Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 18, 2005, 107 – 125

Per K. Jakobsen and V.V. Lychagin
UNIVERSAL SEMIGROUPS

Abstract. In this paper we introduce the notion of a universal semigroup and its dual, the universal cosemigroup. We show that the class of universal semigroups include the class of monoids and is included in the class of semigroups with a product that is an epimorphism. Both inclusions are proper. Semigroups in the category of Banach spaces are Banach algebras and we show that all Banach algebras with an approximate unit are universal and construct a ﬁnite dimensional Banach algebra that has no unit but is universal. The property of being universal is thus a generalized unit property.

Contents
1.   Introduction
2.  Universal semigroups
   2.1.  Examples from

S e t s

and general monoidal categories
   2.2.   Examples from the category of Banach spaces
3.  Universal cosemigroups
   3.1.  Examples from

S e t s

and general monoidal categories
References

1. Introduction

The notions of monoids and semigroups are well known and appear naturally in many contexts. In this paper we introduce a new algebraic structure, the universal semigroup. Its deﬁnition is inspired by the categorical generalization of monoids and semigroups [1]. In a monoidal category a semigroup is a pair $〈 A, μ 〉$ where $A$ is an object in the category and $μ : A \otimes A \to A$ is a morphism such that the following diagram commute

A⊗(A⊗ A) 1A⊗-μA A⊗ A -μA-- A
|
| /
αA,A,A| / /
| / μA⊗ 1A
| /
|/
(A⊗ A)⊗ A

where

α

is the associativity constraint in the category. This deﬁnition does not only include the usual algebraic notion of semigroups and associative algebras, but also many other algebraic structures. What algebraic structure it represents depends on the choice of monoidal category. Let

S_{A}

be the diagram we get by removing the right part of the previous diagram. Thus

S_{A}

is the diagram

1A⊗ μA
A⊗(A⊗ A) ------ A⊗ A
|
αA,A,A| / /
| /
| / / μA⊗ 1A
|/
(A⊗ A)⊗ A

Observe that from a categorical point of view

〈 A, μ 〉

is a semigroup if and only if it is a cocone[1] on the diagram

S_{A}

. In general

〈 A, μ 〉

is not a universal cocone on

S_{A}

. Recall that

〈 B, g 〉

is a universal cocone on

D_{A}

if it is a cocone and if for any cocone

〈 C, f 〉

, there exists a unique map

ϕ :

B \to C

such that

ϕ \circ g = f

$1A⊗ μA g A⊗(A⊗ A) ------ A⊗ A ----- B| | / \ | αA,A,A| / \ ∃!φ | | / \ ∀f | | / / μA⊗ 1A \\ | |/ \ | (A⊗ A)⊗ A C$

The universal cocone could be though of as the ”smallest” cocone on the diagram, the one that gives the best ”commutative ﬁt” to the diagram. In this paper we deﬁne $〈 A, μ 〉$ to be a universal semigroup if it is a universal cocone on the diagram $S_{A}$ and show that the class of universal semigroups includes the class of monoids, but that there are universal semigroups that are not monoids. For any monoid the product is an epimorphism. We show that this also holds for any universal semigroup but that there are semigroups that are not universal and where the product is an epimorphism. Thus the class of universal semigroups includes all monoids and is included in the class of semigroups with products that are epimorphisms.

The category of Banach spaces is a monoidal category where the monoidal structure is determined by the projective tensor product of Banach spaces. In this category semigroups are Banach algebras. We show that all Banach algebras with an approximate unit are universal and that there are universal Banach algebras that does not have an approximate unit. All $C^{*} -$ algebras admit an approximate unit[2] and are thus universal as are $L_{1}$ group algebras of locally compact topological groups since they all admit an approximate unit[3].

These results leads to the idea that the universal cocone property for semigroups is a generalized unit condition. Universal semigroups could be expected to share properties with monoids that general semigroups does not.

Note that the property of being a universal semigroup is a typical categorical property in that it depends on which category the semigroup is placed in. One can thus not state that a semigroup is universal without specifying which category we refer to. This may appear a little strange if one is not familiar with the categorical point of view. It is however a fact that even such well known properties as associativity and commutativity of algebraic structures actually depends on the categorical context. It has for instance been shown that both quaternions and Cayley numbers are commutative, associative algebras[6] if placed in the right categories.

2. Universal semigroups

In order to give a categorical deﬁnition of universal semigroups or even of semigroups we need the notion of a category with an associative product, or a semimonoidal category.

Deﬁnition 1. A semimonoidal category is a triple $〈 𝒞, \otimes, α 〉$ , where $\otimes : 𝒞$ $\times 𝒞$ $\to$ $𝒞$ is a bifunctor on $𝒞$ and $α$ is a natural isomorphism

α : \otimes \circ (1_{𝒞} \times \otimes) \to \otimes \circ (\otimes \times 1_{𝒞}),

such that the following MacLane coherence conditions

α_{A \otimes B, C, D} \circ α_{A, B, C \otimes D} = (α_{A, B, C} \otimes 1_{D}) \circ α_{A, B \otimes C, D} \circ (1_{A} \otimes α_{B, C, D}),

are satisﬁed for all objects $A, B, C, D$

Recall that naturality means that the following identities

((f \otimes g) \otimes h) \circ α_{A, B, C} = α_{A^{'}, B^{'}, C^{'}} \circ (f \otimes (g \otimes h)),

holds for all choices of arrows $f : A \to A^{'}$ , $g : B \to B^{'}$ , $h : C \to C^{'}$ .

The product $\otimes$ is thus not strictly associative but associative up to isomorphism. A semimonoidal category contains the structures we need to deﬁne semigroups[1].

Deﬁnition 2. A semigroup $〈 A, μ_{A} 〉$ in a semimonoidal category is an object $A$ and a morphism $μ_{A} : A \otimes A \to A$ such that the following diagrams commute

Let $S_{A}$ be the previous diagram with the right-hand node removed

A⊗(A⊗ A) 1A⊗-μA A⊗ A
|
αA,A,A| /
| / /
| / μA⊗ 1A
|/ /
|
(A⊗ A)⊗ A

Deﬁnition 3. A universal semigroup in a semimonoidal category $〈 𝒞, \otimes, α 〉$ is a universal cocone $〈 A, μ_{A} 〉$ on the diagram $S_{A}$ .

2.1. Examples from $S e t s$ and general monoidal categories. Let us consider a few examples from the category $S e t s$ with the usual monoidal structure. A monoid $〈 A, μ, e 〉$ in $S e t s$ is a semigroup $〈 A, μ 〉$ with a neutral element $e$ such that $μ (e, x) = μ (x, e) = x$ for all $x$ in $A$ . Let $〈 C, f 〉$ be any cocone on $S_{A}$ . This means that $f \circ (1 \times μ) = f \circ (μ \times 1)$ or $f (x, y z) = f (x y, z)$ for all $x, y, z$ in $A$ . Here we write $x y = μ (x, y)$ .The condition for $〈 A, μ 〉$ to be a universal cocone on $S_{A}$ is therefore that the equation

ϕ \circ μ = f,

has a unique solution for all $f$ such that $f (x, y z) = f (x y, z)$ . Let $ϕ$ and $ψ$ be two solutions. Then we have $ϕ (x) = ϕ (x e) = f (x, e) = ψ (x e) = ψ (x)$ for all $x$ in $A$ . So there can be at most one solution. Deﬁne a map $ϕ :$ $A \to C$ by $ϕ (x) = f (x, e)$ . Then we have

(ϕ \circ μ) (x, y) = ϕ (x y) = f (x y, e) = f (x, y e) = f (x, y)

The map $ϕ$ is therefore a solution and we have proved the following proposition

Proposition 4. Let $〈 A, μ, e 〉$ be a monoid in $S e t s$ . Then $〈 A, μ 〉$ is a universal semigroup.

The class of Universal semigroups in $S e t s$ therefore includes all monoids. This is a proper inclusion, there are universal semigroups that are not monoids as the following two examples show.

Let $A = \{a, b\}$ be a semigroup with two elements and table of multiplication given by


$μ$	a	b

a	a	b

b	a	b

The conditions on $f$ are for this semigroup given by $f (a, a) = f (b, a)$ , $f (a, b) = f (b, b)$ . Let $ϕ_{1}$ and $ϕ_{2}$ be two solutions of $ϕ \circ μ = f$ . Then we must have $ϕ_{1} (a) = ϕ_{1} (μ (a, a)) = f (a, a) = ϕ_{2} (μ (a, a)) = ϕ_{2} (a)$ and $ϕ_{1} (b) = ϕ_{1} (μ (b, b)) = f (b, b) = ϕ_{2} (μ (b, b)) = ϕ_{2} (b)$ . So $ϕ_{1} = ϕ_{2}$ and there is at most one solution. Deﬁne a map $ϕ : A \to B$ by $ϕ (a) = f (a, a), ϕ (b) = f (b, b)$ . Then we have

\begin{array}{rcl} ϕ (μ (a, a)) & = & ϕ (a) = f (a, a) \\ ϕ (μ (a, b)) & = & ϕ (b) = f (b, b) = f (a, b) \\ ϕ (μ (b, a)) & = & ϕ (a) = f (a, a) = f (b, a) \\ ϕ (μ (b, b)) & = & ϕ (b) = f (b, b) \end{array}

This proves that $ϕ$ is the unique solution to the equation $ϕ \circ μ = f$ and as a consequence $〈 A, μ 〉$ is a universal semigroup. On the other hand $〈 A, μ 〉$ does not have a neutral element and is therefore not a monoid.

As a second example of a universal semigroup in $S e t s$ that is not a monoid let us consider a set $A$ and the projection on the ﬁrst factor $μ : A \times A \to A$ . Thus $μ (x, y) = x$ . Then we evidently have

[μ \circ (μ \times 1)] (x, y, z) = μ (x, z) = x

[μ \circ (1 \times μ)] (x, y, z) = μ (x, y) = x

so $〈 A, μ 〉$ is a semigroup.The multiplication clearly does not have a unit, so $〈 A, μ 〉$ is not a monoid. Let now $〈 C, f 〉$ be a cocone on $S_{A}$ . Then $f : A \times A \to C$ and we have the condition $f \circ (1 \times μ) = f \circ (μ \times 1)$ . This means that $f (x, y) = f (x, z)$ for all $x, y, z$ . If $ϕ$ is any solution of $ϕ \circ μ = f$ we must have $ϕ (x) = ϕ (μ (x, y)) = f (x, y)$ , so there can be at most one solution. Let $x_{0}$ be any element in $A$ . Deﬁne the map $ϕ : A \to C$ by $ϕ (x) = f (x, x_{0})$ . Then we have

ϕ (μ (x, y)) = ϕ (x) = f (x, x_{0}) = f (x, y)

so $ϕ$ is a solution. Therefore $〈 A, μ 〉$ is a universal semigroup that is not a monoid. Projection on the second factor would in a similar way produce a universal semigroup.

We will now prove a few simple results that holds for universal semigroups in any semimonoidal category.

Proposition 5. Let $〈 A, μ_{A} 〉$ be a universal semigroup. Then the arrow $μ_{A}$ is epi.

Proof. Let $B$ be any object in the category and let $f, g : A \to B$ be two arrows and assume that $f \circ μ_{A} = g \circ μ_{A}$ . Let $h = f \circ μ_{A}$ . Then the equation

ϕ \circ μ_{A} = h

has both $f$ and $g$ as solutions. Since $〈 A, μ_{A} 〉$ is a universal semigroup we must have $f = g$ and this proves that $μ_{A}$ is epi. □

Any semigroup in $S e t s$ with a nonsurjective product is therefore not a universal semigroup in $S e t s$ .The semigroup $A = \{a, b, c\}$ with table of multiplication given by


$μ$	a	b	c

a	b	c	c

b	c	c	c

c	c	c	c

does not have a surjective multiplication and is therefore not universal.

We will next show that monoids are universal semigroups in any semimonoidal category. In order to deﬁne the notion of a monoid in a semimonoidal category we need a neutral object for the product $\otimes$ . This leads to the well known notion of a monoidal category.

Deﬁnition 6. Let $𝒞$ be a category and let $e$ be a object in $𝒞$ . Deﬁne two functors $L_{e}, R_{e} : 𝒞 \to 𝒞$ on objects and morphisms by

\begin{array}{rcl} L_{e} (A) & = & e \otimes A, \\ L_{e} (f) & = & 1_{e} \otimes f, \\ R_{e} (A) & = & A \otimes e, \\ R_{e} (f) & = & f \otimes 1_{e} . \end{array}

A monoidal category is a 6 tuple $〈 𝒞, \otimes, e, α, β, γ 〉$ where $〈 𝒞, \otimes, α 〉$ is a semimonoidal category and $β : L_{e} \to 1_{𝒞}$ and $γ : R_{e} \to 1_{𝒞}$ are natural isomorphisms such that the following identities

\begin{array}{rcl} (γ_{A} \otimes 1_{B}) \circ α_{A, e, B} & = & (1_{A} \otimes β_{A}), \\ β_{e} & = & γ_{e}, \end{array}

holds for all objects $A$ and $B$ .

Naturality of $β, γ$ means that the following identities

\begin{array}{rcl} f \circ β_{a} & = & β_{B} \circ (1_{e} \otimes f), \\ f \circ γ_{A} & = & γ_{B} \circ (f \otimes 1_{e}), \end{array}

holds for all morphism $f : A \to B$ .

These are the MacLane coherence conditions for a monoidal category. They ensure that all diagrams generated using the functors $\otimes, L_{e}, R_{e}$ and the natural isomorphisms $α, β, γ$ will commute.

Deﬁnition 7. A monoid in a monoidal category is a triple $〈 A, μ_{A}, u_{A} 〉$ , where $〈 A, μ_{A} 〉$ is a semigroup in the category and where $u_{A} : e \to A$ is a morphism such that the following unit condition holds

\begin{array}{rcl} μ_{A} \circ (1_{A} \otimes u_{A}) & = & γ_{A} \\ μ_{A} \circ (u_{A} \otimes 1_{A}) & = & β_{A} \end{array}

We can now prove that monoids are universal semigroups in any monoidal category.

Proposition 8. Let $〈 A, μ_{A}, u_{A} 〉$ be a monoid in a monoidal category $𝒞$ . Then $〈 A, μ_{A} 〉$ is a universal semigroup.

Proof. We want to show that $〈 A, μ 〉$ is a universal cocone on the diagram $S_{A}$ .

This means that the equation

ϕ \circ μ_{A} = f

should have a unique solution for all $f : A \otimes A \to A$ such that $f \circ (1_{A} \otimes μ_{A}) = f \circ (μ_{A} \otimes 1_{A}) \circ α_{A, A, A}$ . Deﬁne a map $δ_{A} : A \to A \otimes A$ by $δ_{A} = (u_{A} \otimes 1_{A}) \circ β_{A}^{- 1}$ . Then we have

μ_{A} \circ δ_{A} = μ_{A} \circ (u_{A} \otimes 1_{A}) \circ β_{A}^{- 1} = β_{A} \circ β_{A}^{- 1} = 1_{A}

This identity show that the equation $ϕ \circ μ = f$ can have at most one solution and this solution must be $ϕ = f \circ δ_{A}$ . The proof is complete if we can show that this $ϕ$ really is a solution. From the naturality of $α, β$ and the MacLane coherence conditions we have the following identities

\begin{array}{rcl} (u_{A} \otimes (1_{A} \otimes 1_{A})) \circ α_{e, A, A}^{- 1} & = & α_{A, A, A}^{- 1} \circ ((u_{A} \otimes 1_{A}) \otimes 1_{A}), \\ β_{A}^{- 1} \circ μ_{A} & = & (1_{e} \otimes μ_{A}) \circ β_{A \otimes A}^{- 1}, \\ β_{A \otimes A}^{- 1} & = & α_{e, A, A}^{- 1} \circ (β_{A}^{- 1} \otimes 1_{A}) . \end{array}

But then we have

\begin{array}{rcl} ϕ \circ μ_{A} & = & f \circ δ_{A} \circ μ_{A} \\ = & f \circ (u_{A} \otimes 1_{A}) \circ β_{A}^{- 1} \circ μ_{A} \\ = & f \circ (u_{A} \otimes 1_{A}) \circ (1_{e} \otimes μ_{A}) \circ β_{A \otimes A}^{- 1} \\ = & f \circ (u_{A} \otimes μ_{A}) \circ β_{A \otimes A}^{- 1} \\ = & f \circ (1_{A} \otimes μ_{A}) \circ (u_{A} \otimes (1_{A} \otimes 1_{A})) \circ β_{A \otimes A}^{- 1} \\ = & f \circ (1_{A} \otimes μ_{A}) \circ (u_{A} \otimes (1_{A} \otimes 1_{A})) \circ α_{e, A, A}^{- 1} \circ (β_{A}^{- 1} \otimes 1_{A}) \\ = & f \circ (1_{A} \otimes μ_{A}) \circ α_{A, A, A}^{- 1} \circ ((u_{A} \otimes 1_{A}) \otimes 1_{A}) \circ (β_{A}^{- 1} \otimes 1_{A}) \\ = & f \circ (μ_{A} \otimes 1_{A}) \circ ((u_{A} \otimes 1_{A}) \otimes 1_{A}) \circ (β_{A}^{- 1} \otimes 1_{A}) \\ = & f \circ (μ_{A} \otimes 1_{A}) \circ (δ_{A} \otimes 1_{A}) \\ = & f, \end{array}

so $ϕ = f \circ δ_{A}$ is a solution □

Proposition 5 and 8 show that in any monoidal category the class of universal semigroups includes all monoids and is included in the class of semigroups where the product is an epimorphism. The following proposition show that one can construct simple universal semigroups in most monoidal categories.

Proposition 9. Let $A$ be a object and assume that there is an arrow $ε_{A} : A \to e$ from $A$ to the neutral object $e$ in the category. Deﬁne an arrow $μ_{A} : A \otimes A \to A$ by

μ_{A} = β_{A} \circ (ε_{A} \otimes 1_{A})

Then $〈 A, μ_{A} 〉$ is a semigroup. It is a universal semigroup if $μ_{A}$ has a right inverse

Proof. It is easy to show, using the monoidal structure and naturality, that the following set of identities holds

\begin{array}{rcl} (β_{A} \otimes 1_{A}) \circ ((1_{e} \otimes ε_{A}) \otimes 1_{A}) & = & (ε_{A} \otimes 1_{A}) \circ (β_{A} \otimes 1_{A}), \\ ((ε_{A} \otimes ε_{A}) \otimes 1_{A}) \circ α_{A, A, A} & = & α_{e, e, A} \circ (ε_{A} \otimes (ε_{A} \otimes 1_{A})), \\ (β_{A} \otimes 1_{A}) \circ α_{e, e, A} & = & (1_{e} \otimes β_{A}) . \end{array}

Using these identities we have

\begin{array}{rcl} μ_{A}^{l} \circ (μ_{A}^{l} \otimes 1_{A}) \circ α_{A, A, A} \\ = & β_{A} \circ (ε_{A} \otimes 1_{A}) \circ (β_{A} \circ (ε_{A} \otimes 1_{A}) \otimes 1_{A}) \circ α_{A, A, A} \\ = & β_{A} \circ (ε_{A} \otimes 1_{A}) \circ (β_{A} \otimes 1_{A}) \circ ((ε_{A} \otimes 1_{A}) \otimes 1_{A}) \circ α_{A, A, A} \\ = & α_{A} \circ (β_{e} \otimes 1_{A}) \circ ((1_{e} \otimes ε_{A}) \otimes 1_{A}) \circ ((ε_{A} \otimes 1_{A}) \otimes 1_{A}) \circ α_{A, A, A} \\ = & β_{A} \circ (β_{e} \otimes 1_{A}) \circ ((ε_{A} \otimes ε_{A}) \otimes 1_{A}) \circ α_{A, A, A} \\ = & β_{A} \circ (β_{e} \otimes 1_{A}) \circ α_{e, e, A} \circ (ε_{A} \otimes (ε_{A} \otimes 1_{A})) \\ = & β_{A} \circ (1_{e} \otimes β_{A}) \circ (ε_{A} \otimes (ε_{A} \otimes 1_{A})) \\ = & β_{A} \circ (ε_{A} \otimes β_{A} \circ (ε_{A} \otimes 1_{A})) \\ = & β_{A} \circ (ε_{A} \otimes 1_{A}) \circ (1_{A} \otimes β_{A} \circ (ε_{A} \otimes 1_{A})) \\ = & μ_{A}^{l} \circ (1_{A} \otimes μ_{A}^{l}), \end{array}

and this proves that $〈 A, μ_{A} 〉$ is a semigroup. In order to prove that it is universal when $μ_{A}$ has a right inverse we have to prove that the equation $ϕ \circ μ_{A} = f$ has a unique solution $ϕ : A \to B$ for all $f : A \otimes A \to B$ such that $f \circ (1_{A} \otimes μ_{A}) = f \circ (μ_{A} \otimes 1_{A}) \circ α_{A, A, A}$ . Let the right inverse for $μ_{A}$ be $δ_{A}$ . We thus have the identity $μ_{A} \circ δ_{A} = 1_{A}$ . But then the only possible solution of the equation $ϕ \circ μ_{A} = f$ is $ϕ = f \circ δ_{A}$ . We must now show that this in fact is a solution. The following identities follow from unit coherence and naturality

\begin{array}{rcl} δ_{A} \circ β_{A} & = & β_{A \otimes A} \circ (1_{e} \otimes δ_{A}), \\ α_{e, A, A} \circ (ε_{A} \otimes (1_{A} \otimes 1_{A})) & = & ((ε_{A} \otimes 1_{A}) \otimes 1_{A}) \circ α_{A, A, A}, \\ (β_{A} \otimes 1_{A}) \circ α_{e, A, A} & = & β_{A \otimes A} . \end{array}

But then we have

\begin{array}{rcl} f \circ δ_{A}^{l} \circ μ_{A}^{l} \\ = & f \circ δ_{A}^{l} \circ β_{A} \circ (ε_{A} \otimes 1_{A}) \\ = & f \circ β_{A \otimes A} \circ (1_{e} \otimes δ_{A}^{l}) \circ (ε_{A} \otimes 1_{A}) \\ = & f \circ β_{A \otimes A} \circ (ε_{A} \otimes δ_{A}^{l}) \\ = & f \circ β_{A \otimes A} \circ (ε_{A} \otimes (1_{A} \otimes 1_{A})) \circ (1_{A} \otimes δ_{A}^{l}) \\ = & f \circ (β_{A} \otimes 1_{A}) \circ α_{e, A, A} \circ (ε_{A} \otimes (1_{A} \otimes 1_{A})) \circ (1_{A} \otimes δ_{A}^{l}) \\ = & f \circ (β_{A} \otimes 1_{A}) \circ ((ε_{A} \otimes 1_{A}) \otimes 1_{A}) \circ α_{A, A, A} \circ (1_{A} \otimes δ_{A}^{l}) \\ = & f \circ (μ_{A}^{l} \otimes 1_{A}) \circ α_{A, A, A} \circ (1_{A} \otimes δ_{A}^{l}) \\ = & f \circ (1_{A} \otimes μ_{A}^{l}) \circ (1_{A} \otimes δ_{A}^{l}) \\ = & f . \end{array}

□

A special case of the previous proposition is

Corollary 10. Assume that there exists a map $u_{A} : e \to A$ such that $ε_{A} \circ u_{A} = 1_{e}$ . Then $〈 A, μ_{A} 〉$ is a universal semigroup.

Proof. Deﬁne an arrows $δ_{A} : A \to A \otimes A$ by

δ_{A} = (u_{A} \otimes 1_{A}) \circ β_{A}^{- 1}

Then we have

\begin{array}{rcl} μ_{A} \circ δ_{A} & = & β_{A} \circ (ε_{A} \otimes 1_{A}) \circ (u_{A} \otimes 1_{A}) \circ β_{A} \\ = & β_{A} \circ (ε_{A} \circ u_{A} \otimes 1_{A}) \circ β_{A}^{- 1} \\ = & β_{A} \circ β_{A}^{- 1} = 1_{A} . \end{array}

But this show that $δ_{A}$ is the right inverse of $μ_{A}$ and the result follows from the previous proposition. □

In a similar way we ﬁnd that $〈 A, μ_{A} 〉$ is a universal semigroup if we deﬁne

μ_{A} = β_{A} \circ (1_{A} \otimes ε_{A})

and assume that $μ_{A}$ has a right inverse. Note that in general the universal semigroups described in these propositions are not monoids.

Let us consider a few examples of the previous construction. Let us ﬁrst consider the case of $S e t s$ with the cartesian product as monoidal structure and neutral object given by the set $T = {*}$ . Since $T$ is the terminal object in $S e t s$ there exists a unique map $ε_{A} : A \to T$ from any set $A$ to the terminal $T$ . This map is obviously deﬁned by $ε_{A} (x) = *$ for all $x \in A$ . In $S e t s$ the natural isomorphism $β_{A} : T \times A \to A$ is given by $β_{A} (*, x) = x$ . Let $x_{0}$ be any element in $A$ and deﬁne a map $u_{A} : T \to A$ by $u_{A} (*) = x_{0}$ . Then $ε_{A} \circ u_{A} = 1_{e}$ and we can conclude that $〈 A, μ_{A} 〉$ is a universal semigroups. The product is explicitly given by

μ_{A} (x, y) = [β_{A} \circ (ε_{A} \times 1_{A})] (x, y) = β_{A} (*, y) = y

so $μ_{A}$ is the projection on the second factor. We have previously shown directly that these maps gives universal semigroups.

Let $V e c t_{ℝ}$ be the category of real vector spaces with linear maps as arrows.In this category tensor product of vector spaces $\otimes = \otimes_{ℝ}$ is a monoidal structure. The neutral object is $ℝ$ and the arrows $β_{A}$ and its inverse $β_{A}^{- 1}$ are given by

\begin{array}{rcl} β_{A} (r \otimes a) & = & r a \\ β_{A}^{- 1} (a) & = & 1 \otimes a \end{array}

where $r$ is any real number, $a$ a element in $A$ and $1$ is the unit in $ℝ$ . Note that a semigroup in this category is a real associative algebra.

Let $A$ be a vector space with a positive deﬁnite inner product $I_{A} : A \otimes A \to ℝ$ and let $a \in A$ . Deﬁne a map $ε_{A} : A \to ℝ$ by

ε_{A} (b) = I_{A} (a \otimes b)

This is clearly a linear map. Deﬁne now the map $μ_{A} : A \otimes A \to A$ by $μ_{A} = β_{A} \circ (ε_{A} \otimes 1_{A})$ . Explicitly we have

μ_{A} (x \otimes y) = β_{A} (I_{A} (a \otimes x) \otimes y) = I_{A} (a \otimes x) y

Then the general theory show that $〈 A, μ_{A} 〉$ is a semigroup in $V e c t_{ℝ}$ .. Now deﬁne a map $u_{A} : ℝ \to A$ by

u_{A} (r) = \frac{r a}{I_{A} (a, a)}

Then $u_{A}$ is a linear map and

(ε_{A} \circ u_{A}) (r) = ε_{A} (\frac{r a}{I_{A} (a, a)}) = \frac{r}{I_{A} (a, a)} ε_{A} (a) = r

so we have $ε_{A} \circ u_{A} = 1$ and the semigroup $〈 A, μ_{A} 〉$ is in fact a universal semigroup. This example be generalized in the following way. We consider a subcategory of topological vector spaces over a ﬁeld $ℱ$ that is closed with respect to tensor product. Such subcategories certainly exists. Let $α$ be a continuous linear functional $α : A \to ℝ$ . The morphism $ε_{A} : A \to ℱ$ is now deﬁned by

ε_{A} (a) = α (a)

and the product in the semigroup $〈 A, μ_{A} 〉$ is

μ_{A} (a \otimes a^{'}) = α (a) b

The resulting semigroup is a universal semigroup because the continuous linear map $u_{A} : ℱ \to A$ deﬁned by $u_{A} (r) = r {(α (b))}^{- 1} b$ ,where $b \in A$ and $α (b) \neq 0$ , satisfy $ε_{A} \circ u_{A} = 1_{A}$ .

We have seen that in general the class of universal semigroups is strictly larger than the class of monoids. We also know that the class of universal semigroups is included in the class of semigroups with a product rule that is an epimorphisms. In general this last inclusion is also proper as the following example show.

Let $A$ be a real vector space of dimension three with basis ${i, j, k}$ . Deﬁne a product, $μ$ on $A$ by the following multiplication table


$μ$	i	j	k

i	i	0	k

j	j	0	0

k	0	0	0

By direct calculation one can show that $〈 A, μ 〉$ is a semigroup in $V e c t_{ℝ}$ ,or in other words, a real associative algebra. The product is clearly an epimorphism. The chosen basis for $A$ induce in the usual way a basis for $A \otimes A$ . Let now $V$ be a one dimensional vector space and let $f : A \otimes A \to V$ be a linear map that is zero on all basis vectors except on the vector $j \otimes k$ where it has a nonzero value. It is a straight forward calculation to show that $〈 V, f 〉$ is a cocone on the diagram $S_{A}$ . If $ϕ$ is a solution of the equation $ϕ \circ μ = f$ we must have

\begin{array}{rcl} ϕ (i) & = & ϕ (μ (i \otimes i)) = f (i \otimes i) = 0, \\ ϕ (j) & = & ϕ (μ (j \otimes i)) = f (j \otimes i) = 0, \\ ϕ (k) & = & ϕ (μ (i \otimes k)) = f (i \otimes k) = 0, \end{array}

so $ϕ = 0$ . But $ϕ = 0$ is not a solution of the equation $ϕ \circ μ = f$ because $f \neq 0$ . The equation $ϕ \circ μ = f$ therefore have no solution for the cocone $〈 V, f 〉$ and then by deﬁnition $〈 A, μ 〉$ is not a universal algebra.

2.2. Examples from the category of Banach spaces. Let $ℬ$ be the category where objects are Banach spaces and where morphisms are bounded linear maps. The Banach spaces are vector spaces over a ﬁeld $F$ where $F$ is $ℝ$ or $ℂ$ . We introduce a monoidal structure in this category by deﬁning $X \otimes Y$ to be the projective tensor product[4] of Banach spaces. The unit object for this product is the Banach space $F$ and the natural isomorphisms $α$ , $β$ and $γ$ are the standard ones. The standard interpretation of the projective tensor product in terms of properties of bilinear maps[5] shows that $〈 A, μ 〉$ is a semigroups in $ℬ$ iﬀ it is a Banach algebra. Recall that an approximate unit[3] in a Banach algebra $A$ is a net ${u_{λ}}_{λ \in Λ}$ such that $∥ u_{λ} ∥ \leq 1$ for all $λ \in Λ$ and such that for all elements $a \in A$ we have

\begin{array}{rcl} {lim}_{λ} u_{λ} a & = & a, \\ {lim}_{λ} a u_{λ} & = & a . \end{array}

Recall that a morphism $f : X \to Y$ in any category is a epimorphism iﬀ for all morphisms $g, h : Y \to Z$ with $f \circ g = f \circ h$ we have $g = h$ . In the category of $S e t s$ epimorphisms are exactly the surjective maps. In the category of Banach spaces it is easy to show that any morphism with a dense image is a epimorphism. Using this observation we ﬁrst prove the following result

Proposition 11. Let $〈 A, μ 〉$ be a Banach algebra with an approximate unit. Then $μ : A \otimes A \to A$ is a epimorphism.

Proof. Let $m = μ (A \otimes A) \subset A$ be the image of $μ$ . We must show that the image is dense in $A$ . Let $a \in A$ and let ${u_{λ}}$ be an approximate unit in $A$ . Deﬁne a net $a_{λ} = u_{λ} a$ in $A$ . Then we have by the deﬁning property of an approximate unit that

{lim}_{λ} a_{λ} = a .

But this means that the closure of $m$ is $A$ and thus that $m$ is dense in $A$ . □

Therefore Banach algebras with an approximate unit is included in the class of algebras where the product is an epimorphism. We now use this result to prove our main result in this section.

Theorem 12. Let $〈 A, μ 〉$ be a Banach algebra with an approximate unit. Then $〈 A, μ 〉$ is a universal semigroup in the category of Banach spaces.

Proof. Let ${u_{λ}}_{λ \in Λ}$ be the approximate unit. For each $λ \in Λ$ deﬁne a linear map $δ_{λ} : A \to A \otimes A$ by

δ_{λ} (a) = u_{λ} \otimes a .

Then since the projective norm is a cross norm we have

∥ δ_{λ} (a) ∥ = ∥ u_{λ} \otimes a ∥ = ∥ u_{λ} ∥ ∥ a ∥ \leq ∥ a ∥,

and therefore each $δ_{λ}$ is bounded with $∥ δ_{λ} ∥ \leq 1$ . Let now $B$ be any other Banach space and let $f : A \otimes A \to B$ be a bounded linear map that satisfy

f \circ (μ \otimes 1_{A}) = f \circ (1_{A} \otimes μ) .

Note that because of linearity and continuity of $f$ this condition holds iﬀ

f (a b \otimes c) = f (a \otimes b c),

holds for all elements $a, b$ and $c$ in $A .$

We must show that the equation $ϕ \circ μ = f$ has one and only one solution. We know from proposition 11 that $μ$ is a epimorphism so there can be at most one solution. For each $λ \in Λ$ deﬁne $ϕ_{λ} : A \to B$ by

ϕ_{λ} = f \circ δ_{λ} .

This map is continuous and

∥ ϕ_{λ} ∥ \leq ∥ f ∥ ∥ δ_{λ} ∥ \leq ∥ f ∥,

so the family of maps ${ϕ_{λ}}_{λ \in Λ}$ is uniformly bounded in $λ$ by $∥ f ∥$ . Let

m = {\sum_{n} a_{n} b_{n} ∣ a_{n}, b_{n} \in A} \subset μ (A \otimes A) .

Then for $a \in m$ we have

ϕ_{λ} (a) = ϕ_{λ} (\sum_{n} a_{n} b_{n}) = f (u_{λ} \otimes \sum_{n} a_{n} b_{n}) = f (\sum_{n} u_{λ} a_{n} \otimes b_{n}),

and therefore by continuity

{lim}_{λ} ϕ_{λ} (a) = {lim}_{λ} f (\sum_{n} u_{λ} a_{n} \otimes b_{n}) = f (\sum_{n} {lim}_{λ} (u_{λ} a_{n}) \otimes b_{n}) = f (\sum_{n} a_{n} \otimes b_{n}) .

Thus ${lim}_{λ} ϕ_{λ} (a)$ exists for all $a \in m$ . We know that

R = {\sum_{n} a_{n} \otimes b_{n} ∣ a_{n}, b_{n} \in A},

is dense in $A \otimes A$ . Therefore $m = μ (R)$ is dense in $A$ and so $m$ is dense in $A$ . Let now $a \in A$ . Then there exists a sequence ${a_{n}}$ in $m$ with $a_{n} \to a$ and if $λ, β \in Λ$ we have

\begin{array}{rcl} ∥ ϕ_{λ} (a) - ϕ_{β} (a) ∥ & = & ∥ ϕ_{λ} (a) - ϕ_{λ} (a_{n}) + ϕ_{λ} (a_{n}) - ϕ_{β} (a) - ϕ_{β} (a_{n}) + ϕ_{β} (a_{n}) ∥ \\ \leq & ∥ ϕ_{λ} (a - a_{n}) ∥ + ∥ ϕ_{λ} (a_{n}) - ϕ_{β} (a_{n}) ∥ + ∥ ϕ_{β} (a) - ϕ_{β} (a_{n}) ∥ \\ \leq & (∥ ϕ_{λ} ∥ + ∥ ϕ_{β} ∥) ∥ a - a_{n} ∥ + ∥ ϕ_{λ} (a_{n}) - ϕ_{β} (a_{n}) ∥ \\ \leq & 2 ∥ f ∥ ∥ a - a_{n} ∥ + ∥ ϕ_{λ} (a_{n}) - ϕ_{β} (a_{n}) ∥ . \end{array}

Let now $n$ be ﬁxed and so large that $2 ∥ f ∥ ∥ a - a_{n} ∥ < \frac{1}{2} ε$ . Since $a_{n} \in m$ we have

a_{n} = \sum_{j = 1}^{p} a_{j}^{n} b_{j}^{n},

and therefore

\begin{array}{rcl} ϕ_{λ} (a_{n}) - ϕ_{β} (a_{n}) & = & f (u_{λ} \otimes \sum_{j = 1}^{p} a_{j}^{n} b_{j}^{n}) - f (u_{β} \otimes \sum_{j = 1}^{p} a_{j}^{n} b_{j}^{n}) \\ = & f (\sum_{j = 1}^{p} (u_{λ} a_{j}^{n} - u_{β} a_{j}^{n}) \otimes b_{j}^{n}), \end{array}

and so

∥ ϕ_{λ} (a_{n}) - ϕ_{β} (a_{n}) ∥ \leq ∥ f ∥ M_{n} \sum_{j = 1}^{p} ∥ u_{λ} a_{j}^{n} - u_{β} a_{j}^{n} ∥,

where we have assumed without loss of generality that $∥ b_{j}^{n} ∥ \leq M_{n}$ for $j = 1 . . . p$ . But $u_{λ}$ is an approximate unit and therefore $u_{λ} a_{j}^{n}$ converges $\forall j$ and is thus Cauchy since $A$ is a Banach space. But then there exists $λ_{j} \in Λ$ $j = 1 . . p$ such that

∥ u_{λ} a_{j}^{n} - u_{β} a_{j}^{n} ∥ < \frac{1}{2 p ε ∥ f ∥ M_{n}} .

Since $Λ$ is a directed set there exists a element $λ_{0}$ such that all these inequalities holds when $λ, β > λ_{0}$ . But then for such $λ$ and $β$ we have using the previously derived inequalities that

∥ ϕ_{λ} (a) - ϕ_{β} (a) ∥ < ε,

and this means that the net ${ϕ_{λ} (a)}_{λ \in Λ}$ is Cauchy and therefore converges. We can now deﬁne a map $ϕ : A \to B$ by

ϕ (a) = {lim}_{λ} ϕ_{λ} (a) .

This map is linear and bounded because

∥ ϕ (a) ∥ = {lim}_{λ} ∥ ϕ_{λ} (a) ∥ \leq ∥ f ∥ ∥ a ∥ .

But for any element $ξ = \sum_{j} a_{j} \otimes b_{j}$ in the dense set $R \subset A \otimes A$ we have

ϕ (μ (ξ)) = {lim}_{λ} ϕ_{λ} (\sum_{j} a_{j} b_{j}) = {lim}_{λ} f (\sum_{j} u_{λ} a_{j} \otimes b_{j}) = f (\sum_{j} a_{j} \otimes b_{j}) = f (ξ) .

Therefore the bounded maps $ϕ \circ μ$ and $f$ agree on a dense set and we can conclude that

ϕ \circ μ = f,

and this shows that the algebra $A$ is universal. □

Universal Banach algebras thus include all Banach algebras with an approximate unit. On the other hand it is not hard to construct Banach algebras without unit that are universal, so the inclusion is proper. Let $〈 A, μ 〉$ be a three dimensional algebra with basis ${i, j, k}$ and table of multiplication given by


$μ$	i	j	k

i	i	0	0

j	j	0	0

k	0	0	k

By direct calculation one can easily show that this algebra is universal but does not have a unit (or approximate unit).

3. Universal cosemigroups

To any categorical concept described in terms of diagrams there is a dual concept that we get by reversing all arrows. For the notion of a universal semigroup this procedure leads to the notion of a universal cosemigroup. Let $〈 𝒞, \otimes, α 〉$ be a semimonoidal category. A cosemigroup in $𝒞$ is a pair $〈 A, δ_{A} 〉$ where $A$ is a object in $𝒞$ and $δ_{A} : A \to A \otimes A$ is a arrow in $𝒞$ and where the following diagram commute.

A⊗(A⊗ A) 1A⊗-δA- A⊗ A δA----A
|
αA,A,A| //
| /
| / δA ⊗ 1A
|/ /
(A⊗ A)⊗ A

Let $C S_{A}$ be the diagram we get by removing the right-hand node.

1A⊗ δA
A⊗(A⊗ A) ------ A⊗ A
| /
αA,A,A| /
| /
| / / δA ⊗ 1A
|/
(A⊗ A)⊗ A

Evidently $〈 A, δ_{A} 〉$ is a cosemigroup iﬀ $〈 A, δ_{A} 〉$ is a cone on the diagram $C S_{A}$ . In general a cosemigroup does not give a universal cone.

By deﬁnition $〈 B, f 〉$ is a universal cone on the diagram $C S_{A}$ if it is a cone and if for all cones $〈 C, g 〉$ there exists a unique arrow $ϕ : C \to B$ in $𝒞$ such that

f \circ ϕ = g

Deﬁnition 13. A universal cosemigroup in a semimonoidal category $〈 𝒞, \otimes, α 〉$ is a cosemigroup $〈 A, δ_{A} 〉$ such that $〈 A, δ_{A} 〉$ is a universal cocone on the diagram $C S_{A}$ .

3.1. Examples from $S e t s$ and general monoidal categories. We have the following result that show that not all cosemigroups are universal.

Proposition 14. Let $〈 A, δ_{A} 〉$ be a universal cosemigroup. Then $δ_{A}$ is a monomorphism.

Proof. Let $B$ be any object in the category and let $f, g : B \to A$ be two arrows and assume that $δ_{A} \circ$ $f = δ_{A} \circ g$ . Let $h = δ_{A} \circ f$ . Then the equation

δ_{A} \circ ϕ = h

has both $f$ and $g$ as solutions. Since $〈 A, δ_{A} 〉$ is a universal cosemigroup we must have $f = g$ and this proves that $δ_{A}$ is a monomorphism. □

We will now illustrate this deﬁnition with several examples. Let us ﬁrst work in the category $S e t s$ . This is a monoidal category with cartesian product as product bifunctor and with trivial associativity constraint. Let $A$ be any set and deﬁne a map $δ_{A} : A \to A \times A$ by $δ_{A} (x) = (x, x)$ . This is the diagonal map in $S e t s$ . We have

\begin{array}{rcl} [(1_{A} \times δ_{A}) \circ δ_{A}] (x) & = & (1_{A} \times δ_{A}) (x, x) = (x, x, x) \\ [(δ_{A} \times 1_{A}) \circ δ_{A}] (x) & = & (δ_{A} \times 1_{A}) (x, x) = (x, x, x) \end{array}

so $〈 A, δ_{A} 〉$ is a cone. Let $〈 B, f 〉$ be any cone and consider the system

δ_{A} \circ ϕ = f

Since $〈 B, f 〉$ is a cone we have $(1_{A} \times δ_{A}) \circ f = (δ_{A} \times 1_{A}) \circ f$ . We have $f : B \to A \times A$ so we can write $f (x) = (f_{1} (x), f_{2} (x))$ . The cone condition then gives

(f_{1} (x), f_{2} (x), f_{2} (x)) = (f_{1} (x), f_{1} (x), f_{2} (x))

so we must have $f_{1} (x) = f_{2} (x)$ for all $x$ in $A$ . Let $ϕ$ be any solution to $δ_{A} \circ ϕ = f$ . Then we have $(ϕ (x), ϕ (x)) = (f_{1} (x), f_{2} (x))$ or $ϕ = f_{1}$ . So there is at most one solution. Let $ϕ = f_{1}$ then we ﬁnd

[δ_{A} \circ ϕ] (x) = δ_{A} (f_{1} (x)) = (f_{1} (x), f_{1} (x)) = (f_{1} (x), f_{2} (x)) = f (x)

so $ϕ = f_{1}$ is a solution. We have therefore proved

Proposition 15. Let $δ_{A} : A \to A \times A$ be the diagonal map of sets. Then $〈 A, δ_{A} 〉$ is a universal cosemigroup in $S e t s$ .

As our next example let us consider a pointed set. This is a set $A$ with a chosen point $x_{0} \in A$ . Deﬁne a map $δ_{A} : A \to A \times A$ by $δ_{A} (x) = (x_{0}, x)$ . Then we have

\begin{array}{rcl} [(1_{A} \times δ_{A}) \circ δ_{A}] (x) & = & (1_{A} \times δ_{A}) (x_{0}, x) = (x_{0}, x_{0}, x) \\ [(δ_{A} \times 1_{A}) \circ δ_{A}] (x) & = & (δ_{A} \times 1_{A}) (x_{0}, x) = (x_{0}, x_{0}, x) \end{array}

so $〈 A, δ_{A} 〉$ is a cone. Let $〈 B, f 〉$ be any cone. This means that $f : B \to A \times A$ and $(1_{A} \times δ_{A}) \circ f = (δ_{A} \times 1_{A}) \circ f$ . As above we write $f (x) = (f_{1} (x), f_{2} (x))$ and the cone condition becomes

(f_{1} (x), x_{0}, f_{2} (x)) = (x_{0}, f_{1} (x), f_{2} (x))

so we must have $f_{1} (x) = x_{0}$ . But then we have

[δ_{A} \circ ϕ] (x) = δ_{A} (ϕ (x)) = (x_{0}, ϕ (x)) = (f_{1} (x), f_{2} (x)) = f (x)

if we choose $ϕ (x) = f_{2} (x)$ . This is clearly the only solution. We therefore have

Proposition 16. Let $〈 A, x_{0} 〉$ be a pointed set. Deﬁne $δ_{A} : A \to A \times A$ by $δ_{A} (x) = (x_{0}, x)$ . Then $〈 A, δ_{A} 〉$ is a universal cosemigroup.

The map $δ_{A} : A \to A \times A$ deﬁned by $δ_{A} (x) = (x, x_{0})$ will similarly give a universal cosemigroup. If $〈 A, μ_{A}, e 〉$ is a monoid we get two universal cosemigroups by choosing $x_{0} = e$ .

Finite sets oﬀer many examples of universal cosemigroups. Consider the set $A = {a, b, c}$ . Deﬁne $δ_{A} : A \to A \times A$ by

\begin{array}{rcl} δ_{A} (a) & = & (a, a) \\ δ_{A} (b) & = & (b, a) \\ δ_{A} (c) & = & (c, c) \end{array}

Then we have

\begin{array}{rcl} [(δ_{A} \times 1_{A}) \circ δ_{A}] (a) & = & (δ_{A} \times 1_{A}) (a, a) = (a, a, a) \\ [(1_{A} \times δ_{A}) \circ δ_{A}] (a) & = & (1_{A} \times δ_{A}) (a, a) = (a, a, a) \\ [(δ_{A} \times 1_{A}) \circ δ_{A}] (b) & = & (δ_{A} \times 1_{A}) (b, a) = (b, a, a) \\ [(1_{A} \times δ_{A}) \circ δ_{A}] (b) & = & (1_{A} \times δ_{A}) (b, a) = (b, a, a) \\ [(δ_{A} \times 1_{A}) \circ δ_{A}] (c) & = & (δ_{A} \times 1_{A}) (c, c) = (c, c, c) \\ [(1_{A} \times δ_{A}) \circ δ_{A}] (c) & = & (1_{A} \times δ_{A}) (c, c) = (c, c, c) \end{array}

so $〈 A, δ_{A} 〉$ is a cone. Next let $〈 B, f 〉$ be any cone. This implies that $(δ_{A} \times 1_{A}) \circ f = (1_{A} \times δ_{A}) \circ f$ . Direct computation show that $(δ_{A} \times 1_{A}) (x, y) = (1_{A} \times δ_{A}) (x, y)$ iﬀ $(x, y)$ is a element in the range of $δ_{A}$ . We therefore can conclude that $〈 B, f 〉$ is a cone if the image of $f$ is included in the range of $δ_{A}$ . Since $δ_{A}$ is injective there is at most one solution to the equation $δ_{A} \circ ϕ = f$ . Let $b$ be any element in $B$ . Then there exists a unique element $a$ in $A$ such that $δ_{A} (a) = f (b)$ . This is true since the range of $f$ is equal to the range of $δ_{A}$ and $δ_{A}$ is injective. Deﬁne $ϕ (b) = a$ , then $ϕ$ is well deﬁned and clearly a solution to the equation $δ_{A} \circ ϕ = f$ . This proves that $〈 A, δ_{A} 〉$ is a universal cosemigroup.

We have seen that a necessary condition for a cosemigroup $〈 A, δ_{A} 〉$ to be a universal cosemigroup is that $δ_{A}$ is a monomorphism. This is not suﬃcient in general. Let $A = {a, b, c}$ and deﬁne a map $δ_{A} : A \to A \times A$ by

\begin{array}{rcl} δ_{A} (a) & = & (a, a), \\ δ_{A} (b) & = & (b, a), \\ δ_{A} (c) & = & (a, c) . \end{array}

The map $δ_{A}$ is a monomorphism and a direct calculation like the one above show that $〈 A, δ_{A} 〉$ is a cosemigroup but that it is not a universal cosemigroup. In general one can prove by dualizing the proof for monoids that comonoid in a monoidal category are universal cosemigroups. Thus universal cosemigroups is a class that contains all comonoids and is included in all cosemigroups with a coproduct that is a monomorphism.

References

[1] S. MacLane, Categories for the Working Mathematician, Springer 1991.

[2] R. V. Kadison, Fundamentals of The Theory of Operator Algebras, Volume 1, Academic Press,1983.

[3] C. E. Richart,Van Nostrand, Banach Algebras, 1960.

[4] G. K $\ddot{o}$ the, Topological Vector Spaces II, Springer,1979.

[5] A. Grothendieck, Produits tensoriels topologiques et espaces nucléaires, Memoires American Mathematical Society, 16, 1955.

[6] Hilja Lisa Huru, Quantization and associativity constraints of the category of graded modules, Cand. Scient. Thesis in Mathematics,University of Tromso,2002.

University of Tromso,9020 Tromso, Norway

E-mail address: perj@math.uit.no

Received May 24, 2005