Downarrow\colon X\lra\CP X$, being a left adjoint, preserves joins.

For each ordered set $X$, the set $\CP X$ of all lower sets of $X$ ordered by inclusion  is  a completely distributive lattice. That $(\CP X,\subseteq)$ is a complete lattice is trivial. For each $A\in \CP X$, it is clear that  for every $x\in A$,  the lower set $\da x$ is totally below $A$,  and $A$ is the join of $\{\da x\mid x\in A\}$, so, $\CP X$ is a completely distributive lattice.

The complete distributivity law is self dual in the sense that if $X$ is a completely distributive lattice, then so is $X^{\rm op}$.

By Theorem <ref>, it suffices to check that for each  family $\{A_i\}_{i\in I}$ of upper sets in $X$,
\[\inf\bigcap_{i\in I}A_i=\bv_{i\in I}\inf A_i.\]

We only need to check that the left side is smaller than or equal to the right side. To this end, we check that if   $x$ is totally below $\inf\bigcap_{i\in I}A_i$, then $x\leq \inf A_i$ for some $i\in I$. Suppose on the contrary that $x\not\leq \inf A_i$ for any $i\in I$. Take some $z$ such that $x\lhd z\lhd\inf\bigcap_{i\in I}A_i$;   for each $i$  take some $a_i\in A_i$ with $x\not\leq a_i$. Since each $A_i$ is an upper set, it follows that
\[\bv_{i\in I}a_i\in \bigcap_{i\in I}A_i.\]
Thus, $z\not\leq \bv_{i\in I}a_i$, contradicting that $z\leq \inf \bigcap_{i\in I}A_i$.

5pt Domains

Suppose $X$ is an ordered set and $\idl X$ is the set of  ideals (= directed lowers sets) of $X$ ordered by inclusion. Since for each $x\in X$, the principal lower set $\da x$ is an ideal, the Yoneda embedding $\sy\colon X\lra\CP X$ factors through $\idl X$.  We also use the symbol $\sy$ to denote the Yoneda embedding with codomain restricted to   $\idl X$. This will cause no confusion,  it is easily detected from the context which one is meant.

Let $x,y$ be elements of a partially ordered $X$. We say that $x$ is way below $y$, in symbols $x\ll y$, if   for each directed set $D$ of $X$ with a join,
\[y\leq\sup D \implies   x\leq d~\text{for some}~  d\in D.\]

The following  facts are easily verified:
* If $x\ll y$ then $x\leq y$.
* $x\ll y$ if and only if   $x\in \bigcap\{I\in \idl X\mid \text{$I$ has a join and}~y\leq \sup I\}$.
* If $a\leq x\ll y\leq b$  then $a\ll b$.

Suppose $X$ is a partially ordered set. We say that

* $X$ is a directed complete partially ordered set, a dcpo for short, if every directed subset of $X$ has a join.

* $X$ is a continuous partially ordered set if for each $x\in X$, the set $\{y\in X\mid y\ll x\}$ is directed and has $x$ as a join.
* $X$ is a domain if $X$ is a continuous dcpo.

We say that a map $f\colon X\lra Y$ between ordered sets is Scott continuous if it preserves directed joins. With dcpos and Scott continuous maps we have a category
$${\bf dcpo}.$$

Suppose $X$ is a partially ordered set.

* $X$ is a dcpo if and only if  the Yoneda embedding $\sy\colon X\lra\idl X$ has a left adjoint, which is denoted by  $\sup\colon \idl X\lra X$.

* $X$ is continuous if and only if for each $x\in X$, there is a directed set $D$ consisting of elements that are way below $x$ such that $x=\sup D$.

* $X$ is a domain if and only if   the left adjoint $\sup\colon \idl X\lra X$ of  $\sy\colon X\lra\idl X$  has a left adjoint, which is denoted by  $\wayb \colon X\lra\idl X$.
In this case, $\wayb x=\{z\in X\mid z\ll x\}$.

Therefore, a partially ordered set $X$ is a domain if there is a string of adjunctions
\[\wayb \dashv \sup\dashv\sy\colon X\lra\idl X.\]

The way below relation in a continuous partially ordered set $X$ is interpolative in the sense that   $x\ll y$ implies  that $x\ll z\ll y$ for some $z\in X$.

For each ordered set $X$, the set $\idl X$ of all ideals of $X$ ordered by inclusion is a domain. For each directed family $\{I_d\}_{d\in D}$ of  ideals in $X$,  the union $\bigcup_{d\in D}I_d$ is an ideal in $X$, and it is the join of $\{I_d\}_{d\in D}$  in $(\idl X,\subseteq)$, hence $(\idl X,\subseteq)$ is directed complete. For each ideal $I$   of $X$, it is clear that $I$ is the join of the directed family $\{\da x\mid x\in I\}$ of ideals. Since for each $x\in I$, the principal ideal $\da x$ is   way below $I$ in $(\idl X,\subseteq)$, it follows that $\idl X$ is a domain.

(Scott [100]) A partially ordered set is   a continuous lattice if it is  at the same time  a domain and a complete lattice.

It is clear that a subset $A$ of a join semilattice $X$ is an ideal if and only if $A$ is a lower set and closed under binary joins. So, the intersection of a family of ideals in a bounded lattice   is also an ideal, though the (set-theoretic) intersection of a family of ideals in a partially ordered may fail to be so.

For a bounded lattice $X$, the set of ideals $\idl X$ is closed under meets in $\CP X$, so it is a complete lattice and the inclusion $\idl X\lra \CP X$ has a left adjoint. In particular, if $X$ is a completely distributive lattice, then the composite of $\Downarrow\colon X\lra\CP X$ and the left adjoint of the inclusion \(\idl X\lra \CP X\) is   left adjoint to $\sup\colon \idl X\lra X$. This shows that every completely distributive lattice is continuous.

For each complete lattice $X$, the following   are equivalent:

* $X$ is continuous.

* The map \(\sup\colon \idl X\lra X\) is a right adjoint.

* $X$ satisfies the directed distributivity law: for any  family $\{x_{t,s}\mid t\in T, s\in S_t\}$ of elements of $X$ such that  $\{x_{t,s}\mid s\in S_t\}$ is directed for each $t\in T$, it holds that
\begin{equation*}  \label{DDlaw2}\bw_{t\in T}\bv_{s\in S_t}x_{t,s} = \bv_{f\in\prod_{t\in T}S_t}\bw_{t\in T}x_{t,f(t)}.\end{equation*}

Suppose $X$ is a lattice and $x\in X$. We say that
* $x$ is meet-irreducible if,   $ a\wedge b=x\implies x=a$ or $x=b$.

* $x$ is  prime if,   $a\wedge b\leq x  \implies a\leq x$ or $b\leq x$.

* $x$ is join-irreducible if,   $x= a\vee b\implies x=a$ or $x=b$.

* $x$ is  coprime if,   $x\leq a\vee b\implies x\leq a$ or $x\leq b$.
It is clear that $x$ is   meet-irreducible in $X$ if and only if $x$ is join-irreducible in $X^{\rm op}$;   $x$ is prime in $X$ if and only if $x$ is coprime in $X^{\rm op}$. Every prime element is   meet-irreducible,  but the converse is not   true.

Every meet-irreducible element in a distributive lattice is prime.

Suppose   $x$ and $y$ are elements of a continuous lattice $X$ with $x\not\leq y$. Then there is a meet-irreducible element $p$ with $y\leq p$ and $x\not\leq p$.

Since $X$ is a continuous lattice and $x\not\leq y$, there is some $z\ll x$ such that $z\not\leq y$.  By interpolation property of the way below relation, there is a sequence of elements $\{z_n\}_n$ with
\[z\ll\cdots\ll z_n\ll z_{n-1}\ll\cdots\ll z_1=x.\]
Let $U=\bigcap_{n=1}^\infty \ua z_n$. It is clear that   $U$ is an upper set and  closed under finite meets.

Pick a maximal chain $C$ in $X\setminus U$ that contains $y$ and set $p=\sup C$. We claim that $p$ is not in $U$, hence maximal in $X\setminus U$. Otherwise, there is some $n$ such that $z_n\leq p$. Since $z_{n+1}\ll z_n$ and $C$ is a chain, it follows that $z_{n+1}\leq c$ for some $c\in C$, contradicting that $c\notin U$. It is obvious that $y\leq p$ and  $x\not\leq p$, so, it remains to check that $p$ is meet-irreducible. Suppose that $p=a\wedge b$. Since $U$ is closed under finite meets,   either $a\notin U$ or $b\notin U$.  Finally, since $p$ is maximal in $X\setminus U$, it follows that $p=a$ or $p=b$.

The above conclusion shows  that each element in a continuous lattice is the meet of a family of meet-irreducible elements. In other words, every continuous lattice has enough meet-irreducible elements.

We say that a complete lattice $X$ has enough coprimes if every element of $X$ is the join of a set of coprimes.

For each complete lattice $X$, the following are equivalent:
* $X$ is a completely distributive.

* $X$ is distributive and both $X$ and $X^{\rm op}$ are   continuous lattices.

* $X$ is continuous and has enough coprimes.

$(1)\Rightarrow(2)$ First,  it follows from the complete distributivity law that $X$ is distributive. Second, since complete distributivity is self-dual and   every completely distributive lattice is continuous, it follows that both $X$ and $X^{\rm op}$ are   continuous lattices.

$(2)\Rightarrow(3)$ Apply Proposition <ref> to $X^{\rm op}$.

$(3)\Rightarrow(1)$ Since $X$ is continuous and has enough coprimes,   for each $x\in X$,
\[x=\sup\{z\in X\mid z\ll x\}=\sup\{z\in X\mid z~\text{is   coprime and }  z\ll x\}.\]
Since for each  coprime $z$,  $z$  is way below $x$ implies $z$ is totally below $x$,    the conclusion thus follows.

§ CONTINUOUS T-NORM

The interval $[0,1]$ together with a continuous t-norm is our truth-value set. So, the structure of continuous t-norms is of fundamental importance in the development of real-enriched categories. For sake of self-contaiment and convenience of the reader,   proofs of the well-known  characterization of continuous Archimedean t-norms (Theorem <ref>) and the ordinal sum decomposition of continuous t-norms (Theorem <ref>) are included.
Standard references for continuous t-norms are  the monographs: Alsina,  Frank,     Schweizer [4],   and Klement,   Mesiar, Pap [60].

A triangular norm (a t-norm  for short) on an interval $[a,b]$, where $a$ is allowed to be $-\infty$ and $b$ is allowed to be $\infty$, is a binary operator $\&$ such that for all $x,y,z\in[a,b]$,

* $x\with y=y\with x$;
* $(x\with y)\with z=x\with(y\with z)$;
* $x\with y\leq x\with z$ whenever $y\leq z$;
* $x\with b =x$.

From (iii) and (iv) we immediately obtain that $a\with x=a$ for all $x\in [a,b]$.

In the vocabulary of lattice theory, a t-norm on $[a,b]$ is a binary operator $\with$ such that $([a,b],\with,b)$ is a commutative lattice-ordered monoid, see e.g. Birkhoff [16]. In the language of category theory, a t-norm is a binary operator $\with$ such that $([a,b],\with,b)$ is a commutative monoid in the category of ordered sets and order-preserving maps, see e.g. MacLane [87].

Suppose $\with_1$ is a t-norm on $[a_1,b_1]$ and $\with_2$ is a t-norm on $[a_2,b_2]$. We say that $\with_1$ is isomorphic to $\with_2$ if there is an order-preserving bijection $f\colon [a_1,b_1]\lra[a_2,b_2]$ such that for all $x,y\in[a_1,b_1]$,
\[f(x\with_1 y)=f(x)\with_2 f(y).\]

Given a  t-norm $\&$ on $[c,d]$ and an order isomorphism $f\colon [a,b]\lra [c,d]$, the binary operator
\[\with_f\colon [a,b]\times[a,b]\lra[a,b],\quad  x\with_f y=f^{-1}(f(x)\with f(y)) \]
is a  t-norm on $[a,b]$ and   isomorphic to $\with$.  This fact shows that we only need to consider t-norms on the unit interval $[0,1]$.

Some basic   t-norms on $[0,1]$:

* The Gödel t-norm: $x\with y= \min\{x,y\}$.

* The product t-norm: $x\with y=x\cdot y$.

* The Łukasiewicz t-norm:
$x\with y=\max\{x+y-1,0\}$.

* The nilpotent minimum t-norm:
$$x\with  y= \begin{cases} 0  & x+y\leq 1,\\ \min\{x,y\}  & x+y>1.\end{cases}$$

* The drastic product t-norm:
$$x\with  y= \begin{cases} 0  & x,y<1,\\ \min\{x,y\}  & {\rm otherwise}.\end{cases}$$

It is clear that   every  t-norm on $[0,1]$ lies between the drastic product  and the Gödel t-norm.

The addition $+$ is a t-norm on the interval $[-\infty,0]$. The correspondence $x\mapsto \mathrm{e}^x$ defines an isomorphism  $([-\infty,0],+,0)\lra([0,1],\cdot,1)$.

[Truncated addition] For each $a\in(-\infty,0)$, the truncated addition
\[+_a\colon[a,0]\times[a,0]\lra[a,0],\quad x+_a y=\max\{a,x+y\}\]
is a t-norm on $[a,0]$.

[Truncated product] For each $b\in(0,1)$, the truncated product
\[\cdot_b\colon[b,1]\times[b,1]\lra[b,1],\quad x\cdot_b y=\max\{b,x\cdot y\}\]
is a t-norm on $[b,1]$.

For all $a\in(-\infty,0)$ and  $b\in(0,1)$, the following t-norms are isomorphic to each other:

* The truncated addition on $[a,0]$.
* The truncated product on $[b,1]$.
* The Łukasiewicz t-norm on $[0,1]$.
It is clear that the truncated addition on $[a,0]$ is isomorphic to the truncated addition on $[-2,0]$  and the truncated product on $[b,1]$  is isomorphic to the truncated product on $[1/4,1]$, so it suffices to show that both the truncated addition on $[-2,0]$ and the truncated product on $[1/4,1]$  are isomorphic to  the Łukasiewicz t-norm on $[0,1]$.

The map
$$f\colon [0,1]\lra[1/4,1],\quad f(x)=2^{2(x-1)}$$
defines an isomorphism between the Łukasiewicz t-norm on $[0,1]$ and the truncated product on $[1/4,1]$;  the map
$$g\colon [0,1]\lra [-2,0],\quad g(x)=2(x-1)$$
defines an isomorphism between the Łukasiewicz t-norm on $[0,1]$ and the truncated addition on $[-2,0]$.   The proof is thus completed.

Given a  t-norm $\&$ on $[0,1]$ and $x\in[0,1]$, the powers (with respect to $\&$) is defined recursively by
\[x^1=x \quad {\rm and} \quad x^{n+1}=x^n\with x \]
for all positive integers $n$. For all $x$ and $n$, $x^{n+1}\leq x^n$, so $\{x^n\}_{n\in\mathbb{N}}$ is a decreasing sequence. By associativity of $\&$ we have
\[x^{m+n}=x^m\with x^n=x^n\with x^m \quad {\rm and} \quad x^{mn}=(x^m)^n=(x^n)^m\]
for all $m,n\geq 1$.

Suppose $\&$ is a  t-norm on $[0,1]$ and $x\in[0,1]$. We say that
* $x$ is idempotent if $x\with x=x$;

* $x$ is nilpotent  if $x^n=0$ for some $n\geq 1$.

The Gödel t-norm and the product t-norm have only one nilpotent element. Every element in $[0,1)$ is nilpotent for the Łukasiewicz t-norm and the drastic t-norm. For the nilpotent minimum t-norm, every $x\in[0,1/2]$ is   nilpotent.

A  t-norm $\&$ on $[0,1]$ is Archimedean  if for all $x,y\in(0,1)$, there is some $n$ such that $x^n<y$.

Both  the product t-norm and the Łukasiewicz t-norm are Archimedean.

Suppose $\&$ is an  Archimedean t-norm. If $\&$ has a nonzero nilpotent element, then all $x\in[0,1)$ are nilpotent.

Suppose that $a>0$ is a nilpotent element of $\&$. Then $a^m=0$ for some $m\geq 1$. For each $x<1$, since $\&$ is Archimedean, there is some $n$ such that $x^n<a$. Then $x^{nm}\leq a^m=0$.

A t-norm $\&$ on $[0,1]$ is left continuous on the first variable if
for each $y$ and each increasing sequence $\{x_n\}_{n\in\mathbb{N}}$ in $[0,1]$,
\[\sup\limits_{n\in\mathbb{N}}(x_n\with y)=(\sup\limits_{n\in\mathbb{N}}x_n)\with y.\]

By commutativity of $\&$, left continuity on the first variable implies left continuity on the second variable.

Let $\&$ be  a  t-norm on $[0,1]$. We say that
* $\&$ is   left continuous   if it is left continuous on each variable;

* $\&$ is  continuous   if it is a continuous map $[0,1]\times[0,1]\lra[0,1]$ with respect to the usual topology.

Said differently, a left continuous t-norm is a binary operator $\&$ on the unit interval such that  $([0,1],\&,1)$ is a commutative and unital qutantale. Every continuous t-norm is left continuous. The Gödel t-norm, the product t-norm and the Łukasiewicz t-norm are all continuous; the nilpotent minimum t-norm  is left continuous but not continuous;  the drastic product t-norm is not left continuous.

Suppose $\&$ is a left continuous t-norm on $[0,1]$. For each $x\in[0,1]$, the map
$$x\with-\colon [0,1]\lra[0,1]$$
has a right adjoint
$$x\ra-\colon [0,1]\lra[0,1].$$
For all $x,y,z\in[0,1]$ it holds that
\[x\with y\leq z\iff y\leq x\ra z.\]
Because of this property, in fuzzy logic left continuous t-norms  and their right adjoints are   employed to model the logic connectives “conjunction” and   “implication”, respectively. So, the binary operator
\[\ra\colon [0,1]\times[0,1]\lra[0,1],\quad (x,y)\mapsto x\ra y\]
is called the implication of the t-norm $\&$.

* The implication operator of the Gödel t-norm:
\[x\ra y= \begin{cases} 1  & x\leq y,\\ y & x>y.\end{cases}\]

It is continuous on $[0,1]^2\setminus\{(x,x)\mid x<1\}$.

* The implication operator of the product t-norm:
$$x\ra y= \begin{cases}  1  & x\leq y,\\ y/x  & x>y.\end{cases}$$
It is continuous except at $(0,0)$.

* The implication operator of the  Łukasiewicz t-norm:
$$x\ra y= \min\{1, 1-x+y\}. $$
It is continuous on $[0,1]^2$.

* The implication operator of the nilpotent minimum t-norm is given by
$$x\ra y= \begin{cases} 1  & x\leq y,\\ \max\{1-x,y\}  & x>y.\end{cases}$$
It is continuous on $[0,1]^2\setminus\{(x,x)\mid x<1\}$.

(Hájek [40]) For each left continuous t-norm $\&$ on $[0,1]$, the following are equivalent:

* $\&$ is continuous.
* $\&$ is divisible in the sense that if $x\leq y$, then $x=y\with z$ for some $z\in[0,1]$.
* For all $x,y\in[0,1]$, $x\with(x\ra y)=\min\{x,y\}$.

$(1)\Rightarrow(2)$ Suppose that $x\leq y$. Since $y\with 0=0$ and $y\with 1=y$, by continuity of $\&$, the function $y\with-$ maps $[0,1]$ onto $[0,y]$,  then  there is some $z\in[0,1]$ such that $x=y\with z$.

$(2)\Rightarrow(3)$  The equality is trivial if $x\leq y$. Assume that $x>y$. Since $\&$ is divisible, $y=x\with z$ for some $z\in[0,1]$. Then, by definition of $\ra$,    $z\leq x\ra y$, hence
$$y= x\with z\leq x\with(x\ra y)\leq y,$$
and consequently, $x\with(x\ra y)=\min\{x,y\}$.

$(3)\Rightarrow(1)$ It suffices to show that $\&$ is separately continuous. Suppose on the contrary that there is some $y$ such that $y\with -\colon [0,1]\lra[0,1]$ is not continuous. Since $y\with-$ preserves joins, there must exist some $x$ such that
\[y\with x<\inf_{z>x}(y\with z).\]
Pick some $b$ with
$$y\with x<b<\inf_{z>x}(y\with z).$$
Then $b<y$ and   $y\with z\not=b$ for all $z\in[0,1]$,   contradicting $y\with(y\ra b)=b$.

Suppose $\&$ is a continuous  t-norm on $[0,1]$ and   $b$ is an idempotent element. Then

* $y\with x=\min\{x,y\}=x$ whenever $x\leq b\leq y$.
* $y\ra x=x$ whenever $x<b\leq y$.

(i) Since $b\with b=b$, then
\[x=b\with(b\ra x)=b\with b\with(b\ra x)\leq b\with x\leq x, \]
\[ x=b\with x\leq y\with x \leq x,\]
so $x=y\with x$.

(ii) Since $\&$ is continuous, then $y\with(y\ra x)= x<b$, so we must have $y\ra x<b$. Thus, by (i) we have
\begin{align*}x&= y\with(y\ra x)=y\ra x.\qedhere \end{align*}

Let $\&$ be a continuous t-norm on $[0,1]$.  Then for any idempotent elements  $a$ and $b$ with $a<b$, the restriction of $\&$ on $[a,b]$ is a continuous t-norm on the interval $[a,b]$.

5pt Continuous   Archimedean  t-norm

A continuous  t-norm $\&$ is Archimedean if and only if it has no idempotent element other than $0$ and $1$.

Necessity is clear. For sufficiency, we show that $\lim\limits_{n\ra\infty}x^n=0$ for each $x<1$. Since   $\{x^n\}_{n\in\mathbb{N}}$ is a decreasing sequence,  the limit $\lim\limits_{n\ra\infty}x^n$ exists. Let $b=\lim\limits_{n\ra\infty}x^n$. From continuity of $\&$ one infers that $b\with b=b$, so $b$ is idempotent and $b=0$.

Suppose $\&$ is a continuous  Archimedean  t-norm on $[0,1]$.

* If $x^n>0$ then $x^n>x^{n+1}$.

* If $x^n>0$ then $x^n>y^n$ whenever $x>y$.

(i) Suppose on the contrary that $x^n=x^n\with x$. By induction we have $x^n=x^n\with x^m$ for all $m\geq 1$. So,
\[x^n=\lim\limits_{m\ra\infty}(x^n\with x^m)= x^n\with\lim\limits_{m\ra\infty}x^m = x^n\with 0=0, \]
contradicting that $x^n>0$.

(ii) Suppose on the contrary that $x^n\leq y^n$. Since $\&$ is divisible, $y=x\with z$ for some $z<1$. Then
\[x^n\leq y^n=y^{n-1}\with x\with z\leq x^n\with z. \]
\[x^n\leq x^n\with\lim\limits_{m\ra\infty}z^m = x^n\with 0=0,\]
contradicting that $x^n>0$.

Suppose $t\colon [0,1]\lra[-\infty,0]$ is  continuous and strictly increasing and $t(1)=0$. Since $t$ preserves meets, it has a left adjoint
$$t^\dashv\colon  [-\infty,0]\lra [0,1] , \quad  t^\dashv (x) =\begin{cases}  t^{-1}(x)  & x\in[t(0),0], \\
0  & x< t(0).\end{cases}$$

The left adjoint $t^\dashv \colon  [-\infty,0]\lra [0,1] $ satisfies:

* $t^\dashv \circ t(x)=x$ for all $x\in[0,1]$.
* $t\circ t^\dashv (x)=x$ for all $x\in[t(0),0]$.
* $t^\dashv $ is continuous and is strictly increasing on $[t(0),0]$.

Suppose $t\colon [0,1]\lra[-\infty,0]$ is continuous and strictly increasing and $t(1)=0$. Define a binary operator $\&$ on $[0,1]$ by
\begin{equation*} \label{formula 1} x\with y=t^\dashv (t(x)+t(y)),\end{equation*}
where $t^\dashv $ is the left adjoint of $t$. Then

* $\&$ is a continuous Archimedean t-norm.

* $\&$ has a nonzero nilpotent element if and only if $t(0)>-\infty$.

* $\&$ is isomorphic to the truncated addition on $[t(0),0]$.

The function $t\colon [0,1]\lra[-\infty,0]$ is called an additive generator of the t-norm $\&$.

(i) The verification of that $\&$ is a continuous t-norm is routine. To see that $\&$ is Archimedean, suppose   $0<x<y<1$. Since $t$ is strictly increasing, then $0>t(y)>t(x)>-\infty$. Pick some $n\geq 1$ with $nt(y)< t(x)$, then
\[y^n=t^\dashv (nt(y))< t^\dashv (t(x))=x.\]

(ii) Suppose   $t(0)>-\infty$. For each $x<1$, since $t(x)<0$, there exists $n\geq 1$ such that $nt(x)\leq t(0)$, then $x^n=t^\dashv (nt(x))=0$, hence every $x<1$ is a nilpotent element of $\&$. Now suppose  $t(0)=-\infty$. For each $x<1$, since $t(x)>-\infty$ and $t^\dashv $ is strictly increasing on $[-\infty,0]$, it follows that for all $n\geq 1$,
\[x^n=t^\dashv (nt(x))>t^\dashv (-\infty)=0,\]
so $\&$ has no nilpotent element other than $0$.

(iii) By definition.

* The  function
$$t\colon [0,1]\lra[-\infty,0],\quad t(x)=\ln x$$
is an additive generator of the product t-norm.

* The  function
$$t\colon [0,1]\lra[-\infty,0],\quad t(x)=x-1$$
is an additive generator of the Łukasiewicz t-norm.

* For each $p>0$,  the function
$$t\colon [0,1]\lra[-\infty,0],\quad t(x)= x^{p}-1$$
is continuous and strictly increasing and  $t(1)=0$.   The  t-norm $\with_p$ with $t$ as  additive generator is given by
$$x\with_p y= (\max\{x^p+y^p-1,0\})^{\frac{1}{p}}.$$
The Gödel t-norm is the limit of $\with_p$ when $p$ tends to $0$; the drastic product is the limit of $\with_p$ when $p$ tends to infinity.

Every continuous  Archimedean t-norm $\&$  on $[0,1]$ has an additive generator. That means, there exists a continuous and strictly increasing function $t\colon [0,1]\lra[-\infty,0]$ with $t(1)=0$   such that for all $x,y\in[0,1]$,
$$x\with y =t^\dashv (t(x)+t(y)),$$
where $t^\dashv $ is the left adjoint of $t$.

We make some preparations first. Under the assumption of Theorem <ref>, define for each natural number $n\geq 1$ a function
\[f_n\colon [0,1]\lra[0,1]\]
Then $f_n(0)=0$, $f_n(1)=1$,   $f_n$ is continuous and is strictly increasing whenever it is positive. In particular,  $f_n\colon [0,1]\lra[0,1]$ preserves joins, so it has a right adjoint
\[g_n\colon [0,1]\lra[0,1],\]
which sends each $x>0$ to the  unique $y$ with $y^n=x$.

For each $n\geq1$, let
$$a_n=g_n(0)=\sup\{x\in[0,1]\mid f_n(x)=0\}.$$
Then for all $n,m\geq 1$, we have:

* $f_n(a_n)=0$ and $f_n$ is continuous and strictly increasing on $[a_n,1]$.

* $g_n(0)=a_n$ and $g_n$ is continuous and strictly increasing on $[0,1]$.

* For all $x\in[0,1]$, $f_n\circ g_n(x)=x$; for all $x\in[a_n,1]$, $g_n\circ f_n(x)=x$.

* $f_n(x)\geq f_{n+1}(x)$ and $g_n(x)\geq g_{n+1}(x)$.

* $f_n\circ f_m(x)=f_{nm}(x)=f_m\circ f_n(x)$.

* $g_n\circ g_m(x)=g_{nm}(x)=g_m\circ g_n(x)$.

* For each natural number $k\geq 1$, $f_{kn}\circ g_{km}=f_n\circ g_m$

(i)-(v)   are left to the reader. (vi) follows from the fact that all of the functions  $g_n\circ g_m$, $g_{nm}$ and $g_m\circ g_n$ are right adjoint to $f_{nm}$. As for (vii),
since $f_{kn}\circ g_{km}=f_n\circ f_k\circ g_k\circ g_m$ by (v) and (vi), the conclusion follows from (iii) immediately.

For each rational number $r=-n/m<0$, define
$$f_r\colon [0,1]\lra [0,1]$$
\[f_r=f_n\circ g_m.\]

Lemma <ref>(vii) ensures that $f_r$ is well-defined.

Fix some $c\in(0,1)$. Define a function
\[ h\colon \mathbb{Q}\cap(-\infty,0]\lra[0,1]\]
$$h(0)=1,\quad \text{$h(r)=f_r(c)$ for all $r<0$}.$$

* If $r<s$ and $h(s)>0$, then $h(r)<h(s)$.

* For all $r,s\in (-\infty,0]$, $h(r+s)=h(r)\with h(s)$.

* $\lim\limits_{r\ra-\infty}h(r)=0$.

* $h$ is uniformly continuous.

(i) Assume that $r=-m/k, s=-n/k$ and $m>n$. Then
\[h(s) =f_n\circ g_k(c)=g_k(c)^n >g_k(c)^{n+1}\geq g_k(c)^m=f_m\circ g_k(c) =h(r).\]

(ii) Assume that $r=-m/k, s=-n/k$. Then
\[h(r)\with h(s)=(f_m\circ g_k(c))\with  (f_n\circ g_k(c)) =g_k(c)^{m+n}= f_{m+n}\circ g_k(c)=h(r+s).\]

(iii) This follows from (i) and that $\lim\limits_{n\ra\infty}h(-n)= \lim\limits_{n\ra\infty}f_n(c) =\lim\limits_{n\ra\infty}c^n=0$.

(iv) It suffices to show that for each $\epsilon>0$, there is a rational number $l>0$ such that for all  $r\in\mathbb{Q}\cap(-\infty,0]$,
\[ h(r+l)-\epsilon\leq h(r)\leq h(r-l)+\epsilon,\]
where we agree  that $h(r+l)=1$ if $r+l>0$.

Since $\&$ is continuous and $[0,1]^2$ is compact, there exists   $\delta>0$ such that
\[|x\with y-u\with v|<\epsilon\]
whenever $|x-u|<\epsilon$ and $|y-v|<\epsilon$. Since $g_n(c)$ tends to $1$ when $n$ tends to infinity, there is a natural number $N$ such that $1-g_N(c)<\delta$. Let   $l=1/N$. Then
$$1-h(-l)= 1-f_1\circ g_N(c)= 1-g_N(c)<\delta.$$
We claim that $l$ satisfies the requirement.

\[h(r)-h(r-l)=|h(r)\with 1-h(r)\with h(-l)|\leq \epsilon,\]
it follows that $h(r)\leq h(r-l)+\epsilon$.
As for the inequality $h(r+l)-\epsilon\leq h(r)$, we proceed with two cases. If $r+l\leq0$, then
\[h(r+l)-h(r)= |h(r+l)\with 1-h(r+l)\with h(-l)|\leq \epsilon.\]
If $r+l>0$, then
\[h(r+l)-h(r)= 1- h(r) \leq 1-h(-l)\leq \epsilon.\]
In either case we have $h(r+l)-\epsilon\leq h(r)$.

Consider the function
$$h\colon \mathbb{Q}\cap(-\infty,0]\lra[0,1]$$
given in Lemma <ref>. Since $h$ is uniformly continuous and $\lim\limits_{r\ra-\infty}h(r)=0$, then $h$  can be extended to a continuous function
$$\overline{h}\colon   [-\infty,0]\lra[0,1].$$
By continuity of $\&$ and Lemma <ref>(ii), for all $u,v\in[-\infty,0]$  we have
\[\overline{h}(u+v)=\overline{h}(u)\with\overline{h}(v). \]

Moreover, if $u<v$ and $\overline{h}(v)>0$, then $\overline{h}(u)<\overline{h}(v)$ by Lemma <ref>(i).

Since $\overline{h}\colon[-\infty,0] \lra[0,1]$ preserves joins, it has a right adjoint
$$t\colon [0,1]\lra[-\infty,0].$$
We claim that $t$ satisfies the requirement.

By definition $\overline{h}$ is the left adjoint of $t$, i.e. $t^\dashv =\overline{h}$.  It is readily verified that $t(1)=0$ and that $t$ is continuous and strictly increasing, in particular, $t$ is injective. Then by Proposition <ref>, $t^\dashv (t(x))=x$ for all $x\in [0,1]$. Therefore,  for all $x,y\in[0,1]$ we have
\begin{align*}x\with y&=t^\dashv (t(x))\with t^\dashv (t(y)) \\ &=\overline{h}(t(x))\with\overline{h}(t(y))\\ &=\overline{h}(t(x)+t(y))\\ &=t^\dashv (t(x)+t(y)),\end{align*}
which completes the proof.

An immediate consequence of Theorem <ref> is the following conclusion which asserts that there are essentially only two  continuous Archimedean t-norms: the product  and the Łukasiewicz.

Suppose $\&$ is a continuous Archimedean t-norm  on $[0,1]$.

* If $\&$ has a nonzero nilpotent element, then $\&$ is isomorphic to the Łukasiewicz t-norm.

* If $\&$ has no nonzero nilpotent element, then $\&$ is isomorphic to the product t-norm.

By Theorem <ref> there is a continuous and strictly increasing function $t\colon [0,1]\lra[-\infty,0]$ with $t(1)=0$ such that
$$x\with y=t^\dashv (t(x)+t(y))$$
for all $x,y\in[0,1]$.

(i) If $\&$ has a nonzero nilpotent element, we must have $t(0)>-\infty$. Then $\&$ is isomorphic to the truncated addition on $[t(0),0]$, hence to the Łukasiewicz t-norm on $[0,1]$.

(ii) If $\&$ has no nonzero nilpotent element, we must have $t(0)=-\infty$. Then $\&$ is isomorphic to the addition on $[-\infty,0]$, hence to the product t-norm on $[0,1]$.

Suppose $\&$ is a left continuous t-norm on $[0,1]$. We say that $([0,1],\&,1)$ (or simply $\&$) satisfies the law of double negation if
\[(x\ra0)\ra0 =x\]
for all $x\in [0,1]$.

Both the Łukasiewicz t-norm and the nilpotent minimum satisfy the law of double negation.

Let $\&$ be a continuous t-norm on $[0,1]$.  Then $\&$ satisfies the law of double negation if and only if $\&$ is isomorphic to the Łukasiewicz t-norm.

First we show that $\&$ is Archimedean. Otherwise it has a nontrivial idempotent element, say $b$. Then by Corollary <ref>, $b\ra0=0$, hence $(b\ra0)\ra0 =1$, contradicting that $\&$ satisfies the law of double negation. Next, since the product does not satisfy the law of double negation, then by Corollary <ref>, $\&$ must be isomorphic to  the Łukasiewicz t-norm.

5pt Ordinal sum decomposition theorem

Suppose   $\{(a_i,b_i)\}_{i\in I}$ is a countable set of pairwise disjoint open intervals in $[0,1]$ and  for each  $i\in I$,   $\&_{i}$ is a t-norm on the closed interval $[a_{i},b_{i}]$.  It is readily verified that
$$\with \colon [0,1]\times[0,1]\lra[0,1],\quad  x\with y=\begin{cases}
x\with_i y  &  (x,y)\in [a_{i},b_{i}]^{2},     \\
\min\{x,y\}  &   {\rm otherwise}
\end{cases}$$
is  a t-norm on $[0,1]$, called the ordinal sum of $\{\&_i\}_{i\in I}$. Each $\with_i$ is called a summand  of $\&$. The ordinal sum is (left) continuous if and only if so is each of its summands. In the case that the ordinal sum $\&$ is left continuous, its implication operator is given by
\begin{equation*}%\label{implication of sum}
x\ra y=\begin{cases} 1 & x\leq y, \\ x\ra_i y  & a_i\leq y<  x\leq b_i~\text{for some }i,\\ y &{\rm otherwise}, \end{cases} \end{equation*}
where $\ra_i$ is the implication operator of the t-norm $\&_i$ on $[a_i,b_i]$.

A fundamental result says that every continuous t-norm is an ordinal sum of continuous Archimedean t-norms. To see this, let $\&$ be a continuous t-norm on $[0,1]$ and let $E$ be the set of idempotent elements of $\&$. Then $E$ is a closed set by continuity of $\&$. The open set $[0,1]\setminus E$ is a countable union of disjoint open intervals $\bigcup_{i\in I}(a_i,b_i)$. By Corollary <ref>,   the restriction of $\&$ on each $[a_i,b_i]$ is a continuous t-norm without nontrivial idempotent elements, hence a continuous Archimedean t-norm. By Corollary <ref>, $x\with y=\min\{x,y\}$ whenever $(x,y)\notin\bigcup_{i\in I}[a_i,b_i]^2$. This proves the following ordinal sum decomposition theorem for continuous t-norms.

(Mostert and Shields <cit.>) Every continuous  t-norm $\&$ on $[0,1]$ is an ordinal sum of continuous Archimedean t-norms. This means, for each continuous t-norm $\&$ on $[0,1]$, there exists a countable set of disjoint open intervals $\{(a_i,b_i)\}_{i\in I}$ in $ [0,1]$ such that

* for each $i\in I$, both $ a_i$ and $ b_i$ are idempotent and the restriction of $\&$ on the closed interval $ [a_i,b_i] $ is Archimedean;

* $x\with y= \min\{x,y\}$ whenever $(x,y)\notin\bigcup_{i\in I}[a_i,b_i]^2$.

Each $[a_i,b_i]$  with the restriction of $\&$ is called an Archimedean block of $\&$.

Suppose $\&$ is a continuous t-norm on $[0,1]$. For each $p\in[0,1]$, let $p^-$ be the greatest idempotent element in $[0,p]$ and let $p^+$ be the least idempotent element in $[p,1]$. Then the following  are equivalent:

* For each non-idempotent element $p\in[0,1]$, the restriction of $\&$ on $[p^-,p^+]$ is isomorphic to the product t-norm on $[0,1]$ whenever $p^->0$.

* The implication  operator $\ra\colon [0,1]^2\lra[0,1]$ is  continuous  at every point off the diagonal $\{(x,x)\mid x\in[0,1]\}$.

* For each $p\in(0,1]$, the function $p\ra -\colon[0,1]\lra[0,1]$ is continuous  on the interval $[0,p)$.

$(1)\Rightarrow(2)$  It suffices to show that $\ra$ is continuous at each point $(x,y)$ with $x>y$. If there is some idempotent element $b$ satisfying $y<b<x$, then $(b,1]\times[0,b)$ is a neighborhood of $(x,y)$  such that $x'\ra y' =y'$ for all $(x',y')\in (b,1]\times[0,b)$, hence $\ra$ is continuous at $(x,y)$. If there is no idempotent element between $x$ and $y$, then $x$ and $y$ belong to the same block by the ordinal sum decomposition theorem. That means, there exists a non-idempotent $p$ such that $p^-\leq y<x\leq p^+$ and that the restriction of $\&$ on $[p^-,p^+]$ is a continuous Archimedean t-norm.  Now we proceed with two cases.

Case 1. $p^-=0$. Then the conclusion follows from the fact that the implication operator of a continuous Archimedean t-norm on $[0,1]$ is continuous except possibly at $(0,0)$.

Case 2. $p^-\not=0$. In this case, the restriction of $\&$ on $[p^-,p^+]$ is isomorphic to the product t-norm on $[0,1]$.
We only need to show that for all $x>p^-$, the function $x\ra-\colon[0,1]\lra[0,1]$ is continuous at $p^-$. Since $ x\ra z=z$ for each $z<p^-$, the right limit of $ x\ra -$ at $p^-$ is equal to $p^-$,  then $\lim\limits_{z\ra p^-}(x\ra z)= p^-= x\ra p^-$,
hence $x\ra-$ is continuous at $p^-$.

$(2)\Rightarrow(3)$ Obvious.

$(3)\Rightarrow(1)$ Suppose on the contrary that there is some non-idempotent element $p$ for which $p^->0$ and the restriction of $\&$ on $[p^-,p^+]$ is isomorphic to the Łukasiewicz t-norm. Then the function $p\ra-\colon[0,1]\lra[0,1]$ is not continuous at $p^-$, a contradiction.

Thus, for a continuous t-norm $\&$, the implication  operator
$$\ra\colon [0,1]^2\lra[0,1]$$
is  continuous  at every point off the diagonal $\{(x,x)\mid x\in[0,1]\}$ if and only if it is almost Łukasiewicz free in the sense that $\&$ has at most one Archimedean block that is isomorphic to the  Łukasiewicz t-norm and that block  should start at $0$. Such t-norms are first studied by Morsi  [91, 92].

From now on, we always assume that $\with$ is a continuous t-norm on the interval $[0,1]$.

§ REAL-ENRICHED CATEGORIES

Suppose $X,Y$ are sets. A $[0,1]$-relation   $r\colon X\rto Y$ is a map $r\colon X\times Y\lra [0,1]$, the value $r(x,y)$ is interpreted as the degree that $x$ is  related to $y$. For a $[0,1]$-relation $r\colon X\rto Y$, the $[0,1]$-relation
$$r^{\rm op}\colon Y\rto X,\quad r^{\rm op}(x,y)=r(y,x) $$
is called the  opposite of $r$.

$[0,1]$-relations can be composed.
Given $[0,1]$-relations $r\colon X\rto Y$  and $s\colon Y\rto Z$, the composite $s\circ r$ is the $[0,1]$-relation $X\rto Z$   given by
\[s\circ r(x,z)=\sup_{y\in Y}s(y,z)\with r(x,y).\]

For each set $X$, write $\id_X\colon X\rto X$ for the $[0,1]$-relation given by
\[\id_X(x,y)=\begin{cases} 1 & x=y,\\
0 & x\not=y.\end{cases}\]
Then $\id_X$  is a unit  for the composition of $[0,1]$-relations; that is, $r\circ \id_X=r=\id_Y\circ r$ for each $[0,1]$-relation $r\colon X\rto Y$. So, we call $\id_X$ the identity relation on $X$.

The composition of $[0,1]$-relations is associative.
Therefore, with sets as objects and $[0,1]$-relations as arrows we obtain a category
\[[0,1]\text{-}{\sf Rel}.\]
The category $[0,1]\text{-}{\sf Rel}$ has an important property: for all objects $X,Y$,  the  hom-set
$$[0,1]\text{-}{\sf Rel}(X,Y)$$
is a complete lattice under the pointwise order,  the composition of $[0,1]$-relations preserves joins in both variables.

For all sets $X,Y,Z$, the composition
\[\circ\colon [0,1]\text{-}{\sf Rel}(Y,Z)\times[0,1]\text{-}{\sf Rel}(X,Y)\lra [0,1]\text{-}{\sf Rel}(X,Z)\]
preserves joins on each variable.

Therefore, $[0,1]\text{-}{\sf Rel}$ is a quantaloid in the sense of [97, 105].
This fact has far-reaching consequences, it plays a crucial role in the calculus of $[0,1]$-relations, as we see now.

For each $[0,1]$-relation   $r\colon X\rto Y$, since $\circ$ preserves joins in each variable, the map
\[-\circ r\colon [0,1]\text{-}{\sf Rel}(Y,Z)\lra [0,1]\text{-}{\sf Rel}(X,Z)\]
has a right adjoint
\[-\swarrow r\colon [0,1]\text{-}{\sf Rel}(X,Z)\lra[0,1]\text{-}{\sf Rel}(Y,Z).\]
Explicitly, for each   $t\colon X\rto Z$, the $[0,1]$-relation $t\swarrow r\colon  Y\rto Z$ is computed as follows:
\[(t\swarrow r)(y,z)=\inf_{x\in X}(r(x,y)\ra t(x,z)).\]

$$\bfig \ptriangle(0,0)/>`>`>/<520,500>[X`Y`Z;r`t`t\swarrow r]
\place(260,500)[\rotatebox{0}{$\mapstochar$}] \place(260,255)[\rotatebox{45}{$\mapstochar$}] \place(5,255)[\rotatebox{90}{$\mapstochar$}] \place(150,350)[\geq]
\efig$$

Likewise, for each $[0,1]$-relation   $s\colon Y\rto Z$, the map
\[s\circ -\colon [0,1]\text{-}{\sf Rel}(X,Y)\lra [0,1]\text{-}{\sf Rel}(X,Z)\]
has a right adjoint
\[s\searrow -\colon [0,1]\text{-}{\sf Rel}(X,Z)\lra[0,1]\text{-}{\sf Rel}(X,Y) \]
given by
\[(s\searrow t)(x,y)=\inf_{z\in Z}(s(y,z)\ra t(x,z)) \]
for all   $t\colon X\rto Z$ and $x\in X$ and $y\in Y$.

$$\bfig \qtriangle(0,0)/>`<-`<-/<520,500>[Y`Z`X;s`s\searrow t`t]
\place(260,500)[\rotatebox{0}{$\mapstochar$}] \place(255,255)[\rotatebox{-45}{$\mapstochar$}] \place(525,255)[\rotatebox{90}{$\mapstochar$}] \place(350,350)[\leq]
\efig$$

Some basic properties of the $[0,1]$-relation calculus are listed below:

* For all $[0,1]$-relations $r\colon X\rto Y$, $s\colon Y\rto Z$ and $t\colon X\rto Z$,
\[s\leq t\swarrow r\iff s\circ r\leq t\iff r\leq s\searrow t. \]

* For all   $r\colon X\rto Y$ and $s\colon X\rto Y$,
\[\id_X\leq r\searrow s\iff r\leq s\iff \id_Y\leq s\swarrow r.\]

* For all $r\colon X\rto Y$, $s\colon Y\rto Z$ and $t\colon X\rto W$,
\[t\swarrow (s\circ r)= (t\swarrow r)\swarrow s.\]

* For all $r\colon X\rto Y$, $s\colon Y\rto Z$ and $t\colon W\rto Z$,
\[(s\circ r)\searrow t= r\searrow(s\searrow t).\]

* For all $r\colon X\rto Y$, $s\colon W\rto Z$ and $t\colon X\rto Z$,
\[(s\searrow t)\swarrow r= s\searrow(t\swarrow r).\]

A real-enriched category is a pair $(X, \alpha)$, where $X$ is a set and   $\alpha\colon X\times X\lra[0,1]$ is a function  such that
* $\alpha(x,x)= 1$ for all $x\in X$;
* $ \alpha(y,z)\with \alpha(x,y)\leq \alpha(x,z)$ for all $x,y,z\in X$.

If we interpret  the value $\alpha(x,y)$  as the truth degree that $x$ is smaller than or equal to $y$, then  (i) is reflexivity and (ii) is transitivity. So, real-enriched categories can be viewed as many-valued ordered sets.

To ease notations, for a real-enriched category $(X, \alpha)$,   we often omit the symbol $\alpha$ and write $X(x,y)$ for $\alpha(x,y)$.

Two elements $x$ and $y$ of a real-enriched category $X$ are isomorphic if $X(x,y)=1=X(y,x)$. A real-enriched category $X$ is   separated if its isomorphic elements are identical.

* The pair $([0,1],\alpha_L)$ is a separated real-enriched category, where   for all $x,y\in [0,1]$, $\alpha_L (x,y)= x\ra y$.
The opposite  of $([0,1],\alpha_L)$ is  denoted by $([0,1],\alpha_R)$; that means, $\alpha_R (x,y)=y\ra x.$ In the language of enriched category theory,  $\alpha_L$ is the internal hom of (the truth-value set) $([0,1],\with,1)$. In the sequel we  write $\sV$ for   $([0,1],\alpha_L)$, hence $\sV^{\rm op}$ for $([0,1],\alpha_R)$. Both $\sV$ and $\sV^{\rm op}$ play  an important role in the theory of real-enriched categories.

* For each set $X$, $([0,1]^X,\sub_X)$ is a separated real-enriched category, where   for all $\lam,\mu\in [0,1]^X$,

$$\sub_X(\lam,\mu) =\inf_{x\in X}\lam(x)\ra \mu(x). $$

If we, following Zadeh [122], view $\lam$ and $\mu$ as fuzzy subsets of $X$,  then the value $\sub_X(\lam,\mu)$ measures the truth degree that $\lam$ is contained in $\mu$, so the category $([0,1]^X,\sub_X)$ is also known as the enriched powerset of $X$.

Suppose $X, Y$ are real-enriched categories. A functor   $f\colon X\lra Y$  is a map such that  $X(x,y)\leq Y(f(x),f(y))$ for all $x,y\in X$. The category of real-enriched categories and functors is denoted by
\[[0,1]\text{-}{\bf Cat}.\]

Given an ordered set $(X,\leq)$, define a map $\omega(\leq)\colon X\times X\lra[0,1]$ by $\omega(\leq)(x,y)=1$ if $x\leq y$, and $\omega(\leq)(x,y)=0$ otherwise. Then $(X,\omega(\leq))$ is a real-enriched category. In this way we obtain a full and faithful  functor
$$\omega\colon{\bf Ord}\lra\QOrd.$$

The functor $\omega$ has both a left and a right adjoint. We spell out the right adjoint here. For each real-enriched category $X$, define a binary relation $\sqsubseteq$ on $X$ by $x\sqsubseteq y$ if $X(x,y)=1.$ Then $\sqsubseteq$ is  a reflexive and transitive relation, hence an order   on $X$, called the underlying order of  $X$. We write $X_0$ for the ordered set $(X,\sqsubseteq)$. The assignment $X\mapsto X_0$ defines a functor
$$(\text{-})_0\colon\QOrd\lra {\bf Ord},$$
a right adjoint of $\omega$.

Suppose $X,Y$ are real-enriched categories, $f\colon X\lra Y$ and $g\colon Y\lra X$ are functors.  We say that $f$ is left adjoint to $g$, or $g$ is right adjoint to $f$, and write $f\dashv g$, if
for all $x\in X$ and $y\in Y$,

The pair $(f,g)$ is  called an adjunction, or a(n enriched) Galois connection. This is an extension of Galois connections between ordered sets; and a special case of adjunction in the theory of enriched categories.

For each pair of maps $f\colon X\lra Y$ and $g\colon Y\lra X$   between real-enriched categories, the following are equivalent:
* Both  $f$ and $g$ are  functors, and $f$ is left adjoint to $g$.

* Both $f$ and $g$ are  functors, and
$f\colon X_0\lra Y_0$ is left adjoint to $g\colon Y_0\lra X_0$.

* For all $x\in X$ and $y\in Y$, $Y(f(x),y)=X(x,g(y))$.

$(1)\Rightarrow(2)$ Obvious.

Since $f(x)\sqsubseteq f(x)$, then $x\sqsubseteq gf(x)$, hence $1\leq X(x,gf(x))$ for all $x\in X$. Likewise, $1\leq Y(fg(y),y)$ for all $y\in Y$. So  we have

\[X(x,g(y))\leq   Y(fg(y),y)\with Y(f(x),fg(y))\leq
\[Y(f(x),y)\leq
X(gf(x),g(y))\with X(x,gf(x))\leq X(x,g(y)).\]

$(3)\Rightarrow(1)$ It suffices to check that  $f$ and $g$ are  functors. For  all $x_1,x_2\in X$,
\begin{align*}X(x_1,x_2)&\leq Y(f(x_2),f(x_2))\with X(x_1,x_2) \\ &= X(x_2,gf(x_2))\with X(x_1,x_2)\\ &\leq X(x_1,gf(x_2))  \\ &=Y(f(x_1),f(x_2)),\end{align*}
it follows that $f$ is a  functor. Likewise, $g$  is a  functor.

Suppose $X,Y$ are set,  $f\colon X\lra Y$ is a map. Then,
$$f^{-1} \colon ([0,1]^Y,\sub_Y)\lra ([0,1]^X,\sub_X), \quad  f^{-1}(\mu)=\mu\circ f$$
is a functor, it has both a left adjoint and a right adjoint. The left adjoint
$$f_\exists\colon ([0,1]^X,\sub_X)\lra ([0,1]^Y,\sub_Y) $$
is given by
$$ f_\exists(\lam)(y)=\sup\{\lam(x)\mid f(x)=y\};$$
the right  adjoint
$$f_\forall\colon ([0,1]^X,\sub_X)\lra ([0,1]^Y,\sub_Y) $$
is given by
$$ f_\forall(\lam)(y)=\inf\{\lam(x)\mid f(x)=y\}.$$
Therefore, we have a string of adjunctions
$$ f_\exists\dashv f^{-1}\dashv f_\forall.$$

§ THE YONEDA LEMMA

Suppose $X, Y$ are real-enriched categories. A distributor $\phi$ from $X$ to $Y$, written $\phi\colon X\oto Y$, is a map $\phi\colon X\times Y\lra [0,1]$ such that

* for all $x_1,x_2\in X$ and $y\in Y$, $\phi(x_2,y)\with X(x_1,x_2)\leq \phi(x_1,y)$;
* for all $x\in X$ and $y_1,y_2\in Y$,   $Y(y_1,y_2)\with\phi(x,y_1)\leq \phi(x,y_2)$.

If   $\phi$ is a distributor from $X$ to $Y$, then  $\phi^{\rm op}(y,x)\coloneqq\phi(x,y)$ is a distributor from   $Y^{\rm op}$ to  $X^{\rm op}$.  As we have agreed, for each real-enriched category $X$,  the symbol “$X$” also denotes  the real-enriched category structure  on $X$. So,  a distributor $\phi\colon X\oto Y$ is  a $[0,1]$-relation $\phi\colon X\rto Y$ such that $\phi\circ X\leq \phi$ and $Y\circ\phi\leq\phi$, or equivalently, $\phi\circ X=\phi=Y\circ\phi$.

Let $\phi\colon X\oto Y$ be a distributor. Fixing the first argument $x$ gives a functor
$$\phi(x,-)\colon Y\lra \sV, \quad y\mapsto \phi(x,y),$$
where $\sV=([0,1],\alpha_L)$; fixing the second argument $y$ gives a functor
$$\phi(-,y)\colon X^{\rm op}\lra \sV, \quad x\mapsto\phi(x,y).$$

Every functor $f\colon X\lra Y$ induces a pair of distributors:
\[f_*\colon X\oto Y, \quad f_*(x,y)=Y(f(x),y)\]
\[f^*\colon Y\oto X, \quad f^*(y,x)=Y(y,f(x)).\]
The  distributors $f_* $  and $f^* $ are called,  respectively, the  graph  and the  cograph  of $f$.

A  functor $f\colon X\lra Y$ is   fully faithful if  for all $x_1,x_2\in X$,
\[X(x_1,x_2)=Y(f(x_1),f(x_2)).\]
We leave it to the reader to check that a functor  $f\colon X\lra Y$ if is fully faithful   if and only if $f^*\circ f_*=X$.

For every pair of real-enriched categories $X$ and $Y$, the set of distributors from $X$ to $Y$  is a subset of the   $[0,1]$-relations from $X$ to $Y$. The following proposition says that this subset is closed with respect to various operations on $[0,1]$-relations.

Suppose $X,Y,Z$ are real-enriched categories.

* For any distributors $\phi\colon X\oto Y$ and $\psi\colon Y\oto Z$, the composite $\psi\circ\phi$ is a distributor $X\oto Z$.

* For each family $\{\phi_i\}_i$ of distributors $X\oto Y$, both the pointwise join $\sup_i\phi_i$ and the pointwise meet $\inf_i\phi_i$ are distributor  $X\oto Y$.

* For any distributors $\phi\colon X\oto Y$, $\psi\colon Y\oto Z$ and $\xi\colon X\oto Z$, both of the $[0,1]$-relations $\xi\swarrow\phi$ and $\psi\searrow\xi$ are distributors.

With real-enriched categories as objects and distributors as morphisms we have a category, indeed a quantaloid
\[[0,1]\text{-}{\bf Dist},\]
called the category of  distributors.

Let $\phi\colon Y\oto X$ and $\psi\colon X\oto Y$ be a pair of distributors. We say   $\phi$ is right adjoint to   $\psi$,  or $\psi$ is left adjoint to $\phi$, and write $\psi\dashv\phi$, if $X\leq \phi\circ\psi$ and $\psi\circ\phi\leq Y$.

For each  functor $f\colon X\lra Y$, the graph $f_*\colon X\oto Y$ is left adjoint to the cograph $f^*\colon Y\oto X$.

The following lemma, contained in Heymans <cit.>, is very useful in the calculus of distributors.

Suppose   $\psi\colon X\oto Y$ is left adjoint to $\phi\colon Y\oto X$.

* For each distributor $\xi\colon Y\oto Z$, $\xi\circ \psi=\xi\swarrow \phi$. In particular, $\psi=Y\swarrow\phi$.

* For each distributor $\lambda\colon W\oto Y$, $\phi\circ \lambda =\psi\searrow \lambda$. In particular, $\phi=\psi\searrow Y$.

We prove (i)  for example. On the one hand, since $\xi\circ\psi\circ\phi \leq \xi\circ Y= \xi$, then $\xi\circ\psi\leq\xi\swarrow \phi$.  On the other hand, since  $(\xi\swarrow \phi)\circ\phi\circ\psi\leq\xi\circ\psi$, then $\xi\swarrow \phi\leq \xi\circ\psi$.

In particular, a distributor has at most one right (left, resp.) adjoint. So, we shall speak of the right (left, resp.) adjoint of a distributor.

Suppose $f\colon X\lra Y$ and $g\colon Y\lra X$ are functors. The following  are equivalent:

* $f$ is left adjoint to $g$.

* $f_*=g^*$.

* As distributors, $g_*$ is left adjoint to  $f_*$.

* As distributors, $g^*$ is left adjoint to   $f^*$.

Suppose $X$ is a real-enriched category. A weight of $X$  is defined to be a distributor $\phi\colon X\oto \star$ from $X$ to the terminal real-enriched category $\star$. In other words, a weight of $X$ is a functor $\phi\colon X^{\rm op}\lra \sV$, where $\sV=([0,1],\alpha_L)$.  Weights of $X$ constitute a real-enriched category $\mathcal{P}X$ with
$$\mathcal{P}X(\phi_1,\phi_2) =\phi_2\swarrow\phi_1=\sub_X(\phi_1,\phi_2) .$$

$$\bfig \ptriangle(0,0)/>`>`>/<500,500>[X`\star`\star;\phi_1`\phi_2`\phi_2\swarrow \phi_1]
\place(230,500)[\circ] \place(250,250)[\circ] \place(0,275)[\circ] %\place(180,350)[\geq]
\efig$$

Dually, a distributor of the form $\star\oto X$ is called a coweight of $X$. In other words, a coweight of $X$ is  a  functor $\psi\colon X\lra \sV$, where $\sV=([0,1],\alpha_L)$.
All coweights of $X$ constitute a real-enriched category $\CPd X$ with
$$\CPd X(\psi_1,\psi_2) =\psi_2\searrow\psi_1=\sub_X(\psi_2,\psi_1) .$$

$$\bfig \qtriangle(0,0)/>`<-`<-/<500,500>[\star`X`\star;\psi_2`\psi_2\searrow \psi_1`\psi_1]
\place(250,500)[\circ] \place(250,250)[\circ] \place(500,265)[\circ]
\efig$$

Every functor $f\colon X\lra \sV$ determines a distributor  $\star\oto X, ~(\star,x)\mapsto f(x)$; conversely, every distributor  $\psi\colon\star\oto X$ determines a functor $X\lra\sV,~x\mapsto\psi(\star,x)$.  So, sometimes we'll view a functor $X\lra \sV$ as a distributor $\star\oto X$, and vice versa. To avoid confusion of notations, we make an agreement here.
* For each functor $f\colon X \lra\sV$,   $f^\sharp$ denotes the   distributor $\star\oto X, ~(\star,x)\mapsto f(x)$. For each distributor $\psi\colon \star\oto X$,   $\psi^\sharp$   denotes the corresponding functor $X \lra\sV,~x\mapsto\psi(\star,x)$.

* For each functor $g\colon X^{\rm op}\lra\sV$,   $g^\sharp$ denotes the   distributor $X\oto\star,~(x,\star)\mapsto g(x)$. For each distributor $\phi\colon X\oto\star$,   $\phi^\sharp$   denotes the  corresponding functor $X^{\rm op}\lra\sV,x\mapsto\phi(x,\star) $.

For each real-enriched category $X$, the underlying order of $\CP X$ coincides with the (pointwise) order on $[0,1]$-${\bf Dist}(X,\star)$, while the underlying order of  $\CPd  X$ is the reverse order  on  $[0,1]$-${\bf Dist}(\star,X)$. That is to say, for any weights $\phi_1,\phi_2\colon X\oto\star $,
\[\phi_1\sqsubseteq\phi_2~~ {\rm in}~~ (\CP X)_0\iff  \forall x\in X, ~\phi_1(x) \leq\phi_2 (x);
\]
while for any coweights $\psi_1,\psi_2\colon \star\oto X$,
\[\psi_1\sqsubseteq\psi_2~~ {\rm in}~~ (\CPd  X)_0\iff \forall x\in X,~ \psi_1(x) \geq\psi_2(x).
\]

Suppose $X,Y$ are real-enriched categories, $f\colon X\lra Y$ is a functor. The map
\[f^{-1}\colon\CP  Y\lra \CP X,\quad f^{-1}(\gamma)=\gamma\circ f_*\]
is   readily verified to be a functor. The functor $f^{-1}$ has at the same time a left adjoint $f_\exists$ and a right adjoint $f_\forall$. The left adjoint is given by
\[f_\exists\colon\CP  X\lra\CP Y, \quad f_\exists(\phi)=\phi\circ f^*;  \]
the  right adjoint is given by
$$f_\forall\colon \CP X\lra \CP Y, \quad f_\forall(\phi)=  \phi\swarrow f_*.$$
In the vocabulary of category theory,  $f_\exists$ and  $f_\forall$ are called the left and the right Kan extension of $f$, respectively.

Dually, the functor
\[f^{-1}\colon\CPd  Y\lra \CPd X,\quad f^{-1}(\mu)=f^*\circ\mu \]
also has  a left adjoint $f^\dag_\forall$  and a right adjoint $f^\dag_\exists$. Before spelling out the left adjoint and the right adjoint, we would like to remind the reader that the underlying order of $\CPd X$ is opposite to the pointwise order of coweights. The left adjoint of $f^{-1}$ is
\[f^\dag_\forall\colon\CPd  X\lra\CPd Y, \quad f^\dag_\forall(\psi)=f^*\searrow\psi;  \]
the  right adjoint is
$$f^\dag_\exists\colon \CPd X\lra \CPd Y, \quad f^\dag_\exists(\psi)= f_*\circ \psi.$$

For each functor $f\colon X\lra Y$,  we have
$$ f_\exists\dashv f^{-1}\dashv f_\forall\colon \CP X\lra\CP Y$$
$$f^\dag_\forall\dashv f^{-1}\dashv f^\dag_\exists\colon \CPd X\lra \CPd Y.$$

Suppose $f\colon X\lra Y$ is a fully faithful functor.

* For all $\phi\in\CP X$, $f^{-1}\circ f_\exists(\phi) =\phi =f^{-1}\circ f_\forall(\phi)$.
* For all $\psi\in\CPd X$, $f^{-1}\circ f^\dag_\exists(\psi) =\psi =f^{-1}\circ f^\dag_\forall(\psi)$.

(i) Since $f$ is fully faithful,  $f^*\circ f_*=X$, then  for all $\phi\in\CP X$,
$$f^{-1}\circ f_\exists(\phi)=\phi\circ f^*\circ f_*=\phi$$
\begin{align*} f^{-1}\circ f_\forall(\phi)&=(\phi\swarrow f_*)\circ f_*\\ &= (\phi\swarrow f_*)\swarrow f^* &(f_*\dashv f^*)\\ &= \phi\swarrow(f^*\circ f_*)\\ &=\phi.  \end{align*}

(ii) Similar.

Assigning to each $f\colon X\lra Y$ the functor
$$\CP f\coloneqq f_\exists\colon \CP X\lra\CP Y$$
defines a functor
$$\CP\colon\QOrd\lra\QOrd, $$
called the presheaf functor.[The terminology comes from the fact that,  in category theory, a contravariant functor from a category   to the category of sets is called a presheaf of that category.]   This functor plays a central role in the theory of real-enriched categories.
Dually,  we have a copresheaf functor
$$\CPd\colon\QOrd\lra\QOrd $$
that maps $f\colon X\lra Y$ to
\[\CPd f\coloneqq f_\exists^\dag\colon\CPd  X\lra\CPd  Y,  \quad \psi\mapsto f_*\circ\psi.  \]

For an element $a$ of a real-enriched category $X$, we  write $\sy(a)$ for the weight
\[  X\oto\star, \quad x\mapsto X(x,a), \]
write $\syd(a)$ for the coweight
\[ \star\oto X, \quad x\mapsto X(a,x).\]
Weights of the form $\sy(a)$  and coweights of the form $\syd(a)$ are said to be  representable.
As distributors, the representable coweight $\syd(a)\colon \star\oto X$ is left adjoint to the representable weight $\sy(a)\colon X\oto\star$.

The following lemma, a special case of the Yoneda lemma in the theory of enriched categories, implies that for each real-enriched category $X$, both of
\[\sy\colon X\lra\CP X, \quad a\mapsto \sy(a)\]
\[\syd\colon X\lra\CPd  X, \quad a\mapsto \syd(a)\]
are fully faithful functors.

Suppose $X$ is a real-enriched category and $a\in X$.

* For each   $\phi\in\CP X$, $\CP X(\sy(a),\phi)=\phi(a)$.

* For each   $\psi\in\CPd  X$, $\CPd  X(\psi,\syd(a))=\psi(a)$.

(i) Since
\[\phi(a)=\sup_{x\in X}\phi(x)\with X(a,x)=\phi\circ\syd(a)\]
and $\sy(a)$ is right adjoint to $\syd(a)$,    by Lemma <ref> we have
\[\phi(a)=\phi\circ\syd(a) =\phi\swarrow \sy(a)=\CP X(\sy(a),\phi).\]

(ii) Since $\psi(a)= \sy(a)\circ \psi$ and $\syd(a)$ is left adjoint to $\sy(a)$,   by Lemma <ref> we have
\begin{align*}\psi(a)&= \sy(a)\circ \psi=\syd(a)\searrow\psi =\CPd  X(\psi,\syd(a)).\qedhere\end{align*}

From the Yoneda lemma one infers that for each real-enriched category $X$, the functor $\sy\colon X\lra \CP X$ is fully faithful, so, it is called   the Yoneda embedding.
Dually, the fully faithful functor  $\syd\colon X\lra \CPd X$ is called the coYoneda embedding.

Suppose $X$ is a real-enriched category.

* $(\sy_X)_\forall=\sy_{\CP X}$, i.e. $(\sy_X)_\forall(\phi)=\CP X(-,\phi)$   for all $\phi\in\CP X$.

* $(\syd_X)^\dag_\forall=\syd_{\CPd X}$, i.e. $(\syd_X)^\dag_\forall(\psi)=\CPd X(\psi,-)$   for all $\psi\in\CPd X$.

(i) For all $\gamma\in \CP X$,
\begin{align*}\sy_\forall(\phi)(\gamma) &= (\phi\swarrow\sy_*)(\gamma)\\ &= \bw_{x\in X} (\CP X(\sy(x),\gamma)\ra\phi(x))\\ &= \bw_{x\in X}(\gamma(x)\ra\phi(x))
\\ &=\CP  X(\gamma,\phi).  \end{align*}

(ii) Similar to (i).

Yoneda embeddings constitute a natural transformation
$$\sy\colon{\rm id}\lra\CP$$
from the identity functor   to the presheaf functor $\CP$. Dually, coYoneda embeddings constitute a natural transformation
$$\syd\colon{\rm id}\lra\CPd$$
from the identity functor to the copresheaf functor $\CPd$.

There exist natural connections between  $\CP X$ and $\CPd X$. We present two examples here. The first is the Isbell adjunction.
For   $\phi\in\CP X$ and $\psi\in\CPd  X$, let
$$\ub\phi=X\swarrow\phi, \quad \lb\psi=\psi\searrow X.$$

$$\bfig \ptriangle(0,0)/>`>`>/<500,500>[X`\star`X;\phi`X`X\swarrow \phi] \place(250,500)[\circ] \place(250,250)[\circ] \place(0,250)[\circ]
\qtriangle(1000,0)/>`<-`<-/<500,500>[\star`X`X;\psi`\psi\searrow X`X] \place(1250,500)[\circ] \place(1250,250)[\circ] \place(1500,250)[\circ]
\efig$$

The functor  $\ub\colon \CP X\lra\CPd  X $ is left adjoint to the functor $\lb\colon \CPd  X\lra\CP X$.     Because
\begin{align*}\CPd  X(\ub\phi,\psi)&=\psi\searrow(X\swarrow\phi)=(\psi\searrow X)\swarrow\phi=\CP X(\phi,\lb\psi).\qedhere \end{align*}

The second is the distributor
$$\cpt\colon \CPd X\oto\CP X $$
that assigns to each pair  $(\psi,\phi)\in\CPd X\times\CP X$ the real number $\phi\circ\psi$.
A close look at this distributor in a special case may help us understand its role in the theory of real-enriched categories. Let $X$ be a set, viewed as a discrete real-enriched category. Then, $\CP X$ is $([0,1]^X,\sub_X)$ and $\CPd X$ is $([0,1]^X,\sub_X^{\rm op})$.   For each $\phi\in[0,1]^X$, the functor
$$\phi\cpt-\colon (\CPd X)^{\rm op}\lra\sV$$
obtained by fixing the second argument of $\cpt\colon \CPd X\oto\CP X$ turns out to be the functor
$$\phi\cpt-\colon([0,1]^X,\sub_X)\lra ([0,1],\alpha_L), \quad \psi\mapsto  \sup_{x\in X}\phi(x)\with\psi(x).$$
Intuitively, the value $\phi\cpt\psi=\sup_{x\in X}\phi(x)\with\psi(x)$
measures the truth degree that $\phi$ and $\psi$ have a common point.

Suppose $\phi$ is a weight and $\psi$ is a coweight of a real-enriched category $X$. For all $p\in[0,1]$,  write $\psi\ra p$ for the weight $\star_p\swarrow\psi$ and write $\phi\ra p$ for the coweight $\phi\searrow\star_p$ of $X$, where $\star_p$ denotes the distributor $\star\oto\star$ with value $p$.

Let $X$ be a real-enriched category.

* For all $\phi\in\CP X$,  $\psi\in\CPd X$ and  $p\in[0,1]$,
$\CP X(\phi,  \psi\ra p)= \phi\cpt\psi\ra p$.

* For all $\phi\in\CP X$ and  $\psi\in\CPd X$, $\phi\cpt\psi=\inf_{p\in [0,1]}(\CP X(\phi,\psi\ra p)\ra p)$.

* For all $\phi_1,\phi_2\in\CP X$, $\CP X(\phi_1,\phi_2)=\inf_{p\in [0,1]}((\phi_1\cpt(\phi_2\ra p)) \ra p)$.

§ COLIMIT AND LIMIT

A colimit of a weight $\phi\colon X\oto\star $ is an element $\colim\phi$ of $X$ such that for all $x\in X$,
\[X(\colim\phi,x)=\CP X(\phi,\sy(x)).\]

It is clear that each weight has, up to isomorphism, at most one colimit. So, we shall speak of the colimit of a weight.

\[\CP X(\phi,\sy(x))=\sy_X(x)\swarrow\phi = (X\swarrow\phi)(x) \]
for all $x\in X$, it follows that $\phi$ has a colimit if and only if the coweight $X\swarrow\phi$
is representable. A colimit of $\phi$ is then a representation of the coweight $X\swarrow\phi$, i.e. $X(\colim\phi,-)=X\swarrow\phi$.
$$\bfig \ptriangle(0,0)/>`>`>/<500,500>[X`\star`X;\phi`X`X\swarrow \phi]
\place(250,500)[\circ] \place(250,250)[\circ] \place(0,250)[\circ] \place(180,350)[\geq] \efig$$

For each functor $f\colon K\lra X$  and each weight $\phi$  of $K$, the colimit of $\phi\circ f^*$, when exists, is called the colimit of $f$ weighted by $\phi$. In particular, $\colim\phi$ is the colimit of the identity functor $X\lra X$ weighted by $\phi$.

We  write $\colim_\phi f$ for the colimit of $f$ weighted by $\phi$. By definition, $b$ is a colimit of $f\colon K\lra X$  weighted by $\phi\colon K\oto\star$ if and only if for all $x\in X$,
$$X(b,x)= \CP X(\phi\circ f^*,X(-,x))=\CP K(\phi,X(f(-),x)). $$

Suppose $f\colon X\lra\sV$ is a functor, where $\sV=([0,1],\alpha_L)$. Then for each weight $\phi\colon X\oto\star$,  the colimit of $f$ weighted by $\phi$ exists and is equal to the composite $\phi\circ f^\sharp$, i.e. $\colim_\phi f =\phi\circ f^\sharp$, where  $f^\sharp\colon\star\oto X$ is the distributor corresponding to $f$.  To see this we  calculate:  for all $r\in[0,1]$,
\begin{align*}\CP\sV(  \phi\circ f^*,\sV(-,r))& = \CP X(\phi, \sV(f(-),r))
\\   &= \inf_{x\in X}(\phi(x)\ra (f(x)\ra r)) \\ &= \Big(\sup_{x\in X}\phi(x)\with f(x)\Big)\ra r\\ &= \sV(\phi\circ f^\sharp,r),
\end{align*}
hence $\phi\circ f^\sharp$ is a colimit of $f$ weighted by $\phi$.

A real-enriched category $X$ is   cocomplete if every weight of $X$ has a colimit.

It is readily seen that $X$ is cocomplete if and only if all weighted colimits of functors to $X$ exist; that means, for each functor $f\colon K\lra X$ and each weight $\phi$ of $K$, the colimit of $f$ weighted by $\phi$ exists.

A real-enriched category $X$ is cocomplete   if and only if the Yoneda embedding $\sy\colon X\lra\CP X$ has a left adjoint.

If $X$ is cocomplete,   the correspondence $\phi\mapsto\colim\phi$ defines a left adjoint of $\sy\colon X\lra\CP X$. Conversely,   any left adjoint of $\sy\colon X\lra\CP X$ must send  a weight $\phi$ of $X$ to a colimit of $\phi$.

Example <ref> shows that   $\sV=([0,1],\alpha_L)$ is cocomplete. The left adjoint of   $\sy\colon\sV\lra\CP\sV$ is given by
$$\CP \sV\lra\sV,\quad \phi\mapsto \phi(1).$$
Or equivalently, for each weight $\phi$ of $\sV$,  $\colim\phi=\phi(1)$.

In fact, since  for all $z\in[0,1]$, $z=1\ra z  \leq   \phi(z)\ra\phi(1)$, then
\[\phi(1)=\sup_{z\in [0,1]}\phi(z)\with z=\colim\phi,\]

the second equality holds by Example <ref>.

For each real-enriched category $X$, $\CP X$ is cocomplete. For this consider the Yoneda embedding $\sy_X\colon X\lra \CP X$. Since
$$ (\sy_X)_\exists \dashv \sy_X^{-1}\dashv (\sy_X)_\forall $$
and $(\sy_X)_\forall=\sy_{\CP X}$ (<ref> and <ref>), it follows that $\sy_{\CP X}$ has a left adjoint, so $\CP X$ is cocomplete.
For each weight $\Phi\colon \CP X\oto\star $   of $\CP X$, the   colimit of $\Phi$ is $\sy_X^{-1}(\Phi)$, i.e. $\colim\Phi=\Phi\circ(\sy_X)_*$.
$$\bfig \qtriangle(0,0)/>`>`>/<520,500>[X`\CP X`\star ;(\sy_X)_*`\colim\Phi`\Phi]
\place(250,500)[\circ] \place(260,250)[\circ] \place(520,250)[\circ] \efig$$

Since  $\CP \sy_X=(\sy_X)_\exists$,  $\sy_X^{-1}=\colim_{\CP X}$ and $(\sy_X)_\forall=\sy_{\CP X}$, we obtain a string of adjunctions
\[\CP \sy_X\dashv {\colim}_{\CP X}  \dashv \sy_{\CP X}\colon \CP X\lra\CP\CP X.\]

For each real-enriched category $X$, the real-enriched category $\CPd  X$ is cocomplete. We check that for each weight $\Phi$ of $\CPd  X$,  the coweight $(\syd)^*\swarrow\Phi$ of $X$ is a colimit of $\Phi$.

$$\bfig \ptriangle(0,0)/>`>`>/<520,500>[\CPd  X`\star`X;\Phi`(\syd)^\star`\colim\Phi]
\place(275,500)[\circ] \place(260,250)[\circ] \place(0,250)[\circ] \place(160,330)[\geq] \efig$$

First we note that for all $\psi \in\CPd X$,
$$\psi\searrow (\syd)^*  =\bw_{x\in X}(\psi(x)\ra \CPd X(-,\syd(x))) =\CPd X(-,\psi). $$

$$\bfig \qtriangle(0,0)/>`<-`<-/<520,500>[\star`X`\CPd X;\psi`\psi\searrow (\syd)^*`(\syd)^*]
\place(230,500)[\circ] \place(260,250)[\circ] \place(520,275)[\circ] \place(360,360)[\leq] \efig$$
Thus for all $\psi\in\CPd X$,
\begin{align*} \CPd X((\syd)^*\swarrow\Phi, \psi) &= \psi\searrow ((\syd)^*\swarrow\Phi) \\ & = (\psi\searrow (\syd)^*)\swarrow\Phi \\ & = \CP(\CPd X)(\Phi,\CPd X(-,\psi)),\end{align*}
showing that $(\syd)^*\swarrow\Phi$ is a colimit of $\Phi$.

In particular, for each weight $\phi$ of $X$, the colimit of the coYoneda embedding $\syd\colon X\lra \CPd X$ weighted by $\phi$ is $X\swarrow\phi$, because
\begin{align*}{\colim}_\phi\thinspace\syd &=  \colim \phi\circ (\syd)^* \\ &= (\syd)^*\swarrow(\phi\circ(\syd)^*)\\ &= ((\syd)^*\swarrow(\syd)^*)\swarrow\phi \\ &=((\syd)^*\circ(\syd)_*)\swarrow\phi &((\syd)_*\dashv(\syd)^*) \\ &= X\swarrow\phi. &(\text{$\syd$ is fully faithful})\end{align*}

The dual notion of colimit is limit. Precisely, a  limit of a coweight $\psi\colon \star\oto X$ is an element $\lim\psi$ in $X$ such that for all $x\in X$,
\[\CPd  X(\syd(x),\psi)=X(x,\lim\psi).\]
It is obvious that each coweight has at most one limit up to isomorphism.

\[\CPd  X(\syd(x),\psi)= \psi\searrow\syd(x) = (\psi\searrow X)(x),\]
it follows that $\psi$ has a limit if and only if the weight $\psi\searrow X$
is representable.
$$\bfig \qtriangle(0,0)/>`<-`<-/<520,500>[\star`X`X;\psi`\psi\searrow X`X]
\place(230,500)[\circ] \place(260,250)[\circ] \place(520,275)[\circ] \place(360,360)[\leq] \efig$$

Let $X$ be a real-enriched category and let $\psi\colon \star\oto X$ be a coweight of $X$. Then, $a\in X$ is a colimit of $\psi^{\rm op}$ of the weight $\psi^{\rm op}$ of   $X^{\rm op}$  if and only if for all $x\in X$,
\[X^{\rm op}(a,x)=\CP(X^{\rm op})(\psi^{\rm op},X^{\rm op}(-,x)).\]
\[\CP(X^{\rm op})(\psi^{\rm op},X^{\rm op}(-,x))= \inf_{z\in X}(\psi(z)\ra X(x,z)),\]
it follows that $a$ is a colimit of $\psi^{\rm op}$ if and only if
\[X(x,a)=\CPd X(\syd(x),\psi).\]
Therefore, a limit of a coweight $\psi$ of   $X$ is precisely a colimit of the weight $\psi^{\rm op}$ of  $X^{\rm op}$.

For each functor $f\colon K\lra X$  and each coweight $\psi$  of $K$, the limit of $f^\dag_\exists(\psi)$, when exists, is called the limit of $f$ weighted by $\psi$. In particular, $\lim\psi$ is the limit of the identity functor $X\lra X$ weighted by $\psi$.  We  write $\lim_\psi f$ for the limit of $f$ weighted by $\psi$.

Suppose $X$ is a real-enriched category and  $f\colon X\lra\sV$ is a functor.  Then, for each coweight $\psi$  of  $X$, the limit of $f$ weighted by $\psi$ exists and is given by ${\lim}_\psi f=\CPd X(f^\sharp,\psi)$, where  $f^\sharp\colon\star\oto X$ is the distributor corresponding to $f$. To see this we   calculate:  for all $r\in[0,1]$,
\begin{align*}    \CPd \sV(\syd_\sV(r), f_*\circ\psi) & =  \psi\searrow(f_*\searrow \sV(r,-))  \\   &= \inf_{x\in X}(\psi(x)\ra (r\ra f(x))) \\ &= \inf_{x\in X}(r\ra(\psi(x)\ra f(x))) \\ &= \sV(r, \CPd X(f^\sharp,\psi)), \end{align*}
so $\CPd X(f^\sharp,\psi)$ is a limit of $f$ weighted by $\psi$.

A real-enriched category $X$ is complete if each coweight of $X$ has a limit.

It is readily seen that $X$ is complete if and only if for any functor $f\colon K\lra X$ and any coweight $\psi$ of $K$, the limit of $f$ weighted by $\psi$ exists.
Dual to Proposition <ref>, we have the following:
A real-enriched category $X$ is complete  if and only if the coYoneda embedding $\syd\colon X\lra\CPd  X$ has a right adjoint.

The real-enriched category $\sV=([0,1],\alpha_L)$ is  complete. For each coweight $\psi$ of $\sV$, by Example <ref>,  $\inf_{z\in [0,1]}(\psi(z)\ra z)$ is a limit of $\psi$.

For each real-enriched category $X$,  $\CPd  X$ is complete. For this consider the coYoneda embedding $\syd_X\colon X\lra\CPd X$. Since
$$ (\syd_X)^\dag_\forall\dashv (\syd_X)^{-1}\dashv (\syd_X)^\dag_\exists $$
and $(\syd_X)^\dag_\forall=\syd_{\CPd X}$ (<ref> and <ref>), it follows that   $\syd_{\CPd X}$ has a right adjoint, so $\CPd X$ is complete.
For each coweight $\Psi\colon \star\oto\CPd  X$,  the limit of $\Psi$ is $(\syd_X)^{-1}(\Psi)$, i.e.  $\lim\Psi=(\sy_X^\dag)^*\circ\Psi$.
$$\bfig \qtriangle(0,0)/>`>`>/<520,500>[\star`\CPd  X`X;\Psi`\lim\Psi`(\sy_X^\dag)^*]
\place(260,500)[\circ] \place(260,250)[\circ] \place(520,250)[\circ] \efig$$

Thus, we have a string of adjunctions
$$ \syd_{\CPd X} \dashv {\lim}_{\CPd X}\dashv \CPd\syd_X\colon \CPd X\lra\CPd\CPd X . $$

For each real-enriched category $X$,  the real-enriched category $\CP X$ is complete. For each coweight $\Psi$ of $\CP X$,   the weight $\Psi\searrow\sy_*$ of $X$ is a limit of $\Psi$.
$$\bfig \qtriangle(0,0)/>`<-`<-/<520,500>[\star`\CP X`X;\Psi`\lim\Psi`\sy_*]
\place(230,500)[\circ] \place(260,250)[\circ] \place(520,275)[\circ] \place(360,360)[\leq] \efig$$

In particular, for each coweight $\psi$ of $X$, the limit of the Yoneda embedding $\sy\colon X\lra\CP X$ weighted by $\psi$ is $\psi\searrow X$, i.e. $\lim_\psi\sy=\psi\searrow X$.

(Stubbe [105]) A real-enriched category is complete  if and only if it is cocomplete.

Let $X$ be a complete real-enriched category. We show that each weight $\phi$   of $X$ has a colimit, so $X$ is cocomplete. Consider the coweight $X\swarrow\phi$ of $X$. Since $X$ is complete, there is some $a\in X$ such that
\[X(-,a)= (X\swarrow\phi)\searrow X.\]
\[X(a,-)= X\swarrow X(-,a) = X\swarrow ((X\swarrow\phi)\searrow X) =X\swarrow\phi,\]
showing that $\phi$ has a colimit.
Likewise for the converse implication.

In the following we examine the relation between functors, colimits of weights and limits of coweights.
First of all,   for any  functor $f\colon X\lra Y$   and any weight $\phi$   of $X$, if both $\colim \phi$ and $\colim_\phi f   $ exist, then $ {\colim}_\phi f  \sqsubseteq f({\colim} \phi)$, because
\begin{align*}1&= X({\colim}\phi,{\colim}\phi)\\ &=\CP X(\phi, X(-,{\colim} \phi))\\ &\leq \CP X(\phi, Y(f(-),f({\colim}\phi))) \\ &= \CP X(\phi, Y(-,f({\colim}\phi))\circ f_*) \\ & =\CP Y(\phi\circ f^*, Y(-,f({\colim}\phi)) &(-\circ f^*\dashv -\circ f_*)\\ &= Y({\colim}_\phi f  ,f({\colim}\phi)).\end{align*}

Suppose $f\colon X\lra Y$ is a functor. We say that  $f$  preserves the colimit of a weight $\phi$ of $X$, if $f(\colim\phi)$ is a colimit of $f$ weighted by $\phi$, i.e.  $f(\colim\phi)=\colim_\phi f $, whenever $\colim\phi$ exists. Likewise,   $f$ preserves the limit of a coweight $\psi$ of $X$ if $f(\lim\psi)$ is a limit of $f$ weighted by $\psi$ whenever $\lim\psi$ exists.

We say that a functor  $f\colon X\lra Y$    preserves colimits if it preserves all colimits that exist. Likewise,  $f$  preserves limits if it preserves all  limits that exist.

Each left adjoint preserves colimits. Dually, each right adjoint preserves limits.

Suppose that $f\colon X\lra Y$ is left adjoint to $g\colon Y\lra X$. We show that $f$ preserves colimits and $g$ preserves limits.

Assume that $\phi\colon X\oto\star $ has a colimit.   Since for all $y\in Y$,
\begin{align*} Y(f(\colim\phi),y) &= X(\colim\phi,g(y)) &(f\dashv g) \\ &= \CP X(\phi,  X(-,g(y))) &(\text{definition of $\colim\phi$}) \\ &= \CP X(\phi,Y(f(-),y)) &(f\dashv g)\\ &= \CP X(\phi,  Y(-,y)\circ f_*)   \\   &= \CP Y(\phi\circ f^*, Y(-,y)), &(-\circ f^*\dashv -\circ f_*) \end{align*}
it follows that $f(\colim\phi)$ is a colimit of $f$ weighted by $\phi$.

By duality, $g$ preserves limits.

If $f\colon X\lra Y$ preserves colimits and $X$ is cocomplete, then $f$ is a left adjoint.

Since $X$ is cocomplete, the Yoneda embedding $\sy_X\colon X\lra\CP X$ has a left adjoint $\colim_X\colon \CP X\lra X$. Let $g$ be the composite
\[Y\to^{\sy_Y}\CP Y\to^{f^{-1}}\CP X\to^{{\colim}_X}X;\]
that is, $g(y) =\colim_X Y(f(-),y)$. We claim that $g$ is a right adjoint of $f$.

By Theorem <ref>, we need to check that for all $x\in X$ and $y\in Y$, $f(x)\sqsubseteq y$  if and only if $x\sqsubseteq g(y)$.
If $f(x)\sqsubseteq y$, then
\begin{align*}1&\leq Y(f(x),y)\\ &  = \CP X(X(-,x), Y(f(-),y)))& (\text{Yoneda lemma})\\ &\leq X({\colim}_X X(-,x),{\colim}_X Y(f(-),y))) &(\text{${\colim}_X$ is a  functor})\\ & = X(x,g(y)), &(\text{definition of $g$})\end{align*}
showing that $x\sqsubseteq g(y)$.

For the converse implication,   first we show  that   $1\leq Y(f(g(y)),y)$  for all $y\in Y$. Since $f$ preserves colimits and $g(y)=   {\colim}_X(Y(-,y)\circ f_*)$, then
\[f(g(y))= {\colim}_Y(Y(-,y)\circ f_*\circ f^*)\leq {\colim}_Y  Y(-,y) \cong y,\]
so $1\leq Y(f(g(y)),y)$.

Now, suppose that $x\sqsubseteq g(y)$. Then
\[1\leq X(x,g(y))\leq Y(f(g(y)),y)\with Y(f(x),f(g(y)))\leq Y(f(x),y),\]
hence $f(x)\sqsubseteq y$.

The Yoneda embedding $\sy\colon X\lra\CP X$ preserves limits; the coYoneda embedding $\syd\colon X\lra\CPd  X$ preserves colimits.

For the first conclusion, we need to show that for each coweight $\psi$ of $X$, if $\lim\psi$ exists, then $\sy(\lim\psi)$ is the limit of $\sy_*\circ\psi$ in $\CP X$, i.e. $\sy(\lim\psi)={\lim}_{\CP X}(\sy_*\circ\psi)$. To see this, we calculate:

\begin{align*}  \lim(\sy_*\circ\psi)& =(\sy_*\circ\psi)\searrow\sy_*  & (\text{Example \ref{inf in PX}})\\ &=\psi\searrow(\sy_*\searrow \sy_*)\\ & = \psi\searrow (\sy^*\circ \sy_*) &(\sy_*\dashv\sy^*)\\ & = \psi\searrow X  &(\text{$\sy$ is fully faithful})\\ &= X(-,\lim\psi) & (\text{definition of $\lim\psi$})\\ &= \sy(\lim\psi).\end{align*}

The second conclusion follows by duality.

A functor  $f\colon X\lra Y$ is   colimit-dense if for each $y\in Y$, there is a weight $\phi$ of $X$ such that $y$ is a colimit of $f$ weighted by $\phi$. Likewise, $f$ is  limit-dense  if for each $y\in Y$, there is a coweight $\psi$ of $X$ such that $y$ is a limit of $f$ weighted by $\psi$.

There is a nice characterization of colimit-dense and limit-dense functors.

(Lai and Shen [65]) Suppose $f\colon X\lra Y$ is a functor.

* $f$  is colimit-dense if and only if $f_*\swarrow f_*=Y$.
* $f$ is limit-dense if and only if $f^*\searrow f^* =Y$.

(i) Assume that $f$  is colimit-dense. We wish to show   $f_*\swarrow f_*=Y$. For this it suffices to show that for each $y$ of $Y$, $(f_*\swarrow f_*)(y,-)\leq Y(y,-)$. Since $f$ is colimit-dense, $y=\colim_\phi f$ for some weight weight $\phi$ of $X$. So,
\[Y(y,-)=Y\swarrow(\phi\circ f^*) = (Y\swarrow f^*)\swarrow\phi= f_*\swarrow\phi,\]
hence   $\phi\leq   f_*(-,y)$. Therefore,
\[(f_*\swarrow f_*)(y,-)= f_*\swarrow f_*(-,y) \leq f_*\swarrow\phi = Y(y,-).\]
Conversely, suppose that $f_*\swarrow f_*=Y$. Then for each $y$ of $Y$,
\[Y(y,-)=(f_*\swarrow f_*)(y,-) = (Y\swarrow f^*)\swarrow f_*(-,y)=Y\swarrow(f_*(-,y)\circ f^*),\]
showing that $y$ is a colimit of $f$ weighted by $f_*(-,y)$, hence $f$  is colimit-dense.

(ii) Similar to (i).
For each real-enriched category, the Yoneda embedding is colimit-dense and the coYoneda embedding is limit-dense.

For  each $\phi \in\CP X$,  by Example <ref> we have
$${\colim}_\phi\thinspace\sy =\colim\phi\circ\sy^*= \phi\circ\sy^*\circ\sy_*=\phi,$$
the last equality holds since $\sy$ is fully faithful, hence $\sy\colon X\lra\CP X$ is colimit-dense.

(Kelly [58]) For each functor $f\colon X\lra Y$ with $Y$ separated and cocomplete, there is a unique left adjoint $\overline{f}\colon \CP X\lra Y$ such that $f=\overline{f}\circ\sy_X$.

Let $\overline{f}$ be the composite of left adjoints
\[ \CP X\to^{f_\exists}\CP Y\to^{\colim_Y}Y,\]
where $\colim_Y\colon\CP  Y\lra Y$ is the left adjoint of  $\sy_Y\colon Y\lra \CP Y$. Then $\overline{f}$ satisfies the requirement. This proves the existence.
For uniqueness, suppose   $\overline{f}$ satisfies the conditions. For each $\phi\in\CP X$, since $\phi$ is the colimit of the Yoneda embedding $\sy_X\colon X\lra \CP X$ weighted by $\phi$ and $\overline{f}$ preserves colimits, then
\[\overline{f}(\phi)= \overline{f}({\colim}_\phi\thinspace\sy_X)={\colim}_\phi(\overline{f}\circ\sy_X)  ={\colim}_\phi\thinspace f,\]
showing that $\overline{f}$ is unique.

§ TENSOR AND COTENSOR

Cocompleteness of a real-enriched category $X$ requires that every weight of $X$ has a colimit. The aim of this subsection is to show that in order that a real-enriched category $X$ be cocomplete, it suffices to require certain subclass of weights of $X$ have colimits.

Let $X$ be a real-enriched category. For each $x\in X$ and  $r\in [0,1]$, the tensor of $r$ with $x$, denoted by $r\otimes  x$, is an element of $X$ such that for all $y\in X$,
\[X(r\otimes  x,y)= r\ra X(x,y).\]
Tensors are a special kind of colimits. Actually, $r\otimes x$ is the colimit of the functor $s_x\colon\star\lra X$ weighted by $r_\star$, where, $s_x$ is the functor sending the unique object of $\star$ to $x$, and  $r_\star$ is the weight of   $\star$   with value $r$.  Said differently,
\[r\otimes  x=\colim(r\with\sy(x)).\]

For each $y\in X$ and  $r\in [0,1]$, the cotensor of $r$ with $y$, denoted by $r\multimap y$,  is an element of $X$ such that for all $x\in X$,
\[X(x,r\multimap y)=r\ra X(x,y).\]
The cotensor $r\multimap y$ is the limit of the functor $s_y\colon\star\lra X$ weighted by $r_\star$. Said differently,
\[r\multimap y=\lim(r\with\syd(y)).\]

A real-enriched category $X$ is   tensored if the tensor $r\otimes  x$ exists for all $x\in X$ and $r\in [0,1]$; $X$ is   cotensored if the cotensor $r\multimap y$ exists for all $y\in X$ and $r\in [0,1]$.

It is clear that $X$ is tensored if and only if   its opposite $X^{\rm op}$  is cotensored.  Every complete (hence cocomplete) real-enriched category is   both tensored and  cotensored.

If   $X$  is both tensored and cotensored, then for all $r\in[0,1]$ and $x,y\in X$,
\begin{align*}  r\otimes x\sqsubseteq y~~\text{in}~~ X_0&\iff r\leq  X(x,y)\\ &\iff x\sqsubseteq r\multimap y ~~\text{in}~~ X_0.\end{align*}
\[r\otimes -\colon X_0\lra X_0\]
is  left adjoint to
\[r\multimap -\colon X_0\lra X_0.\]
This shows that in a both tensored and cotensored  real-enriched category, tensors and cotensors are determined by each other.

The real-enriched category $\sV=([0,1],\alpha_L)$ is tensored and cotensored. For each $r\in [0,1]$ and each $x\in V$,
\[r\otimes  x=r\with x, \quad r\multimap x=r\ra x.\]

The real-enriched category $\sV^{\rm op}=([0,1],\alpha_R)$ is also tensored and cotensored. For each $r\in [0,1]$ and each $y\in \sV^{\rm op}$,
\[r\otimes  y =r\ra y, \quad r\multimap y=y\with r.\]

Let $X$ be a real-enriched category. In $\CP X$, the tensor and cotensor of $r\in [0,1]$ with $\phi\in\CP X$ are, respectively,   $r\with\phi$ and $r\ra\phi$. In $\CPd X$, the tensor and cotensor of $r\in [0,1]$ with $\psi\in\CPd X$ are  $r\ra  \psi$ and $r\with\psi $, respectively.

Let $\mathcal{B}X$ be the subset of $\CP X$ consisting of  weights of the form $r\with\sy(x)$; that is,
\[\mathcal{B}X=\{r\with\sy(x)\mid x\in X, r\in [0,1]\}. \]
Since for each functor $f\colon X\lra Y$, each $x\in X$ and each $r\in[0,1]$, $f_\exists(r\with\sy_X(x))=r\with\sy_Y(f(x)$, it follows that the assignment $X\mapsto \CB X$ defines a subfunctor of the presheaf functor $\CP$, through which the natural transformation $\sy\colon \id\lra\CP$ factors.

Write $\mathfrak{i}\colon\CB X\lra\CP X$ for the inclusion functor and ${\sf t}\colon X\lra  \mathcal{B}X$ for the Yoneda embedding with codomain restricted to $\mathcal{B}X$. It is clear that $\sy=\mathfrak{i}\circ{\sf t}$.

For each  real-enriched category $X$, the following   are equivalent:

* $X$ is tensored.
* The  functor ${\sf t}  \colon X\lra  \mathcal{B}X$ has a left adjoint.
* For each $x\in X$, the functor $X(x,-)\colon X\lra \sV$  is a right adjoint, where $\sV=([0,1],\alpha_L)$.
In this case, $X(x,-)\colon X\lra \sV$ is right adjoint to $-\otimes  x\colon \sV\lra X$.

$(1)\Leftrightarrow(2)$ This follows from the fact that the tensor $r\otimes  x$ of $r$ with $x$, if exists, is precisely the colimit  of the weight $r\with\sy(x)$.

$(1)\Leftrightarrow(3)$ A restatement of  the equality $X(r\otimes  x,y)= r\ra X(x,y)$.

Dual conclusions hold for cotensored real-enriched categories. We spell them out for later use. Let $\mathcal{B}^\dag X$ be the subset of $\CPd  X$ consisting of  coweights of the form $r\with\syd(y) $, $y\in X, r\in[0,1]$.

Write $\mathfrak{i}^\dag\colon\CB^\dag X\lra\CPd X$ for the inclusion functor and ${\sf t}^\dag\colon X\lra  \mathcal{B}^\dag X$ for   the coYoneda embedding with codomain restricted to $\mathcal{B}^\dag X$.

For each  real-enriched category $X$, the following   are equivalent:
* $X$ is cotensored.
* The map ${\sf t}^\dag\colon X\lra  \mathcal{B}^\dag X$ has a right adjoint.
* For each $y\in X$, the  map $X(-,y)\colon X\lra \sV^{\rm op} $ is a left adjoint, where $\sV^{\rm op}=[0,1],\alpha_R)$.   In this case, $X(-,y)\colon X\lra \sV^{\rm op}$ is left adjoint to $-\multimap  y\colon \sV^{\rm op}\lra X$.

Suppose $f\colon X\lra Y$ is a map between real-enriched categories.

* If both $X$ and $Y$ are tensored, then $f$ is a functor if and only if $f\colon X_0\lra Y_0$ preserves order and $r\otimes  f(x)\sqsubseteq f(r\otimes  x)$ for all $x\in X$ and $r\in [0,1]$.
* If both $X$ and $Y$ are cotensored, then $f$ is a functor if and only if $f\colon X_0\lra Y_0$ is order-preserving and $f(r\multimap y)\sqsubseteq  r\multimap f(y)$ for all $y\in X$ and $r\in [0,1]$.
We prove (i) for example.
Suppose  $f$ is a functor. First, it is trivial that $f\colon  X_0\lra Y_0$ preserves order. Next, let $r\in [0,1]$ and  $x\in X$. Since $X$ is tensored and $k\leq X(r\otimes  x, r\otimes  x)$, it follows that $r\leq X(x,r\otimes  x)$, then $r\leq Y(f(x),f(r\otimes  x))$ and consequently, $r\otimes  f(x)\leq f(r\otimes  x)$. This proves the necessity.

Now we prove the sufficiency. Let  $x, y\in X$. Since $X(x, y)\leq X(x, y)$, it follows that  $X(x, y)\otimes  x\leq y$, then
\[X(x, y)\otimes  f(x)\leq f(X(x, y)\otimes  x)\leq f(y),\]
and consequently $X(x,y)\leq Y(f(x),f(y))$,  so  $f$ is a functor.

A weight $\phi$ of $X$ is   conical if there is a functor $f\colon K\lra X$ such that $\phi=\thinspace!_*\circ f^*$, where $!_*$ is the graph of the unique functor   $!\colon K\lra\star$. Dually, a coweight $\psi$ of $X $ is   conical if there is a functor $f\colon K\lra X$ such that $\psi=  f_*\circ\thinspace !^*$, where $!^*$ is the cograph  the unique functor   $!\colon K\lra\star$. It is easily verified that a weight $\phi$ is conical if and only if
\[\phi=\sup_{a\in A}\sy(a)\]
for a subset $A$ of $X$; a coweight $\psi$ is conical if and only if
\[\psi=\sup_{b\in B}\syd(b)\]
for a subset $B$ of $X$.

(Kelly [58], Stubbe [106]) Suppose  $X$ is a real-enriched category. We say that
* $X$ is  conically cocomplete  if every conical weight of $X$ has a colimit.
* $X$ is  conically  complete  if every conical coweight of $X$ has a limit.
* $X$ is order-complete if the underlying ordered set   $X_0$ is complete.

(Kelly [58], Stubbe [106]) For each real-enriched category $X$, the following statements are equivalent:
* $X$ is cocomplete.

* $X$ is complete.

* $X$ is tensored and  conically cocomplete.
* $X$ is cotensored and  conically  complete.

* $X$ is order-complete, tensored and  cotensored.

By duality and Theorem <ref> we only need to check  that (1), (3) and (5) are equivalent.

$(1)\Rightarrow(3)$ Obvious.

$(3)\Rightarrow(5)$ First we show that $X$ is order-complete. Suppose $A$ is a subset of $X$.  We show that each colimit of the conical weight $\phi= \sup_{x\in A}\sy(x)$   is a join of $A$ in $X_0$.  For all $x\in A$,   we have
\begin{align*}X(x,\colim\phi)&=   X(\colim\phi,-)\searrow X(x,-) & (\text{Yoneda lemma}) \\ &= (X\swarrow\phi)\searrow X(x,-) & (\text{definition of $\colim\phi$}) \\ &\geq (X\swarrow X(-,x))\searrow X(x,-) &(x\in A)\\ &= X(x,-)\searrow X(x,-) \\ &=1,\end{align*}
which shows that $\colim\phi$ is an upper bound of $A$ in $X_0$. Suppose that $y$ is another upper bound of $A$ in $X_0$. Then
\begin{align*}X(\colim\phi,y) &= X(-,y)\swarrow\phi \\ &=\inf_{x\in A}(X(-,y)\swarrow X(-,x))\\ & = \inf_{x\in A} X(x,y)\\ & = 1, \end{align*}
hence $\colim\phi\sqsubseteq y$ in $X_0$. Therefore, $\colim\phi$ is a join of $A$ in $X_0$.

Next we show that $X$ is cotensored. For each $r\in[0,1]$ and $y\in X$, let
$$ A=\{a\in X\mid  r\leq X(a,y)\}.$$
We show that the colimit of the conical weight $\phi=\bv_{a\in A}\sy(a)$ is the cotensor of $r$ with $y$; that is, $X(x,\colim\phi)=r\ra X(x,y)$ for all $x\in X$. For all  $s\in[0,1]$,
\begin{align*} s\leq r\ra X(x,y) &\iff r\leq s\ra X(x,y) \\ &\iff r\leq X(s\otimes x, y) \\ & \iff  s\otimes x\sqsubseteq\colim\phi &(\text{$\colim\phi$ is a join of $A$ in $X_0$})\\ &\iff s\leq X(x,\colim\phi), \end{align*}
hence $\colim\phi$ is the cotensor of $r$ with $y$.

For a weight $\phi$ of $X$, let $a$ be the join in $X_0$ of the set $A=\{\phi(x)\otimes  x\mid x\in X\}$. We   show that $a$ is a colimit of $\phi$, hence $X$ is cocomplete.
For all $y\in X$, since $a$ is a join of $A$ in $X_0$, then
\begin{align*}X(a,y)& =\inf_{z\in A}X(z,y) &(\text{Proposition \ref{chara of cotensored}\thinspace(3)})\\ & = \inf_{z\in A}(X(-,y)\swarrow X(-,z)) &(\text{Yoneda lemma}) \\ &= X(-,y)\swarrow\sup_{z\in A}X(-,z)\\
& =X(-,y)\swarrow\sup_{x\in X}X(-,\phi(x)\otimes  x)\\
& = \inf_{x\in X}(X(-,y)\swarrow X(-,\phi(x)\otimes  x))   \\ & = \inf_{x\in X}   X(\phi(x)\otimes  x,y) &(\text{Yoneda lemma}) \\ & = \inf_{x\in X} (\phi(x)\ra  X(x,y)) &(\text{definition of tensor})\\ &= \CP X (\phi,\sy(y)),\end{align*}
hence $a$ is a colimit of $\phi$.

Let $X,Y$ be cocomplete real-enriched categories. Then  a functor $f\colon X\lra Y$ is a left adjoint if and only if

* $f\colon X_0\lra Y_0$ preserves joins;
* $f$ preserves tensors in the sense that $f(r\otimes  x)=r\otimes  f(x)$ for all $r\in [0,1]$ and $x\in X$.

Necessity is trivial since tensors are a special kind of colimits. For sufficiency, let $g\colon Y\lra X$ be the map such that $g\colon Y_0\lra X_0$ is right adjoint to $f\colon X_0\lra Y_0$. We claim that $g\colon Y\lra X$ is a right adjoint of $f\colon X\lra Y$. By Theorem <ref> and Proposition <ref>, we only need to check that $r\otimes  g(y)\leq g(r\otimes  y)$ for all $r\in [0,1]$ and $y\in Y$. Since
\[f(r\otimes  g(y))=r\otimes  f(g(y))\leq r\otimes  y,\]
it follows that $r\otimes  g(y)\leq g(r\otimes  y)$, as desired.

Let $X$ be a complete real-enriched category. Then for each functor  $f\colon K\lra X$ and each weight $\phi$ of $ K$, the colimit of $f$ weighted by $\phi$  is given by  the join of  $\{\phi(z)\otimes  f(z)\mid z\in K\}$ in $X_0$. Dually,  for each coweight $\psi$ of $K$, the limit of $f$ weighted by $\psi$ is given by  the meet of  $\{\psi(z)\multimap f(z)\mid z\in K\}$ in $X_0$.

Since $\phi\circ f^*$ is the join  in   $(\CP X)_0$ of the subset $\{\phi(z)\with X(-,f(z))\mid z\in K\},$   the functor $\colim\colon\CP X\lra X$ preserves colimits, it follows that
\[{\colim}_\phi f =\colim\phi\circ f^* = \sup_{z\in K}\colim(\phi(z)\with X(-,f(z))) = \sup_{z\in K} \phi(z)\otimes  f(z) . \]

The second conclusion follows by duality.

Suppose $X$ is a real-enriched category and $f\colon K\lra \CP X$ is a functor.

* For each weight $\phi$ of $K$, $\colim_\phi f=\sup_{z\in K}\phi(z)\with f(z)$.

* For each coweight $\psi$ of $K$, $\lim_\psi f=\sup_{z\in K}(\psi(z)\ra f(z))$.

Since $\CP\star=\sV$, Example <ref> and Example <ref> are then   special case of the above corollary.
The following conclusion, proved in  Lai and Zhang [71], extends the  Tarski fixed point theorem, which says that every order-preserving map from a complete lattice to itself has a fixed point, to the enriched context.
If $X$ is a complete real-enriched category and $f\colon X\lra X$ is a functor, then the subcategory
\[{\rm Fix}(f)=\{x\in X\mid f(x)\cong x\}\]
of $X$ composed of fixed points of $f$  is   complete.

First  we show that the subcategory
\[Y=\{x\in X\mid x\sqsubseteq f(x) \}\]
of $X$ is complete. Let $i\colon Y\lra X$ denote the inclusion. Then for each weight $\phi$ of $Y$, the join
\[a=\bv\{\phi(x)\otimes  x\mid x\in Y\}\]
in $X_0$ is a  colimit of the functor $i$ weighted by $\phi$. Since $f$ is a  functor,
\[f(a)\sqsupseteq \bv\{f(\phi(x)\otimes  x)\mid x\in Y\}\sqsupseteq \bv\{ \phi(x)\otimes  f(x)\mid x\in Y\}\sqsupseteq a,\]
which shows that $a\in Y$. Therefore, $Y$ is closed in $X$ under formation of colimits. In particular, $Y$ is cocomplete, hence complete.

Next, restricting the domain and codomain of $f$ to $Y$ we obtain a  functor $f\colon  Y\lra Y$. A similar argument shows that
\[Z=\{y\in Y\mid f(y)\sqsubseteq y\}\]
is closed in $Y$ under formation of limits, hence a complete real-enriched category.
It is clear that ${\rm Fix}(f)=Z$, the conclusion thus follows.

Suppose $X$ is a partially ordered set. A (left) action of $[0,1]$  on $X$ is a map $\otimes\colon[0,1]\times X \lra X$ subject to the following conditions:  for all $x\in X$ and $r,s, \in [0,1]$,

* $1\otimes  x=x$;
* $s\otimes (r\otimes  x) =(s\with r)\otimes  x$;
* $r\otimes-\colon X\lra X$ is a left adjoint;
* $-\otimes  x\colon [0,1]\lra X$ is a left adjoint.

By the condition (iv) one sees that for each $x\in X$, $0 \otimes x$ is a bottom element of $X$. By (iii),  the map $0\otimes-\colon X\lra X$ has a right adjoint, which is denoted by $r\multimap-\colon X\lra X$. Then, for each $x\in X$,  $0\multimap x$ is  a top element of $X$. So, the partially ordered set $X$ is bounded; that is to say, it has both a bottom element and a top element.

A lax morphism $f\colon (X,\otimes )\lra(Y,\otimes )$ between   $[0,1]$-actions is an order-preserving map $f\colon X\lra Y$ such that
\[r\otimes  f(x)\leq f(r\otimes  x)\]
for all $r\in [0,1]$ and $x\in X$.
\[\bfig\Square[{[0,1]}\times X`X`{[0,1]}\times Y`Y;\otimes `1\times f`f`\otimes ] \place(300,240)[\leq]\efig\]

Let $X$ be   a separated, tensored and cotensored real-enriched category  $X$. Assigning to each pair $(r,x)$   the tensor $r\otimes  x$ defines a  $[0,1]$-action on the partially ordered set $X_0$. Proposition <ref> says that   $X\mapsto(X_0,\otimes )$ is a functor from the full subcategory   of $\QOrd $ consisting of separated, tensored and cotensored real-enriched categories to the category of $[0,1]$-actions and lax morphisms. This functor   is indeed an isomorphism of categories.

Suppose $\otimes$ is an action of  $[0,1]$  on  a partially ordered set $(X,\leq)$. For all $x,y\in X$, let
\[\alpha_{\otimes}(x,y)=\bv\{r\in [0,1]\mid r\otimes  x\leq y\}. \]
Then $(X,\alpha_{\otimes})$ is a separated, tensored and cotensored real-enriched category with $\leq$ being its underlying order.

First of all,    it is clear that for all $r\in [0,1]$ and $x,y\in X$,
\[r\leq  \alpha_{\otimes}(x,y)\iff r\otimes  x\leq y.\]

Since $1\otimes  x=x$, then $\alpha_{\otimes}(x,x)\geq 1$ for all $x\in X$. If $r\leq \alpha_{\otimes}(x,y)$ and $s\leq \alpha_{\otimes}(y,z)$, then
\[(s\with r)\otimes  x=s\otimes (r\otimes  x)\leq s\otimes  y\leq z,\]
which shows that $s\with r\leq \alpha_{\otimes}(x,z)$, hence $\alpha_{\otimes}(y,z)\with \alpha_{\otimes}(x,y)\leq \alpha_{\otimes}(x,z)$ and consequently, $(X,\alpha_{\otimes})$ is a real-enriched category. It remains to show that $(X,\alpha_{\otimes})$ is   tensored, cotensored,   separated,  and has $\leq$ as underlying order

For all $r,s\in [0,1]$ and $x,y\in X$, since
\begin{align*} s\leq \alpha_{\otimes}(r\otimes  x, y) &\iff s\otimes (r\otimes  x)\leq y\\ &\iff s\with r\leq \alpha_{\otimes}(x, y) \\ &\iff s\leq r\ra\alpha_{\otimes}(x, y)  ,\end{align*}
then $r\otimes  x$ is a tensor of $r$ with $x$ in $(X,\alpha_{\otimes})$, hence $(X,\alpha_{\otimes})$ is  tensored.

For each $r\in [0,1]$, let $r\multimap-\colon X\lra X$ be the right adjoint of $r\otimes-\colon X\lra X$.
For all $x,y\in X$ and all $s\in [0,1]$, since
\begin{align*} s\leq X(x,r\multimap y)&\iff s\otimes  x\leq r\multimap y\\ &\iff  r\otimes (s\otimes  x)\leq y \\ &\iff (r\with s)\otimes  x\leq y\\ &\iff r\with s \leq X(x,y)\\ &\iff s\leq r\ra X(x,y),\end{align*}
$r\multimap y$ is a cotensor of $r$ with $y$ in $(X,\alpha_{\otimes})$, hence $(X,\alpha_{\otimes})$ is  cotensored.

Finally, since
\[1\leq \alpha_{\otimes}(x, y)\iff 1\otimes  x\leq y\iff x\leq y,\]
it follows that $(X,\alpha_{\otimes})$ is separated with $\leq $ being its underlying order.

The category of  $[0,1]$-actions and lax morphisms is isomorphic to the full subcategory of $\QOrd$ consisting of separated, tensored and cotensored real-enriched categories.

§ CAUCHY COMPLETENESS

A weight $\phi\colon X\oto\star$ of a real-enriched category  $X$ is Cauchy if it is a right adjoint, as a distributor. That means, there is a distributor $\psi\colon\star\oto X$ such that $\phi\circ\psi\geq1$ and $\psi(y)\with\phi(x)\leq X(x,y)$ for all $x,y\in X$.

Let $f\colon X\lra Y$ be a functor between real-enriched categories and let $\phi$ be a weight of $X$.  If $\phi$ is a Cauchy weight,  as a composite of right adjoints, $\phi\circ f^*$ is a right adjoint, hence a Cauchy weight of $Y$. So,  assigning to each real-enriched category $X$ the subcategory $\CC X$ of $\CP X$ composed of Cauchy weights defines a subfunctor
of the presheaf functor $\CP$. Since each representable  weight  is Cauchy, the natural transformation  from the identity functor to $\mathcal{P}$ with Yoneda embeddings as components  factors through the functor $\CC$.

The following conclusion is a special case of the characterization of Cauchy weights of enriched categories in Street [104].

Suppose $X$ is a real-enriched category and $\phi$ is a weight of $X$. The following are equivalent:
* $\phi$ is Cauchy.
* The colimit $\colim_\phi f$ of any functor $f\colon X\lra Y$ weighted by $\phi$, when exists, is preserved by every functor $g\colon Y\lra Z$, i.e.  $g(\colim_\phi f)=\colim_\phi(g\circ f)$.  In particular, every Cauchy weight with a colimit  is representable.

$(1)\Rightarrow(2)$ Since $\colim_\phi f$, when exists, is the colimit of the Cauchy weight $\phi\circ f^*$ of $Y$, it suffices to show that if $b$ is a colimit of a Cauchy weight $\xi$ of $Y$, then for every functor $g\colon Y\lra Z$, $g(b)$ is a colimit of $g$ weighted by $\xi$, i.e. $g(b)=\colim_\xi g$.  By definition of colimit  we have  $Y\swarrow\xi=Y(b,-)$. Since $Y\swarrow\xi$ is   left adjoint to  $\xi$ by Lemma <ref>, then
$$\xi=Y(b,-)\searrow Y =Y(-,b).$$
This  shows  in particular that $\xi$ is representable. Since
\[  \xi\circ g^*= Y(-,b)\circ g^* = Z(-,g(b)),\]
it follows that $g(b)$ is a colimit of $\xi\circ g^*$, i.e. $g(b)=\colim_\xi g$.

$(2)\Rightarrow(1)$ We  show that the coweight $\psi\coloneqq X\swarrow\phi$ is left adjoint to $\phi$. Consider the coYoneda embedding $\syd\colon X\lra\CPd X$ and the functor
$$g \colon \CPd X\lra\sV, \quad  \xi\mapsto\CPd X(\psi,\xi),$$
where $\sV=([0,1],\alpha_L)$. On the one hand, by Example <ref> we have
$${\colim}_\phi\thinspace \syd= X\swarrow\phi=\psi,$$

so $g(\colim_\phi\thinspace \syd)=1$. On the other hand, since for each $x\in X$,
$$g\circ \syd(x) =\CPd X(\psi,X(x,-))=\psi(x),$$
i.e. $g\circ \syd=\psi^\sharp \colon X\lra\sV$, then by Example <ref> we have
$$ {\colim}_\phi(g\circ \syd)={\colim}_\phi\thinspace\psi^\sharp =\phi\circ\psi.$$
Therefore, $\phi\circ\psi=1$. The inequality $\psi\circ\phi\leq X$ is trivial, so $\psi$ is left adjoint to $\phi$, as desired.

The next characterization of Cauchy weights  is a special case of that   in  Kelly and  Schmitt [59].

Suppose $X$ is a real-enriched category and $\phi$ is a weight of $X$. The following are equivalent:
* $\phi$ is Cauchy.
* The functor $\sub_X(\phi,-)\colon \CP X \lra\sV$ preserves  colimits.
* The functor $\phi\circ-\colon (\CPd X)^{\rm op} \lra\sV$ preserves limits.

$(1)\Rightarrow(2)$ If $\psi$ is a left adjoint of $\phi$, then for all $\xi\in\CP X$,
from which one deduces that $\sub_X(\phi,-)\colon \CP X \lra\sV$ preserves colimits.

$(2)\Rightarrow(1)$ Assume that the functor  $\sub_X(\phi,-)$ preserves colimits. We show that $\psi\coloneqq X\swarrow\phi$   is left adjoint to $\phi$, hence $\phi$ is Cauchy. It suffices to check that $\phi\circ\psi\geq1$.   Since $\sub_X(\phi,-)$ preserves colimits and
$$\phi={\colim}_\phi\thinspace\sy=\sup_{x\in X}(\phi(x)\otimes\sy(x)),$$
it follows that
\begin{align*}\phi\circ\psi &=\sup_{x\in X}\phi(x)\with\psi(x)\\ &=\sup_{x\in X} (\phi(x)\with \sub_X(\phi,  \sy(x))\\ & =\sub_X(\phi,\phi)\\ &=1.\end{align*}

$(1)\Rightarrow(3)$ If $\psi$ is a left adjoint of $\phi$, then for all $\lam\in\CPd X$,
$$ \phi\circ\lam =\psi\searrow\lam=\CPd X(\lam,\psi),$$
from which one infers that   $\phi\circ-\colon (\CPd X)^{\rm op} \lra\sV$ preserves limits.

$(3)\Rightarrow(1)$ Assume that the functor $\phi\circ-\colon (\CPd X)^{\rm op} \lra\sV$ preserves limits. We show that $\psi\coloneqq X\swarrow\phi$   is left adjoint to $\phi$. It suffices to check that $\phi\circ\psi\geq1$. Since
$$X\swarrow\phi =\bw_{y\in X}(\phi(y)\ra X(y,-)) $$
and $\phi\circ -\colon (\CPd X)^{\rm op} \lra\sV$ preserves limits, it follows that
\begin{align*}\phi\circ\psi &= \bw_{y\in X}(\phi(y)\ra \phi\circ X(y,-))= \bw_{y\in X}(\phi(y)\ra  \phi(y))=1.  \end{align*}

The proof is completed.

A real-enriched category $X$ is  Cauchy complete  if it is separated and all of its  Cauchy weights are representable.

In other words, $X$ is Cauchy complete if each of its Cauchy weight has a unique colimit. Suppose $\phi\colon X\oto\star$ is a Cauchy weight of $X$, and $\psi\colon\star\oto X$ is a left adjoint of $\phi$. Then $\psi^{\rm op}\colon X^{\rm op}\oto\star$ is a Cauchy weight of $X^{\rm op}$. Since all Cauchy weights of $X^{\rm op}$ arise in this way, it follows that the notion of Cauchy completeness is self-dual; that is, if $X$ is Cauchy complete, then so is its opposite $X^{\rm op}$.

Each  functor $f\colon X\lra Y$ generates an adjunction of distributors  $f_*\dashv f^*$. The following proposition shows that when $Y$ is Cauchy complete,   every such adjunction  arises in this way.

(Lawvere [78]) Suppose $Y$ is a real-enriched category. The following are equivalent:

* $Y$ is Cauchy complete.
* For any pair of distributors $\phi\colon X\oto Y$ and $\psi\colon Y\oto X$, if $\phi$ is left adjoint to $\psi$, then there is a unique functor  $f\colon X\lra Y$ such that $\phi=f_*$ and $\psi=f^*$.

It suffices to observe that if $\phi\colon X\oto Y$ is left adjoint to $\psi\colon Y\oto X$, then for each $x\in X$,   the distributor $\phi(x,-)\colon\star\oto Y$   is left adjoint to  $\psi(-,x)\colon Y\oto\star$.

For a real-enriched category $X$, let $\CC X$ be the subcategory of $\CP X$ composed of Cauchy weights. Let $\mathfrak{i}\colon\CC X\lra\CP X$ be the inclusion functor, and let ${\sf c}\colon X\lra  \mathcal{C}X$ be the Yoneda embedding with codomain restricted to $\mathcal{C}X$. The composite $ \mathfrak{i}\circ{\sf c}$ is  then the Yoneda embedding $\sy\colon X\lra\CP X$.

(Stubbe [105]) Suppose $X$ is a real-enriched category. Then
$${\sf c}_*\circ{\sf c}^*=\CC X\quad \text{and}\quad {\sf c}^*\circ {\sf c}_*=X.$$
Therefore, $X$ is isomorphic to $\CC X$ in the category of distributors.

The equality ${\sf c}^*\circ {\sf c}_*=X$ follows from that ${\sf c}   \colon X\lra \CC X$ is fully faithful. For the equality ${\sf c}_*\circ{\sf c}^*=\CC X$, let $\phi,\psi\in\CC X$. Since  $X\swarrow \phi$ is left adjoint to $\phi$ by Lemma <ref>, then
\begin{align*}{\sf c}_*\circ{\sf c}^*(\phi,\psi)&= \sup_{x\in X} {\sf c}_*(x,\psi)\with{\sf c}^*(\phi,x) \\ &= \sup_{x\in X}\CC X(\sy  (x),\psi)\with\CC X(\phi,\sy  (x))\\ &=\sup_{x\in X}\psi(x)\with(X(-,x)\swarrow \phi) \\ &= \sup_{x\in X}\psi(x)\with(X\swarrow\phi)(x)   \\  &= \psi\circ(X\swarrow\phi)  \\ &= \psi\swarrow\phi
\\ & =\CC X(\phi,\psi).  \qedhere\end{align*}

(Kelly [58]) Suppose  $X$ is a real-enriched category.

* $\CC X$ is Cauchy complete.
* For each functor $f\colon X\lra Y$ with $Y$ Cauchy complete, there is a unique functor $\overline{f}\colon\CC X\lra Y$ that extends $f$. Furthermore, if $f$ is fully faithful, then so is $\overline{f}$.

(i) We  show that for each Cauchy weight $\Phi$ of $\CC X$, the colimit $\phi$ of the inclusion functor  $\mathfrak{i}\colon \CC X\lra\CP X$ weighted by $\Phi$ is a Cauchy weight of $X$, which implies that $X$ is Cauchy complete. In other words, we show that $\CC X$ is closed in $\CP X$ under formation of colimits of Cauchy weights.   Since the composite $\mathfrak{i}\circ{\sf c}$ is the Yoneda embbeding, from Example <ref> it follows that
$$\phi={\colim}_\Phi\mathfrak{i} =\colim \Phi\circ\mathfrak{i}^*= \Phi\circ \mathfrak{i}^*\circ\mathfrak{i}_*\circ {\sf c}_*= \Phi\circ{\sf c}_*.$$

Let $\psi$ be the coweight
\[ {\sf c}^*\circ \Psi\colon *\oto\CC X \oto X,\]
$\Psi\colon \star\oto\CC X$ is the left adjoint of $\Phi\colon\CC X\oto\star$. We claim that $\psi$ is a left adjoint of $\phi$, hence $\phi$ is Cauchy.

On the one hand, since ${\sf c}\colon X\lra\CC X$ is fully faithful, we have
\[\psi\circ\phi = {\sf c}^* \circ\Psi\circ \Phi\circ   {\sf c}_*\leq {\sf c}^*\circ {\sf c}_*= X.  \]
On the other hand, by  Lemma <ref>, we have
\[\phi\circ\psi = \Phi\circ {\sf c}_*\circ {\sf c}^* \circ\Psi=   \Phi\circ  \Psi   \geq 1.\]
Therefore, $\psi$ is   left adjoint to $\phi$, as claimed.

(ii) It is readily verified that the functor
\[\overline{f}\colon \CC X\lra Y, \quad \phi\mapsto {\colim}_\phi f \]
extends $f$. Uniqueness follows from that  each Cauchy weight $\phi$ is the colimit of ${\sf c}\colon X\lra\CC X$ weighted by $\phi$ and that colimits of Cauchy weights are preserved by all functors.

Finally, we check that if $f$ is fully faithful, then  so is $\overline{f}$; that is, $\CC X(\phi_1,\phi_2)= Y(\overline{f}(\phi_1), \overline{f}(\phi_2))$ for all Cauchy weights $\phi_1,\phi_2$ of $X$.  Since
\begin{align*}\CP X(\phi_1,\phi_2) &\leq \CP Y(\phi_1\circ f^*,\phi_2\circ f^*)\\ & \leq \CP X(\phi_1\circ f^*\circ f_*,\phi_2\circ f^*\circ f_*)\\ &=\CP X(\phi_1,\phi_2), \end{align*}
it follows that
$$\CC X (\phi_1,\phi_2) = \CP Y(\phi_1\circ f^*,\phi_2\circ f^*) = Y(\overline{f}(\phi_1), \overline{f}(\phi_2)),$$
where the last equality holds because for $i=1,2$, the weight $\phi_i\circ f^*$ is Cauchy, hence represented by $\overline{f}(\phi_i)$.

An immediate corollary of the above theorem is that the full subcategory of $\QOrd$ composed of Cauchy complete real-enriched categories is reflective, with reflection given by $\sy\colon X\lra\CC X$.
The real-enriched category $\CC X$ is called the  Cauchy completion  of $X$.

Cauchy completion is idempotent in the sense that $\CC X=\CC(\CC X)$ for every real-enriched category $X$.

Since $\CC X$ is Cauchy complete,   all of its Cauchy weights are representable, then the conclusion follows.

The following characterization of Cauchy completion is taken from Hofmann [46], Hofmann and Tholen [51].
For each real-enriched category $X$, the Cauchy completion of $X$ is the equalizer in $\QOrd$  of the parallel pair
\begin{equation*}\bfig
\morphism(0,0)|a|/@{->}@<3pt>/<500,0>[\CP X`\CP\CP X;\CP\sy_X]
\morphism(0,0)|b|/@{->}@<-3pt>/<500,0>[\CP X`\CP \CP X; \sy_{\CP X} ]
\efig\end{equation*}

We show that a weight $\phi$ of $X$ is Cauchy if and only if $\CP\sy_X(\phi)=\sy_{\CP X}(\phi)$.

For sufficiency consider the coweight $\psi=X\swarrow\phi=\sy_X^*\swarrow\CP X(-,\phi)$.   We wish to show that $\phi$ is right adjoint to $\psi$, hence a Cauchy weight.  Since $\CP\sy_X(\phi)=\sy_{\CP X}(\phi)=\CP X(-,\phi)$, then
\begin{align*} \phi\circ\psi &=\phi\circ  (X\swarrow\phi)\\ & =\phi\circ(\sy_X^*\swarrow\CP X(-,\phi))\\ &=  \CP\sy_X(\phi)\circ \CP X(\phi,-) %&(\CP X(-,\phi)\dashv\CP X(-,\phi))
\\ & = \CP X(-,\phi)\circ \CP X(\phi,-)  \\ &=1.\end{align*}
The inequality  $\psi\circ\phi\leq X$ is trivial. Thus, $\phi$ is right adjoint to $\psi$.

For necessity suppose $\phi$ is a Cauchy weight. Then $\psi\coloneqq X\swarrow\phi$ is left adjoint to $\phi$, hence
$$ 1=\phi\circ\psi=\phi\circ \sy_X^*\circ\CP X(\phi,-)= \CP\sy_X(\phi)\circ \CP X(\phi,-),$$
\begin{align*}\sy_{\CP X}(\phi) &= \CP\sy_X(\phi)\circ \CP X(\phi,-)\circ \CP X(-,\phi)= \CP\sy_X(\phi). \qedhere \end{align*}

In the following we   characterize Cauchy completeness in terms of topological property.
Let $X$ be a real-enriched category. For each $x\in X$ and $r<1$, the set
$$B(x,r)\coloneqq\{y\in X\mid X(x,y)>r\}$$
is called the open ball of $X$ with center $x$ and radius $r$. The collection $\{B(x,r)\mid x\in X, r<1\}$  is a base for a topology on $X$, the resulting topology is called the open ball topology of $X$.

For each real-enriched category $(X,\alpha)$, the  symmetrization  of  $(X,\alpha)$  refers to the symmetric (in an evident sense) real-enriched category $(X,S(\alpha))$, where $S(\alpha)(x,y)=\min\{\alpha(x,y),\alpha(y,x)\}$.

It is clear that the open ball topology of the symmetrization of $X$ is the least common refinement of the open ball topology of $X$ and that of $X^{\rm op}$.

A net $\{x_i\}_{i\in D}$ of a real-enriched category $X$ converges to $x$ w.r.t the open ball topology if and only if $\bv_{i\in D}\inf_{j\geq i}X(x,x_j)=1$.

Assigning open ball topology to real-enriched categories yields  a functor from   enriched categories to  topological spaces.

Suppose $\{x_i\}_{i\in D}$ is a net and $a$ is an element of a real-enriched category $X$.  We say that
* $\{x_i\}_{i\in D}$   is   Cauchy (also called  biCauchy) if
\[\sup_{i\in D}\inf_{j,k\geq i}X(x_j,x_k)= 1.\]

* $a$ is a bilimit  of $\{x_i\}_{i\in D}$ if for all $x\in X$,
\[\sup_{i\in D}\inf_{j\geq i}X(x,x_j)=X(x,a),\quad  \sup_{i\in D}\inf_{j\geq i}X(x_j,x)=X(a,x).  \]

It is clear that a net has at most one bilimit up to isomorphism.

Let $\{x_i\}_{i\in D}$ be a Cauchy net and $a$ be an element of a real-enriched category $X$. Then the following   are equivalent:
* $a$ is a bilimit of $\{x_i\}_{i\in D}$.

* $\sup_{i\in D}\inf_{j\geq i}X(a,x_j)= 1$ and $ \sup_{i\in D}\inf_{j\geq i}X(x_j,a)= 1$.

* $\{x_i\}_{i\in D}$ converges to $a$ in the open ball topology of the symmetrization of $X$.

$(1)\Rightarrow(2)$   trivial.

$(2)\Leftrightarrow(3)$ This follows from Proposition <ref> and  the fact  that the requirement in (2) is equivalent to that
$$\sup_{i\in D}\inf_{j\geq i}\min\{X(a,x_j),X(x_j,a)\}= 1.$$

$(2)\Rightarrow(1)$ For all $x\in X$, we calculate:
\begin{align*} X(x,a)&= \Big(\sup_{i\in D}\inf_{j\geq i}X(a,x_j)\Big)\with X(x,a) \\ &\leq  \sup_{i\in D}\inf_{j\geq i} X(a,x_j) \with X(x,a)  \\ &\leq  \sup_{i\in D}\inf_{j\geq i}X(x,x_j)   \\ &=  \Big( \sup_{i\in D}\inf_{j\geq i}X(x_j,a)\Big) \with \Big(\sup_{i\in D}\inf_{j\geq i}X(x,x_j)\Big) \\ &= \sup_{i\in D}\inf_{j\geq i} X(x_j,a)\with X(x,x_j)    \\ &\leq X(x,a),\end{align*}
which implies   $\sup_{i\in D}\inf_{j\geq i}X(x,x_j)=X(x,a)$.  The other equality is verified likewise. So  $a$ is a bilimit of $\{x_i\}_{i\in D}$.