ANOTHER APPROACH TO CORRESPONDENCE FUNCTORS

SERGE BOUC AND JACQUES THÉVENAZ

Abstract. A new construction of correspondence functors is introduced, allowing for a more direct approach to the subject. In particular a description of the evaluations of simple functors is obtained by means of this approach, simplifying the proofs obtained earlier in [BT4].

1. Introduction

A correspondence functor is a functor from the category of finite sets and correspondences to the category of k-modules, where k is a fixed commutative ring. They are studied in [BT2, BT3, BT4]. A main tool is the correspondence functor F_{T} associated to any finite lattice T. In particular, F_{T} is used to prove some of the main properties of simple correspondence functors. Recall that, for any finite set X, the evaluation F_{T} (X) is the free k-module whose basis is the set T^{X} of all maps from X to T. The simple correspondence functors are isomorphic to quotients of the (so-called) fundamental functors, which consequently play an important role in the theory.

It is proved in [BT3] that there is a perfect duality between F_{T} and F_{T^{op}}, where T^{op} denotes the opposite lattice. Moreover, the fundamental functor 𝕊_{E,R^{op}} appears as a quotient of F_{T}, while its dual 𝕊_{E,R} is isomorphic to a subfunctor of F_{T^{op}}. Here E is the set of join-irreducible elements of the lattice T, R is the natural partial order on E viewed as a full subposet of T, and R^{op} denotes the opposite partial order on E. It is also proved that the subfunctor 𝕊_{E,R} ⊆ F_{T^{op}} is generated by an explicit element γ ∈ F_{T^{op}}(E). However, no further study of this subfunctor was carried out.

In the present paper, a new construction of correspondence functors is described directly inside the functor F_{T^{op}}, without using the duality. Swapping the role of T and T^{op}, we actually work inside a functor F_{T}. In particular, E is the set Mirr(T) of all meet-irreducible elements of T and R is the natural partial order on E. As an interesting aspect of these new constructions, some of the main results proved in [BT4] can be proved in a direct fashion, simplifying the whole approach to fundamental and simple correspondence functors.

We first construct a subfunctor Q_{T} of F_{T}, which can be characterized as the intersection of all the kernels of morphisms F_{T} → F_{T′} associated with surjective join-morphisms T → T′ where |T′| < |T |. In order to describe the structure of Q_{T}, a k-basis of each evaluation Q_{T} (X) at a finite set X is obtained by means of a k-linear idempotent endomorphism

F_{T} (X) ⟶ F_{T} (X) , φ ↦ φ̂

whose image is equal to Q_{T} (X).

Date: December 27, 2025.

1991 Mathematics Subject Classification. AMS Subject Classification: 06B05, 08A05, 16B50, 16D90, 16G99, 18A25, 18B05, 18B35, 18E99.

Key words and phrases. Finite set, correspondence, functor category, simple functor, poset, lattice.

1.1. Theorem. For any finite set X, the set

M̂_{X} = { φ̂ | φ ∈ T^{X} such that E ⊆ φ(X) }

is a k-basis of Q_{T} (X).

The subfunctor Q_{T} contains in turn a subfunctor S_{T}, generated by the specific element î ∈ F_{T} (E), where i : E → T denotes the inclusion map. This generator î is precisely the element γ mentioned above. Again, the map φ ↦ φ̂ is the key for a description of each evaluation S_{T} (X).

1.2. Theorem. For any finite set X, the set

B̂_{X} = { φ̂ | φ ∈ T^{X} such that E ⊆ φ(X) ⊆ G }

is a k-basis of S_{T} (X).

Here G denotes a subset of T which plays a prominent role and which already was a main tool in [BT4]. The proof of this theorem is not easy, but it is much simpler than the proof of a similar result obtained in [BT4]. More precisely, we prove that S_{T} is isomorphic to the fundamental functor 𝕊_{E,R^{op}} and therefore we recover the description of a k-basis of each evaluation 𝕊_{E,R^{op}}(X), which was one of the main achievements in [BT4].

The parametrization of simple correspondence functors appears in [BT2, BT3, BT4]. In particular, we know from [BT4] that any simple correspondence functor is isomorphic to a quotient of a fundamental functor (but this requires some tedious technical work). Motivated by this fact, we introduce an additional construction S_{T,V} and prove directly that this is a simple correspondence functor which is isomorphic to a quotient of S_{T}. Here V is a simple left k Aut(E, R)-module and S_{T,V} = S_{T} ⊗_{k Aut(E,R)} V for a suitable right action of Aut(E, R) on S_{T}. This provides another approach to simple correspondence functors and their parametrization by triples (E, R, V).

The present development differs significantly from the previous procedure in [BT4] because it starts with the construction of S_{T} and rather quickly obtains a basis of each evaluation S_{T} (X). This requires some subtle arguments, but involves less pain than in the previous approach. Only afterwards, S_{T} is shown to be isomorphic to a fundamental functor. Also, simple functors are obtained by a direct construction and it is rather straightforward to view them as quotients of fundamental functors. We see that the approach of the present paper has somehow inverted the main sequence of results. As an outcome, the whole exposition can be presented with a considerably shorter number of pages.

Whereas join-irreducible elements were prominent in our previous approach in [BT4], the present paper uses instead meet-irreducible elements in T. In both theorems above, E denotes the set Mirr(T) of all meet-irreducible elements of T, but we actually prove more generally that the same results hold if we use an arbitrary subset E ⊆ Mirr(T), replacing Q_{T} and S_{T} by more general subfunctors Q_{T}^{E} and S_{T}^{E}. The functor Q_{T}^{E} and its evaluations are studied in Section 3, while S_{T}^{E} appears in Section 4. The theorem on the evaluations of S_{T}^{E} is proved in Section 6, using an intermediate result in Section 5. The evaluation S_{T}^{E}(E) of S_{T}^{E} at the set E is a minimal nonzero evaluation, which is studied in Section 7.

The next sections require to use E = Mirr(T) instead of a subset of Mirr(T). First the product of all restriction maps to S_{T} (E) is proved to be injective in Section 8. This injectivity result is a main tool in Section 9, when we prove that S_{T} is isomorphic to the fundamental functor 𝕊_{E,R^{op}}. The simple functors S_{T,V} are constructed and classified in Section 10. Finally Section 11 explains why the parametrizaion of simple functors does not depend on the choice of a lattice T, but only on the poset (E, R).

Finally, in Sections 12, we explore the link between the approach of the present paper and the development obtained in [BT5], where some other correspondence functors are introduced by means of a suitable duality.

2. Preliminaries on finite lattices

Finite lattices are essential tools in our work on correspondence functors. In the present paper, meet-irreducible elements play a central role. We recall a few elementary facts and fix some notation. More details can be found in [BT3, BT4]. If T is a finite lattice, we write ≤ for its (partial) order relation, ∨ for its join, ∧ for its meet, ˆ0_{T}, or simply ˆ0, for its bottom element and ˆ1_{T}, or simply ˆ1, for its top element. We also use the standard notation for intervals (closed intervals, half-open intervals, open intervals). Since T is finite, a join-closed subset U of T is a lattice with respect to the join of T and the induced meet ˜∧, defined by

u ˜∧ u′ = ⋁_{\begin{matrix}t∈U \\ t≤u \\ t≤u′ \end{matrix}} t .

Similarly, a meet-closed subset U of T is a lattice with respect to the meet of T and the induced join ˜∨, defined by

u ˜∨ u′ = ⋀_{\begin{matrix}t∈U \\ t≥u \\ t≥u′ \end{matrix}} t .

An element of T is called join-irreducible if it cannot be written as the join of some subset of strictly smaller elements of T. In particular, ˆ0 is not join-irreducible because it is an empty join. We write Jirr(T) for the set of join-irreducible elements of T, viewed as a full subposet of T. For any u ∈ T, we define

(2.1)  r_{T} (u) = ⋁_{\begin{matrix}t∈T \\ t<u \end{matrix}} t .

Then r_{T} (u) = u if and only if u ∉ Jirr(T), while, if u ∈ Jirr(T), r_{T} (u) is the unique maximal element in [ ˆ0, u [ and the open interval ] r_{T} (u), u [ is empty.

The opposite lattice T^{op} of a finite lattice T is obtained by reversing the partial order, swapping ∨ and ∧ and also swapping ˆ0 and ˆ1. An element of T is called meet-irreducible if it is join-irreducible in T^{op}, that is, if it cannot be written as the meet of some subset of strictly larger elements of T. In particular, ˆ1 is not meet-irreducible because it is an empty meet. We write Mirr(T) for the set of meet-irreducible elements of T, viewed as a full subposet of T. For any u ∈ T, we define

(2.2)  s_{T} (u) = ⋀_{\begin{matrix}t∈T \\ t>u \end{matrix}} t .

Then s_{T} (u) = u if and only if u ∉ Mirr(T), while, if u ∈ Mirr(T), s_{T} (u) is the unique minimal element in ] u, ˆ1 ] and the open interval ] u, s_{T} (u) [ is empty.

Recall that any t ∈ T is a join of join-irreducible elements, hence

t = ⋁_{\begin{matrix}e∈Jirr(T) \\ e≤t \end{matrix}} e ,

and similarly

(2.3)  t = ⋀_{\begin{matrix}f∈Mirr(T) \\ f≥t \end{matrix}} f .

We shall work with the category having all finite lattices as objects and joinmorphisms as morphisms. Recall that a map α : T → T′ between two finite lattices is a join-morphism if, for any (finite) subset U of T, we have

α( ⋁_{u∈U} u) = ⋁_{u∈U} α(u)

and in particular α(ˆ0) = ˆ0 by using an empty join. Equivalently, α is a joinmorphism if and only if α(t ∨ u) = α(t) ∨ α(u) for all t, u ∈ T and α(ˆ0) = ˆ0. A join-morphism is an order-preserving map, but we emphasize that, in general, it does not preserve meets. Consequently, the image Im(α) of a join-morphism α : T → T′ is only join-closed in T′ and may not be a sublattice of T′.

Let us recall the construction of opposite morphisms. If α : T → T′ is a joinmorphism, there is a join-morphism α^{op} : T′^{op} → T^{op}, called the opposite of α, defined by

α^{op}(t′) = ⋁_{\begin{matrix}t∈T \\ α(t)≤t′ \end{matrix}} t .

In case α is surjective, we can also describe α^{op} by the condition

α^{op}(t′) = ⋁_{\begin{matrix}t∈T \\ α(t)=t′ \end{matrix}} t = sup\{t ∈ T | α(t) = t′\} .

It is not hard to prove that α^{op} is a join-morphism from T′^{op} to T^{op} and that (α^{op})^{op} = α, see [BT3] for details.

Given a ∈ Mirr(T), it is elementary to check that T −{a} is meet-closed, hence a lattice with respect to the induced join. Moreover, the map

(2.4)  f_{a} : T → T −\{a\} ,  f_{a}(t) = \{ \begin{matrix}s_{T} (a) & if t = a , \\ t & if t ≠ a , \end{matrix}

is a join-morphism. This plays a important role in the next section.

3. Some subfunctors of the functor associated to a lattice

We first recall from [BT2, BT3, BT4] some basic definitions about correspondence functors. We denote by C the category of finite sets and correspondences. Its objects are the finite sets and the set C(Y, X) of morphisms from X to Y is the set of all correspondences from X to Y, namely all subsets of Y × X (using a reverse notation which is convenient for left actions). Correspondences can be composed in the usual manner. A correspondence from X to X is also called a relation on X and R_{X} := C(X, X) is the monoid of all relations on X. If k is a commutative ring, we let kC be the category whose objects are the finite sets and the set of morphisms from X to Y is the free k-module kC(Y, X) with basis C(Y, X). In particular, kR_{X} is the algebra of the monoid R_{X}.

A correspondence functor over k is a functor from the category C to the category of k-modules, or in other words a k-linear functor from kC to the category of kmodules. If F is a correspondence functor and U ⊆ Y × X is a correspondence, we simply write

Uφ := F (U)(φ) , ∀ φ ∈ F (X) .

Associated to a finite lattice T, there is a correspondence functor F_{T} whose evaluation at a finite set X is the free k-module F_{T} (X) = kT^{X} with basis the set T^{X} of all maps from X to T. The action of a correspondence U ⊆ Y × X on a map φ ∈ T^{X} is the map in T^{Y} defined by

(3.1)  (Uφ)(y) = ⋁_{\begin{matrix}x∈X \\ (y,x)∈U \end{matrix}} φ(x) , ∀ y ∈ Y .

Recall that if Z is a finite set and S ⊆ Z × Y, then S(Uφ) = (SU)φ where SU denotes the usual composition of correspondences. This shows that F_{T} is indeed a functor.

Whenever π : T → T′ is a join-morphism of finite lattices, there is an associated morphism of correspondence functors π : F_{T} → F_{T′} (still writen π for simplicity) which is defined, for every finite set X, by

π_{X} : F_{T} (X) → F_{T′}(X) , π_{X}(φ) = π ∘ φ , ∀ φ ∈ T^{X} ,

extended by k-linearity from this definition on basis elements.

From now on, we fix a finite lattice T with |T | ≥ 2. Our main aim is to study some canonical subfunctors of F_{T}, the first being Ker(f_{a}) where a ∈ Mirr(T) and where f_{a} : T → T −{a} is the surjective join-morphism defined in (2.4). We emphasize that f_{a} also denotes the morphism of correspondence functors f_{a} : F_{T} → F_{T −{a}}. Its kernel is of course a subfunctor of F_{T}.

3.2. Proposition. Let a ∈ Mirr(T). For any finite set X, Ker(f_{a})(X) has a k-basis

A_{X} = { φ − f_{a} φ | φ ∈ T^{X} with a ∈ φ(X) } .

Proof : It is clear that φ − f_{a} φ ∈ Ker(f_{a})(X) because f_{a} is idempotent. Notice that a does not belong to the image of f_{a} φ, so that φ and f_{a} φ belong respectively to two disjoint subsets of the basis T^{X} of F_{T} (X), namely {φ ∈ T^{X} | a ∈ φ(X)} and {ψ ∈ T^{X} | a ∉ ψ(X)}. Since the elements of the subset {φ ∈ T^{X} | a ∈ φ(X)} are linearly independent, so are the elements of A_{X}. It remains to prove that they generate Ker(f_{a})(X).

Let ∑_{φ∈T^{X}} λ_{φ}φ be an element of Ker(f_{a})(X), where λ_{φ} ∈ k. Thus we have

(3.3)  ∑_{φ∈T^{X}} λ_{φ}(f_{a} φ) = 0 .

For any ψ ∈ T^{X} such that a ∉ ψ(X), define

Y_{ψ} = {φ ∈ T^{X} | f_{a} φ = ψ} .

Then (3.3) can be written

∑_{\begin{matrix}ψ∈T^{X} \\ a∉ψ(X) \end{matrix}} ( ∑_{φ∈Y_{ψ}} λ_{φ})ψ = 0 .

By the linear independence of the functions in T^{X}, we deduce that

∑_{φ∈Y_{ψ}} λ_{φ} = 0 , ∀ψ ∈ T^{X} with a ∉ ψ(x) .

Notice that if φ ∈ Y_{ψ} satisfies a ∉ φ(X), then f_{a} φ = φ, hence φ = ψ. Therefore Y_{ψ} = {ψ} ⊔ Z_{ψ} where

Z_{ψ} = {φ ∈ T^{X} | f_{a} φ = ψ and a ∈ φ(X)} .

It follows that we can write

0 = ∑_{φ∈Y_{ψ}} λ_{φ} = λ_{ψ} + ∑_{φ∈Z_{ψ}} λ_{φ} ,

and therefore λ_{ψ} = − ∑_{φ∈Z_{ψ}} λ_{φ}. Now we obtain

∑_{φ∈T^{X}} λ_{φ}φ = ∑_{\begin{matrix}ψ∈T^{X} \\ a∉ψ(X) \end{matrix}} ∑_{φ∈Y_{ψ}} λ_{φ}φ = ∑_{\begin{matrix}ψ∈T^{X} \\ a∉ψ(X) \end{matrix}} (λ_{ψ}ψ+ ∑_{φ∈Z_{ψ}} λ_{φ}φ) = ∑_{\begin{matrix}ψ∈T^{X} \\ a∉ψ(X) \end{matrix}} ∑_{φ∈Z_{ψ}} λ_{φ}(φ−ψ) .

Since φ ∈ Z_{ψ}, we have ψ = f_{a} φ and therefore the latter sum is a linear combination of elements of the form φ − f_{a} φ with a ∈ φ(X), as was to be shown.

Our next main subfunctor of F_{T} is

Q_{T} = ⋂_{a∈Mirr(T)} Ker(f_{a}) .

3.4. Remark. Let T′ be a lattice with |T′| < |T |. It is not hard to prove that any surjective join-morphism π : T → T′ is a composite ρf_{a} for some a ∈ Mirr(T), where ρ : T −{a} → T′ is the restriction of π. Consequently Ker(f_{a}) ⊆ Ker(π). Since any such f_{a} : T → T −{a} is a surjective join-morphism, it follows that

Q_{T} = ⋂_{π} Ker(π) ,

where π runs over the set of all surjective join-morphisms from T to a lattice T′ with |T′| < |T |. This is a pleasant description of Q_{T}, but we shall use only the previous definition.

Many results of this paper hold for a subset of Mirr(T) instead of the whole of Mirr(T). For simplicity of notation, we now fix a subset E ⊆ Mirr(T) and we use a superscript E whenever we need to emphasize a dependence on E. In particular, we define the subfunctor

(3.5)  Q_{T}^{E} = ⋂_{a∈E} Ker(f_{a}) .

Notice that Q_{T}^{∅} = F_{T}, Q_{T}^{{a}} = Ker(f_{a}), and Q_{T}^{Mirr(T)} = Q_{T}, as defined above.

For any subset A of Mirr(T), we also generalize the map s_{T} defined in (2.2) and introduce an operator s_{T}^{A} : T → T defined by

(3.6)  s_{T}^{A}(t) = \{ \begin{matrix}t & if t ∉ A , \\ s_{T} (t) & if t ∈ A . \end{matrix}

For every φ ∈ T^{X}, we now define a map φ_{A} : X → T as the composite

(3.7)  φ_{A} = s_{T}^{A} ∘ φ ,

or explicitly

φ_{A}(x) = \{ \begin{matrix}φ(x) & if φ(x) ∉ A, \\ s_{T} (φ(x)) & if φ(x) ∈ A, \end{matrix}

Notice that φ_{∅} = φ and φ_{{a}} = f_{{a}}φ for any a ∈ Mirr(T). Next we define

(3.8)  φ^{E} = ∑_{A⊆E}(−1)^{|A|} φ_{A} .

Note that φ^{{a}} = φ − f_{{a}}φ was used in Proposition 3.2. The element φ^{E} belongs to F_{T} (X) but we have more.

3.9. Lemma. For any φ ∈ T^{X}, we have φ^{E} ∈ Q_{T}^{E}(X).

Proof : We need to prove that f_{a} φ^{E} = 0 for every a ∈ E. We first write

f_{a} φ^{E} = ∑_{\begin{matrix}A⊆E \\ a∈A \end{matrix}}(−1)^{|A|}f_{a} φ_{A} + ∑_{\begin{matrix}B⊆E \\ a∉B \end{matrix}}(−1)^{|B|}f_{a} φ_{B} = ∑_{B⊆E−\{a\}}(−1)^{|B|}f_{a}(φ_{B} −φ_{B∪\{a\}}) ,

and we claim that f_{a}(φ_{B} − φ_{B∪{a}}) = 0 for any B ⊆ E−{a}.

To prove the claim, we now fix x ∈ X and we consider the various possibilities for φ(x). If φ(x) ∉ B ∪ {a}, then

f_{a} φ_{B}(x) = f_{a} φ(x) = f_{a}φ_{B∪{a}}(x) .

If φ(x) ∈ B, then

f_{a} φ_{B}(x) = f_{a} s_{T} (φ(x)) = f_{a}φ_{B∪{a}}(x) .

If φ(x) = a, then

f_{a} φ_{B}(x) = f_{a} φ(x) = f_{a}(a) = s_{T} (a) = f_{a}(s_{T} (a)) = f_{a} s_{T} (φ(x)) = f_{a} φ_{B∪{a}}(x) .

This shows that f_{a} φ_{B} = f_{a} φ_{B∪{a}}, proving the claim.

In view of its important role in what follows, it is convenient to define

M_{X}^{E} = {φ ∈ T^{X} | E ⊆ φ(X)} .

This comes into play in the following result.

3.10. Lemma. Let φ ∈ T^{X}.

(a) If φ ∉ M_{X}^{E}, then φ^{E} = 0.

(b) For any nonempty subset A of E, φ_{A} ∉ M_{X}^{E}. In particular φ̂_{A}^{E} = 0.

Proof : (a) Since φ ∉ M_{X}^{E} by assumption, there exists a ∈ E such that a ∉ φ(X). For any B ⊆ E−{a} and any x ∈ X, we have

φ_{B∪\{a\}}(x) = \{ \begin{matrix}φ(x) & if φ(x) ∉ B ∪ \{a\}, \\ s_{T} (φ(x)) & if φ(x) = a, \\ s_{T} (φ(x)) & if φ(x) ∈ B. \end{matrix}

The second case does not occur because a ∉ φ(X). Therefore

φ_{B∪\{a\}}(x) = \{ \begin{matrix}φ(x) & if φ(x) ∉ B, \\ s_{T} (φ(x)) & if φ(x) ∈ B. \end{matrix}

But this is precisely the definition of φ_{B}(x), so φ_{B∪{a}}(x) = φ_{B}(x). It follows that the two functions φ_{B} and φ_{B∪{a}} are equal and therefore cancel in the alternating sum defining φ^{E}. This holds for any B ⊆ E−{a}, so all the terms cancel in the alternating sum and φ^{E} = 0.

(b) Let a be a minimal element in A, so that a cannot have the form a = s_{T} (b) for some b ∈ A. Then the definition of φ_{A} shows that a is not in the image of φ_{A}. Thus φ_{A} ∉ M_{X}^{E}, as required. Applying (a) to the function φ_{A}, we get in particular φ̂_{A}^{E} = 0.

3.11. Corollary. Let M̂_{X}^{E} = {φ^{E} | φ ∈ M_{X}^{E}}. The map

M_{X}^{E} ⟶ M̂_{X}^{E} , φ ↦ φ^{E}

is a bijection which preserves linear independence.

Proof : Since φ_{A} ∉ M_{X}^{E} for every nonempty subset A ⊆ E, the definition of φ^{E} shows that every φ^{E} ∈ M_{X}^{E} has the form

φ^{E} = φ + φ′ ,

where φ′ is a linear combination of functions outside M_{X}^{E}. Since the elements of M_{X}^{E} are linearly independent (because they form a subset of the basis T^{X} of F_{T} (X)), we deduce that so are the elements of M̂_{X}^{E}. In particular, the map φ ↦ φ^{E} is a bijection.

Our main result in this section provides an explicit description of the subfunctor Q_{T}^{E}.

3.12. Theorem. For any finite set X, the set M̂_{X}^{E} = {φ^{E} | φ ∈ M_{X}^{E}} is a k-basis of Q_{T}^{E}(X).

The proof is based on the following technical lemma.

3.13. Lemma. Let ω ∈ Q_{T}^{E}(X), ω ≠ 0, and write ω = ∑_{i∈I} λ_{i}ψ_{i} for some index set I, with ψ_{i} ∈ T^{X}, λ_{i} ∈ k, and λ_{i} ≠ 0. Then there exists i_{0} ∈ I such that ψ_{i_{0}} ∈ M_{X}^{E}.

Proof : Choose i_{0} such that |ψ_{i_{0}}(X) ∩ E| is maximal among all |ψ_{i}(X) ∩ E|, i ∈ I. Suppose that

(3.14)  a ∉ ψ_{i_{0}}(X) for some a ∈ E .

Since ω ∈ Q_{T}^{E}(X), we have ω ∈ Ker(f_{a})(X), so ω can be written as a linear combination of the basis

A_{X} = { φ − f_{a} φ | φ ∈ T^{X} with a ∈ φ(X) } ,

see Proposition 3.2. Since ψ_{i_{0}} is one of the functions appearing in the given sum ω = ∑_{i∈I} λ_{i}ψ_{i}, we see that ψ_{i_{0}} must be equal to some f_{a} φ for some φ ∈ T^{X} with a ∈ φ(X) (noticing that it cannot be equal to φ because a ∈ φ(X) while a ∉ ψ_{i_{0}}(X)). But f_{a} φ always comes together with some φ in the basis A_{X}, so φ must also appear in the given expression of ω.

Now the functions φ and f_{a} φ = ψ_{i_{0}} take the same values in E−{a} whereas, in addition, φ also takes the value a. Thus |φ(X) ∩ E| = |ψ_{i_{0}}(X) ∩ E| + 1. This contradicts the maximality in the choice of i_{0}. Therefore our supposition (3.14) is impossible and so E ⊆ ψ_{i_{0}}(X). Hence ψ_{i_{0}} ∈ M_{X}^{E}.

Proof of Theorem 3.12. By Corollary 3.11, we know that the elements of M̂_{X}^{E} are linearly independent. We need now to prove that they generate Q_{T}^{E}(X). Let ω ∈ Q_{T}^{E}(X) and write

ω = ∑_{φ∈T^{X}} λ_{φ}φ = ∑_{φ∈M_{X}^{E}} λ_{φ}φ + ∑_{ψ∉M_{X}^{E}} λ_{ψ}ψ .

As in the proof of Corollary 3.11, we can write φ = φ^{E} − φ′ where φ′ is a linear combination of functions outside M_{X}^{E}. Thus we get

ω = ∑_{φ∈M_{X}^{E}} λ_{φ}(φ^{E} − φ′) + ∑_{ψ∉M_{X}^{E}} λ_{ψ}ψ = ∑_{φ∈M_{X}^{E}} λ_{φ}φ^{E} + ω′ ,

where ω′ is a linear combination of functions outside M_{X}^{E}. Since ω ∈ Q_{T}^{E}(X) and φ̂ ∈ Q_{T}^{E}(X) by Lemma 3.9, we obtain ω′ ∈ Q_{T}^{E}(X). By Lemma 3.13, ω′ cannot be a linear combination of functions outside M_{X}^{E}, unless ω′ = 0. It follows that ω = ∑_{φ∈M_{X}^{E}} λ_{φ}φ^{E}, proving that M̂_{X}^{E} generates Q_{T}^{E}(X).

3.15. Remark. A subfunctor H_{T} of F_{T} is introduced in [BT3] such that a k-basis of H_{T} (X) is the set of all functions φ : X → T satisfying the condition Jirr(T) ⊈ φ(X). The subfunctor Q_{T^{op}} = Q_{T^{op}}^{Mirr(T^{op})} of the present paper is such that a k-basis of Q_{T^{op}}(X) is the set of all functions φ : X → T^{op} satisfying the condition Mirr(T^{op}) = Jirr(T) ⊆ φ(X). It is not hard to prove that H_{T} and Q_{T^{op}} are orthogonal to each other with respect to the duality between F_{T} and F_{T^{op}} mentioned in the introduction.

We continue with the fixed subset E ⊆ Mirr(T) and we write Σ_{E} for the symmetric group of all permutations of the set E.

3.16. Corollary.

(a) If |X| < |E|, then Q_{T}^{E}(X) = {0}.

(b) Q_{T}^{E}(E) has a k-basis M̂_{E}^{E} = { îσ | σ ∈ Σ_{E}}, where i denotes the inclusion map E → T.

Proof : (a) Whenever |X| < |E|, the condition E ⊆ φ(X) is impossible, so M_{X}^{E} is empty. Therefore M̂_{X}^{E} is empty too and Q_{T}^{E}(X) = {0}.

(b) When X = E, the condition E ⊆ φ(X) implies that φ must map E onto E. This forces also the injectivity of φ and so φ = i ∘ σ for some σ ∈ Σ_{E}. Thus the basis M̂_{E}^{E} is as stated.

We end this section with another property of Q_{T}^{E}. The definition of φ^{E} in (3.8) extends k-linearly to a k-linear map F_{T} (X) → F_{T} (X) specified on the basis by φ ↦ φ^{E} and still written ω ↦ ω^{E} for any ω ∈ F_{T} (X).

3.17. Proposition. The k-linear endomorphism ω ↦ ω^{E} of F_{T} (X) is idempotent and its image is equal to Q_{T}^{E}(X). In particular, ω^{E} = ω for every ω ∈ Q_{T}^{E}(X).

Proof : Let φ ∈ T^{X}. By Lemma 3.10, φ̂_{A}^{E} = 0 for any nonempty subset A of E. It follows that

φ̂^{E}^{E} = ∑_{A⊆E}(−1)^{|A|} φ̂_{A}^{E} = φ̂_{∅}^{E} = φ^{E} .

Extending this by k-linearity, we see that the map ω ↦ ω^{E} is idempotent. Since φ^{E} belongs to Q_{T}^{E}(X) if φ ∈ M_{X}^{E} (by Lemma 3.9) and is zero otherwise (by Lemma 3.10), the image of the idempotent is exactly Q_{T}^{E}(X) (by Theorem 3.12).

It should be noted that the image of the idempotent happens to be a subfunctor of F_{T}, but the idempotent map is not a morphism of functors.

4. More subfunctors

We now introduce more subfunctors of F_{T} which will be our main objects of study. We let T be a finite lattice with |T | ≥ 2 and we let E be a fixed subset of Mirr(T), as before. Since E is fixed, we can simplify notation and write simply φ̂ instead of φ^{E}, as defined by (3.8). We let S_{T}^{E} be the subfunctor of F_{T} generated by î ∈ F_{T} (E), where i denotes the inclusion map E → T. Explicitly, for any finite set X, S_{T}^{E}(X) is k-linearly generated by all elements of F_{T} (X) of the form U î where U ⊆ X × E is a correspondence. Since î ∈ Q_{T}^{E}(E) by Lemma 3.9, S_{T}^{E} is a subfunctor of Q_{T}^{E}.

We need to study U î. Since this belongs to Q_{T}^{E}(X), it can be expressed in the basis M̂_{X}^{E} = {φ̂ | φ ∈ T^{X}, E ⊆ φ(X)} (see Theorem 3.12).

4.1. Lemma. Let U ⊆ X × E be a correspondence. Then

U î = ∑_{A⊆E}(−1)^{|A|} Ui_{A} = ∑_{A⊆E}(−1)^{|A|} Ûi_{A} = ∑_{\begin{matrix}A⊆E \\ Ui_{A}∈M_{X} \end{matrix}}(−1)^{|A|} Ûi_{A} .

Proof : The first equality follows from the definition of î and k-linearity. Since U î belongs to Q_{T}^{E}(X) (because Q_{T}^{E} is a subfunctor), it is fixed under the map ω ↦ ω̂ (by Proposition 3.17), proving the second equality. Finally Ûi_{A} = 0 if Ui_{A} ∉ M_{X}^{E} (by Lemma 3.10), proving the third equality.

In this section and the next, our purpose will be to prove that S_{T}^{E}(X) has a basis

(4.2)  B̂_{X}^{E} = {φ̂ | φ ∈ B_{X}^{E}} ,

where φ̂ = φ^{E} and B_{X}^{E} = {φ ∈ T^{X} | E ⊆ φ(X) ⊆ G^{E}}, for a suitable subset G^{E} of T, which we need to define. Our previous approach to simple correspondence functors in [BT4] used a suitable lattice T′, the set E = Jirr(T′), the subset ⋀E of all meets of elements of E and a certain subset G = ⋀E ⊔ G̃. Since we have swapped the role of T and T^{op}, we need here the opposite version of this construction. We start with an arbitrary subset E ⊆ Mirr(T) and we let ⋁E be the set of all joins of elements of E. For any t ∈ T, we define

(4.3)  ρ(t) = ⋁_{\begin{matrix}e∈E \\ e<t \end{matrix}} e and inductively ρ^{j}(t) = ρρ^{j−1}(t) for any j > 1 ,

and we set ρ^{∞}(t) = ρ^{n}(t) where n is such that ρ^{n}(t) = ρ^{n+1}(t). Notice that ρ depends on E and should be written ρ^{E}, but we keep the notation simple for the clarity of reading. Using the map s_{T}^{E} defined in (3.6), we write inductively (s_{T}^{E})^{j}(t) = s_{T}^{E}(s_{T}^{E})^{j−1}(t) for any j > 1, and (s_{T}^{E})^{∞}(t) = (s_{T}^{E})^{n}(t) where n is such that (s_{T}^{E})^{n}(t) = (s_{T}^{E})^{n+1}(t). Then we define

(4.4) G^{E} = ⋁E ⊔ G̃^{E} where G̃^{E} = {h ∈ T | (s_{T}^{E})^{∞}ρ^{∞}(h) = h, h ∉ ⋁E} .

As the set G̃^{E} plays a prominent role throughout this paper, it is important to note its properties. The following lemma is Lemma 2.9 in [BT4], applied to the opposite lattice T^{op} and to the fixed subset E ⊆ Mirr(T).

4.5. Lemma. For any h ∈ G̃^{E}, there is a unique chain

b_{n} < b_{n−1} < … < b_{1} < b_{0} = h (with n ≥ 1)

such that

(a) s_{T}^{E}(b_{i+1}) = b_{i} for 0 ≤ i ≤ n − 1 and s_{T}^{E}(b_{0}) = b_{0}.

(b) ρ(b_{i}) = b_{i+1} for 0 ≤ i ≤ n − 1 and ρ(b_{n}) = b_{n}.

Part (a) implies that b_{1}, …, b_{n} belong to E, but b_{0} = h ∉ E. In particular b_{0} = (s_{T}^{E})^{∞}(b_{n}). Also, (b) implies that ρ^{∞}(b_{0}) = b_{n}. Note that the condition h ∉ ⋁E forces ρ(h) < h, that is b_{1} < b_{0}, so that n ≥ 1.

We write kB̂_{X}^{E} for the k-linear span of the set B̂_{X}^{E} defined in (4.2). Notice that B̂_{X}^{E} is a k-linearly independent set, because it is a subset of the basis M̂_{X}^{E} of Q_{T}^{E}(X).

4.6. Theorem. For any finite set X, S_{T}^{E}(X) ⊆ kB̂_{X}^{E}.

We first need a technical lemma.

4.7. Lemma. Let x ∈ X and U ⊆ X × E. Suppose that {e ∈ E | (x, e) ∈ U}, viewed as a subposet of E, has at least two maximal elements. Then (Ui)(x) = (Ui_{A})(x) for every subset A ⊆ E. Moreover (Ui)(x) > e for every element e of the poset {e ∈ E | (x, e) ∈ U}.

Proof : Let f, g be two distinct maximal elements of {e ∈ E | (x, e) ∈ U}. Then, by Definition 3.1,

(Ui)(x) = ⋁_{(x,e)∈U} i(e) = ⋁_{(x,e)∈U} e ≥ f, g .

Therefore (Ui)(x) ≥ f ∨ g > f and so (Ui)(x) ≥ s_{T}^{E}(f) > f. Similarly (Ui)(x) ≥ s_{T}^{E}(g) > g for every maximal element g of the poset {e ∈ E | (x, e) ∈ U}.

If e ∈ {e ∈ E | (x, e) ∈ U} and g ∈ E is maximal with e ≤ g, then s_{T}^{E}(e) ≤ s_{T}^{E}(g) ≤ (Ui)(x). Therefore

(Ui_{E})(x) = ⋁_{(x,e)∈U} i_{E}(e) = ⋁_{(x,e)∈U} s_{T}^{E}(e) ≤ (Ui)(x) .

It follows that, for every subset A ⊆ E, we get

(Ui)(x) ≤ (Ui_{A})(x) ≤ (Ui_{E})(x) ≤ (Ui)(x) ,

hence (Ui_{A})(x) = (Ui)(x).

Proof of Theorem 4.6. Since S_{T}^{E}(X) is k-linearly generated by all elements of F_{T} (X) of the form U î where U ⊆ X × E, we need to prove that U î belongs to kB̂_{X}^{E}. In the expression

U î = ∑_{\begin{matrix}A⊆E \\ Ui_{A}∈M_{X} \end{matrix}}(−1)^{|A|} Ûi_{A}

of Lemma 4.1, we will show that the function Ui_{A} belongs to B_{X}^{E}, that is, E ⊆ (Ui_{A})(X) ⊆ G^{E}, where G^{E} is defined by (4.4). Since Ui_{A} ∈ M_{X}^{E}, we know that E ⊆ (Ui_{A})(X). We let x ∈ X and we need to show that (Ui_{A})(x) ∈ G^{E}.

If {e ∈ E | (x, e) ∈ U} has at least two maximal elements, then by Lemma 4.7

(Ui_{A})(x) = (Ui)(x) = ⋁_{(x,e)∈U} e ∈ ⋁E ⊆ G^{E} .

If {e ∈ E | (x, e) ∈ U} is empty, then (Ui_{A})(x) is an empty join (i.e. ˆ0), which belongs to ⋁E, hence to G^{E}. So we can assume that {e ∈ E | (x, e) ∈ U} has a unique maximal element a. Then

(Ui_{A})(x) = ⋁_{(x,e)∈U} i_{A}(e) = \{ \begin{matrix}a & if a ∉ A , \\ s_{T} (a) & if a ∈ A . \end{matrix}

Now a ∈ E ⊆ ⋁E ⊆ G^{E}, so we are left to prove that s_{T} (a) ∈ G^{E}. This is again clear if s_{T} (a) ∈ ⋁E, so we now assume that s_{T} (a) ∉ ⋁E for some a ∈ A and we have to prove that s_{T} (a) ∈ G̃^{E}.

Let b = b_{0} = s_{T} (a) and b_{1} = a. We have ρ(b) ≤ b, ρ(b) ∈ ⋁E, and b ∉ ⋁E, hence ρ(b) < b. Now b = s_{T} (b_{1}), so the interval ]b_{1}, b[ is empty. Thus the strict inequality b_{1} ≤ ρ(b) < b forces ρ(b) = b_{1}. For any integer i ≥ 1, we set b_{i} = ρ^{i}(b) and we let n be minimal such that ρ^{n}(b) = ρ^{∞}(b), so that we have a chain

b_{n} < b_{n−1} < … < b_{1} < b_{0} = b with ρ(b_{i}) = b_{i+1} (0 ≤ i ≤ n − 1) .

We now prove by induction that

(4.8)  b_{r+1} ∈ A and s_{T} (b_{r+1}) = b_{r} for 0 ≤ r ≤ n − 1 .

This is clear if r = 0 because b_{1} = a ∈ A and s_{T} (b_{1}) = s_{T} (a) = b_{0}. So assume that b_{i} ∈ A and s_{T} (b_{i}) = b_{i−1} for some 1 ≤ i ≤ n − 1. Since Ui_{A} ∈ M_{X}^{E}, we have

b_{i} ∈ A ⊆ E ⊆ (Ui_{A})(X) ,

hence b_{i} = (Ui_{A})(x_{i}) for some x_{i} ∈ X. Let E_{i} = {e ∈ E | (x_{i}, e) ∈ U}. If E_{i} had at least two maximal elements, then Lemma 4.7 would imply

b_{i} = (Ui_{A})(x_{i}) = (Ui)(x_{i}) = ⋁_{e∈E_{i}} e ,

with each such e strictly smaller than b_{i}. Therefore we would get ρ(b_{i}) = b_{i} contrary to the fact that ρ(b_{i}) = b_{i+1} (since i ≤ n − 1). If now E_{i} was empty, then we would obtain b_{i} = ˆ0, contrary to the fact that ρ(b_{i}) = b_{i+1} < b_{i}. It follows that E_{i} has a unique maximal element c_{i} and we get

b_{i} = (Ui_{A})(x_{i}) = ⋁_{e∈E_{i}} i_{A}(e) = i_{A}(c_{i}) = \{ \begin{matrix}c_{i} & if c_{i} ∉ A , \\ s_{T} (c_{i}) & if c_{i} ∈ A . \end{matrix}

The case c_{i} ∉ A is impossible because we would get b_{i} = c_{i}, but we know that b_{i} ∈ A (by induction). Thus c_{i} ∈ A and b_{i} = s_{T} (c_{i}). Now we have c_{i} ∈ E and c_{i} < b_{i}, so that

c_{i} ≤ ⋁_{\begin{matrix}f∈E \\ f<b_{i} \end{matrix}} f = ρ(b_{i}) = b_{i+1} < b_{i} = s_{T} (c_{i}) .

Since the interval ]c_{i}, s_{T} (c_{i})[ is empty, we obtain c_{i} = ρ(b_{i}) = b_{i+1}. Therefore s_{T} (b_{i+1}) = s_{T} (c_{i}) = b_{i} as required for the proof of (4.8). Moreover, we have seen that b_{i+1} = c_{i} ∈ A, completing the induction proof of (4.8). Note also that s_{T} (b_{0}) = b_{0} because b_{0} = b ∉ E.

Now Definition 3.6 implies that s_{T} (b_{i}) = s_{T}^{E}(b_{i}) when 0 ≤ i ≤ n − 1, because b_{i} ∈ E. So we get a chain

b_{n} < b_{n−1} < … < b_{1} < b_{0} = b with ρ(b_{i}) = b_{i+1} and s_{T}^{E}(b_{i+1}) = b_{i} (0 ≤ i ≤ n−1) .

Therefore b = (s_{T}^{E})^{∞}ρ^{∞}(b). Since we started with b = s_{T} (a) ∉ ⋁E, we obtain s_{T} (a) ∈ G̃^{E}, as required. This completes the proof of Theorem 4.6.

5. The element θ̂

By Theorem 4.6, we know that S_{T}^{E}(X) ⊆ kB̂_{X}^{E}. Our main task in Section 6 will be to show that the opposite inclusion also holds. In other words, we will need to prove that φ̂ ∈ S_{T}^{E}(X) for any function φ ∈ T^{X} such that E ⊆ φ(X) ⊆ G^{E}, where we write again φ̂ = φ^{E} for simplicity of notation. In this section, we prove that this holds for the specific element θ̂, where θ : G^{E} → T denotes the inclusion map and G^{E} = ⋁E ⊔ G̃^{E} as in (4.4). In fact, θ̂ plays a universal role.

The proof that θ̂ ∈ S_{T}^{E}(G^{E}) uses a subset P of E which we now define. By Lemma 4.5, for each h ∈ G̃^{E}, there is a totally ordered subset

b_{n} < b_{n−1} < … < b_{1} < b_{0} = h with ρ(b_{i}) = b_{i+1} and s_{T}^{E}(b_{i+1}) = b_{i} (0 ≤ i ≤ n − 1)

and we write P_{h} = {b_{n}, b_{n−1}, …, b_{1}}. Since s_{T}^{E}(b_{i}) = b_{i−1} > b_{i}, we have b_{i} ∈ E for each 1 ≤ i ≤ n, hence P_{h} ⊆ E. However, the top element b_{0} = h does not belong to E. Note that s_{T}^{E}(ρ(b_{i})) = b_{i} for 0 ≤ i ≤ n − 1, while the bottom element b_{n} = ρ^{∞}(h) is in P_{h} and is fixed under ρ. We then define

(5.1)  P = ⨆_{h∈G̃^{E}} P_{h}

and we note that this is a disjoint union because any b_{i} ∈ P_{h} determines uniquely h via h = (s_{T}^{E})^{i}(b_{i}) = (s_{T}^{E})^{∞}(b_{i}). We also set

(5.2)  P^{∞} = ρ^{∞}(P) = {ρ^{∞}(h) | h ∈ G̃^{E}}

and we recall that ρ^{∞}(h) is the bottom element b_{n} in the notation above. Finally, for any subset C ⊆ P, we define a correspondence

U_{C} = {(g, e) ∈ G^{E} × E | e < g} ⊔ {(e, e) | e ∈ E−C} ⊆ G^{E} × E ,

which we can also write as

(5.3)  U_{C} = {(g, e) | g ∈ (G^{E}−E) ⊔ C , e<g} ∪ {(g, e) | g ∈ E−C , e≤g} .

Note that C ⊆ D implies U_{C} ⊇ U_{D}.

5.4. Theorem. Let θ : G^{E} → T be the inclusion map. Then

θ̂ = ∑_{C⊆P}(−1)^{|C|} U_{C} î.

In particular, θ̂ belongs to S_{T} (G^{E}).

The proof is rather involved. We will start with the right hand side and use several reductions to show that it is equal to θ̂. We first decompose U_{C} î according to Lemma 4.1:

(5.5)  U_{C} î = ∑_{A⊆E}(−1)^{|A|} { \atop U_{C}i_{A}} .

We need to describe the values of the function U_{C}i_{A} : G^{E} → T. The set G^{E} = ⋁E ⊔ G̃^{E} is a disjoint union

G^{E} = (E−C) ⊔ C ⊔ (⋁E−E) ⊔ G̃^{E} ,

and we will consider the cases successively.

5.6. Lemma. Let C ⊆ P and A ⊆ E. Let g ∈ G^{E}.

(a) If g ∈ E−C, then (U_{C}i_{A})(g) = \{ \begin{matrix}g & if g ∉ A , \\ s_{T} (g) & if g ∈ A . \end{matrix}

(b) If g ∈ C and g ∉ P^{∞}, then (U_{C}i_{A})(g) = \{ \begin{matrix}ρ(g) & if ρ(g) ∉ A , \\ g & if ρ(g) ∈ A . \end{matrix}

(c) If g ∈ C and g ∈ P^{∞}, then (U_{C}i_{A})(g) = g.

(d) If g ∈ ⋁E−E, then (U_{C}i_{A})(g) = g.

(e) If g ∈ G̃^{E}, then (U_{C}i_{A})(g) = \{ \begin{matrix}ρ(g) & if ρ(g) ∉ A , \\ g & if ρ(g) ∈ A . \end{matrix}

Moreover, in all cases (b)-(e), we have (U_{C}i_{A})(g) = ⋁_{e<g} i_{A}(e).

Proof : Recall that (U_{C}i_{A})(g) = ⋁_{\begin{matrix}e∈E \\ (g,e)∈U_{C} \end{matrix}} i_{A}(e).

(a) Let g ∈ E−C. The definition of U_{C} shows that (g, g) ∈ U_{C}, so that g is the unique maximal element of {e ∈ E | (g, e) ∈ U_{C}}. Therefore

(U_{C}i_{A})(g) = i_{A}(g) = \{ \begin{matrix}g & if g ∉ A , \\ s_{T} (g) & if g ∈ A . \end{matrix}

(b) Let g ∈ C with g ∉ P^{∞}. The definition of U_{C} shows that (g, e) ∈ U_{C} if and only if e < g, hence (U_{C}i_{A})(g) = ⋁_{e<g} i_{A}(e). Since P is a disjoint union of totally ordered subsets and g ∉ P^{∞}, ρ(g) is the unique maximal element of {e ∈ E | e < g} and s_{T} (ρ(g)) = g. Therefore

(U_{C}i_{A})(g) = i_{A}(ρ(g)) = \{ \begin{matrix}ρ(g) & if ρ(g) ∉ A , \\ s_{T} (ρ(g)) = g & if ρ(g) ∈ A . \end{matrix}

(c) Let g ∈ C with g ∈ P^{∞}. Then ρ(g) = g. Moreover, e < g implies g ≥ s_{T} (e) ≥ i_{A}(e) ≥ e and therefore

g ≥ ⋁_{e<g} s_{T} (e) ≥ ⋁_{e<g} i_{A}(e) ≥ ⋁_{e<g} e = ρ(g) = g .

It follows that (U_{C}i_{A})(g) = ⋁_{e<g} i_{A}(e) = g.

(d) Let g ∈ ⋁E−E. The definition of U_{C} implies that (U_{C}i_{A})(g) = ⋁_{e<g} i_{A}(e). Moreover, g is a join of elements of E, all strictly smaller than g since g ∉ E. Thus g = ⋁_{e<g} e. We obtain again

g ≥ ⋁_{e<g} s_{T} (e) ≥ ⋁_{e<g} i_{A}(e) ≥ ⋁_{e<g} e = g ,

so that (U_{C}i_{A})(g) = ⋁_{e<g} i_{A}(e) = g.

(e) Let g ∈ G̃^{E}. The definition of U_{C} implies that (U_{C}i_{A})(g) = ⋁_{e<g} i_{A}(e). Moreover, ρ(g) is in P, hence in particular in E, so that ρ(g) is the unique maximal element of {e ∈ E | e < g}. It follows that

(U_{C}i_{A})(g) = i_{A}(ρ(g)) = \{ \begin{matrix}ρ(g) & if ρ(g) ∉ A , \\ s_{T} (ρ(g)) = g & if ρ(g) ∈ A . \end{matrix}

This completes the proof.

We now show that many terms { \atop U_{C}i_{A}} vanish.

5.7. Lemma. Let C ⊆ P and A ⊆ E. If A ⊈ P, then { \atop U_{C}i_{A}} = 0.

Proof : By Lemma 3.10, all we have to show is that U_{C}i_{A} ∉ M_{G^{E}}^{E}, that is, E ⊈ (U_{C}i_{A})(G^{E}). Since A ⊈ P, we can choose a ∈ A minimal such that a ∉ P and we claim that a ∉ (U_{C}i_{A})(G^{E}). Looking for a contradiction, we assume that a = (U_{C}i_{A})(g) for some g ∈ G^{E} and we treat the cases of Lemma 5.6.

(a) Suppose g ∈ E−C. If g ∉ A, Lemma 5.6 asserts that a = g, hence a ∉ A, a contradiction. If g ∈ A, we get a = s_{T} (g) > g and by minimality of a, we deduce that g ∈ P. Since P is a disjoint union of totally ordered subsets, we must have s_{T} (g) ∈ P or s_{T} (g) ∈ G̃^{E} by Lemma 4.5. The case a = s_{T} (g) ∈ P is impossible because a ∉ P. The case a = s_{T} (g) ∈ G̃^{E} is impossible because the elements of G̃^{E} are not in E, while a ∈ A ⊆ E.

(b) Suppose g ∈ C and g ∉ P^{∞}. The case a = ρ(g) cannot hold because a ∈ A and ρ(g) ∉ A. The case a = g cannot hold either because g ∈ C, hence g ∈ P, whereas a ∉ P.

(c) Suppose g ∈ C and g ∈ P^{∞}. Then a = g and this contradicts again the fact that g ∈ P and a ∉ P.

(d) Suppose g ∈ ⋁E−E. We obtain again a = g. This is a contradiction because g ∉ E and a ∈ A ⊆ E.

(e) Suppose g ∈ G̃^{E}. The first case a = ρ(g) cannot hold because a ∈ A and ρ(g) ∉ A. The second case a = g cannot hold either because g ∈ G̃^{E} and this is not in E while a ∈ E.

We have obtained a contradiction in all cases, proving the claim a ∉ (U_{C}i_{A})(G^{E}). Therefore E ⊈ (U_{C}i_{A})(G^{E}) and this completes the proof of Lemma 5.7.

5.8. Corollary. U_{C} î = ∑_{A⊆P}(−1)^{|A|} { \atop U_{C}i_{A}}.

Proof : Instead of running over all subsets A of E, the sum (5.5) can be restricted to run only over all subsets A of P, by Lemma 5.7.

We will now consider the alternating sum over C as in the statement of Theorem 5.4 and we want to show that further cancelations occur. To this end, we fix a subset A of P which we write A = ⋃_{h∈G̃^{E}} A_{h} where A_{h} = A ∩ P_{h}. Whenever P_{h}−A_{h} is nonempty, we let d_{h} be the smallest element of P_{h}−A_{h} (which exists because P_{h} is totally ordered). Then we define

D = {d_{h} | h ∈ G̃^{E} such that P_{h}−A_{h} ≠ ∅} .

5.9. Lemma. Let A ⊆ P and let D be defined as above. If C is a subset of P and S = C−D, then U_{C}i_{A} = U_{S}i_{A}.

Proof : Again we need to treat the cases of Lemma 5.6. Let g ∈ G^{E}. Assume first that g ∈ E−C. Since S ⊆ C, g also belongs to E−S. Case (a) of Lemma 5.6 applies and we see that (U_{C}i_{A})(g) = (U_{S}i_{A})(g).

Suppose now that g ∈ S, so that g ∈ C. The additional statement in Lemma 5.6 yields (U_{C}i_{A})(g) = ⋁_{e<g} i_{A}(e) = (U_{S}i_{A})(g).

Next assume that g ∈ E−S and g ∈ C. Since S = C−D, we deduce that g ∈ D and in particular g ∉ A. Then Case (a) of Lemma 5.6 implies that (U_{S}i_{A})(g) = g because g ∉ A. Since g ∈ D, we can write g = d_{h} and d_{h} is the smallest element of P_{h} − A_{h}, for some h ∈ G̃^{E} (namely h = (s_{T}^{E})^{∞}(g)). For the computation of (U_{C}i_{A})(g), we note that g ∈ C, so either (b) or (c) in Lemma 5.6 applies, depending on whether or not g ∈ P^{∞}.

If g ∉ P^{∞}, we have ρ(g) < g and both lie in the same totally ordered set P_{h}. Since g = d_{h} is the smallest element of P_{h} − A_{h}, we must have ρ(g) ∈ A_{h} and in particular ρ(g) ∈ A. By Case (b) in Lemma 5.6, we see that (U_{C}i_{A})(g) = g. If now g ∈ P^{∞}, then Case (c) in Lemma 5.6 implies that (U_{C}i_{A})(g) = g. This shows that (U_{C}i_{A})(g) = g = (U_{S}i_{A})(g).

Finally, if g ∈ ⋁E−E or if g ∈ G̃^{E}, then (U_{C}i_{A})(g) and (U_{S}i_{A})(g) are equal because they are both equal to ⋁_{e<g} i_{A}(e).

5.10. Corollary. ∑_{C⊆P}(−1)^{|C|} U_{C} î = (−1)^{|P |} ∑_{C⊆P}(−1)^{|C|} { \atop U_{C}i_{P}} .

Proof : By Corollary 5.8,

∑_{C⊆P}(−1)^{|C|} U_{C} î = ∑_{\begin{matrix}C⊆P \\ A⊆P \end{matrix}}(−1)^{|C|+|A|} { \atop U_{C}i_{A}} = ∑_{A⊆P}(−1)^{|A|} ∑_{C⊆P}(−1)^{|C|} { \atop U_{C}i_{A}} .

Every subset C determines uniquely a subset S = C−D contained in P −D, so the inner sum can be written

∑_{C⊆P}(−1)^{|C|} { \atop U_{C}i_{A}} = ∑_{\begin{matrix}S \\ S⊆P −D \end{matrix}} ∑_{\begin{matrix}C \\ S⊆C⊆S∪D \end{matrix}}(−1)^{|C|} { \atop U_{C}i_{A}} .

Since all subsets C in the second sum satisfy S = C−D, we have U_{C}i_{A} = U_{S}i_{A} by Lemma 5.9. Thus all functions U_{C}i_{A} are equal when C runs over all subsets of S ∪ D containing S and therefore

∑_{S⊆C⊆S∪D}(−1)^{|C|} { \atop U_{C}i_{A}} = ( ∑_{S⊆C⊆S∪D}(−1)^{|C|}) { \atop U_{S}i_{A}} .

Note that the latter sum has 2^{|D|} terms and is zero if |D| ≥ 1, that is, if D ≠ ∅. If A is a proper subset of P, then P −A ≠ ∅, hence D ≠ ∅, and it follows that

∑_{C⊆P}(−1)^{|C|} { \atop U_{C}i_{A}} = ∑_{\begin{matrix}S \\ S⊆P −D \end{matrix}} ∑_{\begin{matrix}C \\ S⊆C⊆S∪D \end{matrix}}(−1)^{|C|} { \atop U_{C}i_{A}} = ∑_{\begin{matrix}S \\ S⊆P −D \end{matrix}} 0 = 0 .

Thus the sum over A at the beginning of the proof reduces to a single term for A = P and we are left with (−1)^{|P |} ∑_{C⊆P} (−1)^{|C|} { \atop U_{C}i_{P}}, as was to be shown.

Now we come to the last step in our series of reductions.

5.11. Lemma. Let θ : G^{E} → T be the inclusion map. For every subset C of P, we have U_{C}i_{P} = θ\overline{C}, where \overline{C} = P −C.

Proof : We apply Lemma 5.6 in the special case when A = P.

(a) If g ∈ E−C, then (U_{C}i_{P})(g) = \{ \begin{matrix}g & if g ∉ P , \\ s_{T} (g) & if g ∈ P . \end{matrix}

(b) If g ∈ C and g ∉ P^{∞}, then ρ(g) ∈ P and so (U_{C}i_{P})(g) = g.

(c) If g ∈ C and g ∈ P^{∞}, then (U_{C}i_{P})(g) = g.

(d) If g ∈ ⋁E−E, then (U_{C}i_{P})(g) = g.

(e) If g ∈ G̃^{E}, then ρ(g) ∈ P, hence (U_{C}i_{P})(g) = g. Notice that E−C = (E−P) ⊔ \overline{C}. It follows that

(U_{C}i_{P})(g) = \{ \begin{matrix}g & if g ∉ \overline{C} , \\ s_{T} (g) & if g ∈ \overline{C} . \end{matrix}

These are precisely the values of the function θ\overline{C}.

Proof of Theorem 5.4. Starting from Corollary 5.10 and replacing each U_{C}i_{P} by θ\overline{C} (by Lemma 5.11), we obtain

∑_{C⊆P}(−1)^{|C|} U_{C} î = (−1)^{|P |} ∑_{C⊆P}(−1)^{|C|} { \atop U_{C}i_{P}} = (−1)^{|P |} ∑_{C⊆P}(−1)^{|C|} θ̂\overline{C} .

But θ̂\overline{C} = 0 if \overline{C} ≠ ∅ by Lemma 3.10. So we are left with the case \overline{C} = ∅, that is, C = P, and we get

∑_{C⊆P}(−1)^{|C|} U_{C} î = (−1)^{|P |}(−1)^{|P |} θ̂_{∅} = θ̂,

as was to be shown.

5.12. Corollary. The subfunctor of Q_{T}^{E} generated by θ̂ is equal to the subfunctor S_{T}^{E} generated by î.

Proof : Let U = {(e, e) | e ∈ E} ⊆ E × G^{E}. It is trivial to check that, for any A ⊆ E, we have Uθ_{A} = i_{A}. Therefore Uθ̂ = î, so that S_{T}^{E} is contained in the subfunctor of Q_{T}^{E} generated by θ̂. The reverse inclusion follows from the fact that θ̂ belongs to S_{T}^{E}(G^{E}), by Theorem 5.4.

6. A basis for each evaluation

Having achieved the largest part of the work with the universal element θ̂ in Section 5, it is now easy to prove our main result about the evaluation S_{T}^{E}(X) at some finite set X. We need to prove that φ̂ ∈ S_{T}^{E}(X) for any function φ ∈ T^{X} such that E ⊆ φ(X) ⊆ G^{E}, where we write again φ̂ = φ^{E} for simplicity of notation. We first start with two lemmas.

6.1. Lemma. Let η ∈ T^{Y} and let ψ : X → Y be a map. Then { \atop η ∘ ψ} = η̂ ∘ ψ.

Proof : For any subset A ⊆ E, we have

(η ∘ ψ)_{A} = s_{T}^{A} ∘ (η ∘ ψ) = (s_{T}^{A} ∘ η) ∘ ψ = η_{A} ∘ ψ .

Taking the alternating sum over A, we get { \atop η ∘ ψ} = η̂ ∘ ψ.

6.2. Lemma. Let η be a k-linear combination of maps in T^{Y}. Let ψ : X → Y be a map and let V_{ψ} be the correspondence V_{ψ} = {(x, ψ(x)) | x ∈ X} ⊆ X × Y. Then η ∘ ψ = V_{ψ}η.

Proof : By linearity, it suffices to prove the result when η is a map in T^{Y}. We have (V_{ψ}η)(x) = ⋁_{(x,y)∈V_{ψ}} η(y). But the join has a single term for y = ψ(x), by the

definition of V_{ψ}. Therefore (V_{ψ}η)(x) = η(ψ(x)).

6.3. Theorem. For any finite set X, let B_{X}^{E} = {φ ∈ T^{X} | E ⊆ φ(X) ⊆ G^{E}} and B̂_{X}^{E} = {φ̂ | φ ∈ B_{X}^{E}}.

(a) For any φ ∈ B_{X}^{E},

φ̂ = ∑_{C⊆P}(−1)^{|C|} (V_{φ}U_{C}) î,

where V_{φ} = {(x, φ(x)) | x ∈ X} ⊆ X × G^{E} and where P and U_{C} are defined by (5.1) and (5.3). In particular, φ̂ ∈ S_{T}^{E}(X).

(b) B̂_{X}^{E} is a k-basis of S_{T}^{E}(X). In other words S_{T}^{E}(X) = kB̂_{X}^{E}.

(c) The k-module S_{T}^{E}(X) is free of rank

rk_{k}(S_{T}^{E}(X)) = |B_{X}^{E}| = ∑_{i=0}^{|E|}(−1)^{i}\binom{|E|}{i}(|G^{E}| − i)^{|X|} .

Proof : (a) Since φ(X) ⊆ G^{E}, we can write φ = θ ∘ ψ where θ : G^{E} → T is the inclusion map and ψ : X → G^{E} denotes the map φ with its codomain restricted to G^{E}. By Lemma 6.1, we have

φ̂ = { \atop θ ∘ ψ} = θ̂∘ ψ .

By Theorem 5.4, we know that θ̂ belongs to S_{T} (X) with an explicit expression, namely

φ̂ = θ̂∘ ψ = ∑_{C⊆P}(−1)^{|C|} (U_{C} î) ∘ ψ .

Now (U_{C} î) ∘ ψ = V_{ψ}(U_{C} î) by Lemma 6.2, where V_{ψ} = V_{φ} = {(x, φ(x)) | x ∈ X}. Therefore

φ̂ = ∑_{C⊆P}(−1)^{|C|} V_{φ}(U_{C} î) = ∑_{C⊆P}(−1)^{|C|} (V_{φ}U_{C}) î.

This shows that φ̂ belongs to the subfunctor of F_{T} generated by î. In other words, φ̂ ∈ S_{T} (X).

(b) Recall that B̂_{X}^{E} is linearly independent because it is a subset of the basis M̂_{X}^{E} of Q_{T}^{E}(X). Moreover, we know that S_{T}^{E}(X) ⊆ kB̂_{X}^{E} by Theorem 4.6. By (a), φ̂ ∈ S_{T}^{E}(X) for any φ ∈ B_{X}^{E}, so kB̂_{X}^{E} ⊆ S_{T}^{E}(X).

(c) The formula follows from Lemma 6.5 in [BT4].

7. Minimal nonzero evaluation

We continue to work with a finite lattice T and a fixed subset E ⊆ Mirr(T). By Corollary 3.16, we know that Q_{T}^{E}(X) = {0} if |X| < |E| and that Q_{T}^{E}(E) has a k-basis { îσ | σ ∈ Σ_{E}}, where i : E → T is the inclusion map. We first note that the same holds for the subfunctor S_{T}^{E}.

7.1. Lemma. S_{T}^{E}(X) = {0} if |X| < |E|. Moreover, S_{T}^{E}(E) has a k-basis

{ îσ | σ ∈ Σ_{E}} = {Δ_{τ} î | τ ∈ Σ_{E}} ,

where Δ_{τ} = {(τ(e), e) | e ∈ E}. In particular, S_{T}^{E}(E) = Q_{T}^{E}(E).

Proof : Since Q_{T}^{E}(X) = {0} if |X| < |E|, it is clear that S_{T}^{E}(X) = {0} if |X| < |E|. Let îσ be a basis element of Q_{T}^{E}(E). Lemmas 6.1 and 6.2 show that

îσ = îσ = U_{σ} î = Δ_{σ^{−1}} î,

where U_{σ} = Δ_{σ^{−1}} = {(e, σ(e)) | e ∈ E}. Since each Δ_{σ^{−1}} î belongs to S_{T}^{E}(E), we get a k-basis of S_{T}^{E}(E) and S_{T}^{E}(E) = Q_{T}^{E}(E).

We aim to describe the minimal nonzero evaluation S_{T}^{E}(E), viewed as a left kR_{E}- module, where R_{E} denotes the monoid of all relations on E (i.e. correspondences in E × E) and kR_{E} is the monoid algebra of R_{E} (where k is a commutative ring).

Associated to a poset (E, R) (i.e. R is a partial order on the set E) there is a specific kR_{E}-module M_{E,R} which we call the fundamental kR_{E}-module and which plays a major role in all our work on correspondence functors. The module M_{E,R} appears in Section 7 of [BT1] as a left ideal Pf_{R}, where f_{R} is a suitable idempotent in some quotient algebra P of kR_{E}. The action (7.2) below appears in Proposition 8.5 of [BT1].

We define M_{E,R} to be the free k-module with a basis {m_{σ} | σ ∈ Σ_{E}}, where Σ_{E} is the group of all permutations of the set E. We let Δ_{τ} = {(τ(e), e) | e ∈ E} ⊆ E ×E and ^{σ}R = Δ_{σ}RΔ_{σ^{−1}}, while Δ = Δ_{id} is the diagonal subset of E × E. Note that Δ_{σ}Δ_{τ} = Δ_{στ} and that Δ_{σ} is invertible with inverse Δ_{σ^{−1}}. The left kR_{E}-module structure is entirely determined by specifying the action of each relation U ∈ R_{E} on the k-basis :

(7.2)  U · m_{σ} = \{ \begin{matrix}m_{τσ} & if ∃ τ ∈ Σ_{E} such that Δ ⊆ Δ_{τ^{−1}}U ⊆ ^{σ}R , \\ 0 & otherwise. \end{matrix}

Moreover, if τ exists with this condition, then it is unique. Notice that M_{E,R} is generated by m := m_{id} and moreover Δ_{τ} · m = m_{τ}.

7.3. Lemma. Let (E, R) be a finite poset and let M be a left kR_{E}-module such that:

(a) M is a free k-module with a basis {Δ_{σ} · m | σ ∈ Σ_{E}} for some m ∈ M,

(b) U · m = 0 if U ∈ R_{E} does not contain any permutation,

(c) U · m = m if U ∈ R_{E} is reflexive and U ⊆ R,

(d) U · m = 0 if U ∈ R_{E} is reflexive and U ⊈ R.

Then M is isomorphic to the fundamental module M_{E,R}, generated by m.

Proof : Let m_{σ} = Δ_{σ} · m and let U ∈ R_{E}. Assume first that U does not contain any permutation. Then the inclusion Δ ⊆ Δ_{τ^{−1}}U cannot hold. On the other hand, UΔ_{σ} does not contain any permutation, so UΔ_{σ} ·m = 0 by (b), that is, U ·m_{σ} = 0.

Assume now that U contains a permutation Δ_{τ}. Then Δ_{τ^{−1}}U is reflexive, that is, Δ ⊆ Δ_{τ^{−1}}U. We fix σ ∈ Σ_{E} and we need to compute U · m_{σ}.

Suppose that Δ_{τ^{−1}}U ⊆ ^{σ}R. Conjugating on the right by Δ_{σ}, we get

Δ ⊆ Δ_{σ^{−1}}Δ_{τ^{−1}}UΔ_{σ} ⊆ R .

By (c) applied to the reflexive relation Δ_{σ^{−1}}Δ_{τ^{−1}}UΔ_{σ}, we obtain

Δ_{σ^{−1}}Δ_{τ^{−1}}UΔ_{σ} · m = m .

Multiplying on the left by Δ_{τσ}, we have UΔ_{σ} · m = Δ_{τσ} · m, that is, U · m_{σ} = m_{τσ}. Suppose now that Δ_{τ^{−1}}U ⊈ ^{σ}R. Then

Δ ⊆ Δ_{σ^{−1}}Δ_{τ^{−1}}UΔ_{σ} ⊈ R .

By (d), we have Δ_{σ^{−1}}Δ_{τ^{−1}}UΔ_{σ} · m = 0. Multiplying on the left by Δ_{τσ}, we obtain UΔ_{σ} · m = 0, that is, U · m_{σ} = 0.

This completes the proof that the defining properties (7.2) hold.

Since E is a subset of the lattice T, it is naturally partially ordered and we write

R = {(e, f) ∈ E × E | e ≤ f} .

It is actually the opposite partial order R^{op} which comes into play.

7.4. Theorem. Let E ⊆ Mirr(T) and let R^{op} be the opposite order of the natural partial order on E. The kR_{E}-module S_{T}^{E}(E), generated by î, is isomorphic to the fundamental module M_{E,R^{op}}, generated by m.

Proof : We will apply Lemma 7.3 to the R_{E}-module S_{T}^{E}(E), its specific element î, and the partial order R^{op}. By Lemma 7.1, we know that S_{T}^{E}(E) has a k-basis

{Δ_{σ} î | σ ∈ Σ_{E}} ,

so condition (a) in Lemma 7.3 holds.

Let U ∈ R_{E} be a relation. Suppose first that U does not contain any permutation. Then Theorem 3.2 in [BT1] asserts that U is inessential, that is, U factors through a set X of cardinality strictly smaller than |E|. In other words, U = V W where V ⊆ E × X and W ⊆ X × E. It follows that U î = V W î = 0, because W î ∈ S_{T}^{E}(X) = {0} by Lemma 7.1.

Suppose next that U is reflexive and U ⊆ R^{op}. Then, for any subset A of E and any f ∈ E, we have

(Ui_{A})(f) = ⋁_{(f,e)∈U} i_{A}(e) .

Since U ⊆ R^{op}, the condition (f, e) ∈ U implies that (f, e) ∈ R^{op}, that is, e ≤ f, hence i_{A}(e) ≤ i_{A}(f). On the other hand, (f, f) ∈ U by reflexivity of U. It follows that the join above is equal to the single term i_{A}(f) and so (Ui_{A})(f) = i_{A}(f). This shows that Ui_{A} = i_{A}, hence U î = î.

Suppose finally that U is reflexive and U ⊈ R^{op}. The reflexivity of U implies that, for any subset A of E,

(Ui_{A})(f) = ⋁_{(f,e)∈U} i_{A}(e) ≥ i_{A}(f) ≥ f ,

hence Ui_{A} ≥ i. We claim that if a function φ : E → T satisfies φ > i (i.e. φ(x) ≥ x for every x ∈ E and φ(e) > e for some e ∈ E), then φ̂ = 0.

Postponing the proof of this claim, we apply it to the function Ui_{A}. If A ≠ ∅, then Ui_{A} ≥ i_{A} > i, because for any e ∈ A we have i_{A}(e) = s_{T} (e) > e. Thus Ûi_{A} = 0. If A = ∅, then i_{A} = i. Since U ⊈ R^{op}, there exists (f, g) ∈ U such that (f, g) ∉ R^{op}, that is, g ≰ f. Then

(Ui)(f) = ⋁_{(f,e)∈U} i(e) ≥ i(f) ∨ i(g) = f ∨ g > f

because g ≰ f. It follows that Ui > i, hence Ûi = 0. We have now proved that Ûi_{A} = 0 for all A and, by Lemma 4.1, this implies that

U î = ∑_{A⊆E}(−1)^{|A|} Ûi_{A} = 0 .

We see that all the assumptions of Lemma 7.3 are satisfied, so that S_{T}^{E}(E) is isomorphic to the fundamental module M_{E,R^{op}}.

We are left with the proof of the claim. Let φ ∈ T^{E} such that φ > i. Let e ∈ E be minimal such that φ(e) ≠ e. If we had e ∈ φ(E), then we would have e = φ(f) for some f ∈ E. We must have f ≠ e because φ(f) ≠ φ(e). Moreover, e = φ(f) ≥ i(f) = f, hence e > f. By minimality of e, we deduce that φ(f) = f, hence e = f, a contradiction. It follows that e ∉ φ(E), so that φ ∉ M_{E} and φ̂ = 0 by Lemma 3.10.

8. An embedding property

In this section, we show that the functors Q_{T} and S_{T} have an important injectivity property which will be used repeatedly, in particular as the main ingredient in our next section on fundamental functors. In order to obtain the results, we cannot use an arbitrary subset E of Mirr(T), so we assume from now on that E = Mirr(T). Consequently, we ignore the superscript E and in particular we have Q_{T} = Q_{T}^{E} and S_{T} = S_{T}^{E}.

Let X be a finite set. Any correspondence W ∈ C(E, X) induces a k-linear map

Q_{T} (X) ⟶ Q_{T} (E) , ω ↦ W ω .

The product of all these maps when W runs over correspondences in E × X yields a map

μ_{X} : Q_{T} (X) ⟶ ∏_{W ∈C(E,X)} Q_{T} (E) , ω ↦ (W ω)_{W ∈C(E,X)} .

8.1. Theorem. The map μ_{X} above is injective. Explicitly, if ω ∈ Q_{T} (X) satisfies W ω = 0 for any correspondence W ∈ E × X, then ω = 0.

Proof : By Theorem 3.12, we can write

ω = ∑_{φ∈M_{X}} λ_{φ}φ̂ ,

where λ_{φ} ∈ k, M_{X} = M_{X}^{E} and φ̂ = φ^{E}. Therefore 0 = W ω = ∑_{φ∈M_{X}} λ_{φ}W φ̂ for any correspondence W ⊆ E × X and we now make a specific choice. For any function π ∈ M_{X}, we define

W_{π} = {(e, x) ∈ E × X | e ≥ π(x)} .

It follows that

(8.2)  0 = { \atop W_{π} ω} = ∑_{φ∈M_{X}} λ_{φ}{ \atop W_{π} φ̂} .

Now φ̂ = ∑_{A⊆E}(−1)^{|A|}φ_{A} and, by Lemma 3.10, we have W_{π} φ_{A} ∈ M_{X} whenever

{ \atop W_{π} φ_{A}} ≠ 0. The definition of M_{X} yields E ⊆ W_{π} φ_{A}(E) and therefore W_{π} φ_{A} = i ∘ σ_{A} for some permutation σ_{A} ∈ Σ_{E}, where i : E → T denotes the inclusion map. By Lemma 7.1 and its proof, we get

{ \atop W_{π} φ_{A}} = î σ_{A} = Δ_{σ_{A}^{−1}} î,

and therefore the equation (8.2) above becomes

0 = ∑_{φ∈M_{X}} λ_{φ} ∑_{σ∈Σ_{E}} ( ∑_{\begin{matrix}A⊆E \\ σ_{A}=σ \end{matrix}}(−1)^{|A|} î σ_{A})

= ∑_{σ∈Σ_{E}} ( ∑_{φ∈M_{X}} λ_{φ} ∑_{\begin{matrix}A⊆E \\ W_{π} φ_{A}=iσ \end{matrix}}(−1)^{|A|})Δ_{σ^{−1}} î.

because σ_{A} = σ if and only if W_{π} φ_{A} = iσ. Since {Δ_{τ} î | τ ∈ Σ_{E}} is a k-basis of Q_{T} (E) (by Lemma 7.1), the coefficients vanish. In particular, when σ = id, we get

∑_{φ∈M_{X}} λ_{φ} ∑_{\begin{matrix}A⊆E \\ W_{π} φ_{A}=i \end{matrix}}(−1)^{|A|} = 0 .

Defining a square matrix (indexed by M_{X} × M_{X}) by

ε_{π,φ} = ∑_{\begin{matrix}A⊆E \\ W_{π} φ_{A}=i \end{matrix}}(−1)^{|A|} ,

we obtain

∑_{φ∈M_{X}} ε_{π,φ} λ_{φ} = 0 .

In other words, the product of the matrix with the column vector (λ_{φ}) is equal to zero.

The condition W_{π} φ_{A} = i gives, for any f ∈ E,

f = i(f) = W_{π} φ_{A}(f) = ⋁_{\begin{matrix}y∈X \\ (f,y)∈W_{π} \end{matrix}} φ_{A}(y) = ⋁_{\begin{matrix}y∈X \\ f≥π(y) \end{matrix}} φ_{A}(y) ≥ ⋁_{\begin{matrix}y∈X \\ f≥π(y) \end{matrix}} φ(y) ≥ φ(x)

for any x ∈ X such that f ≥ π(x). Since every element of T is a meet of meetirreducible elements, see (2.3), we obtain

π(x) = ⋀_{\begin{matrix}f∈E \\ f≥π(x) \end{matrix}} f ≥ φ(x) ,

showing that π ≥ φ. Therefore, if π ≱ φ, the condition W_{π} φ_{A} = i cannot hold, ε_{π,φ} = 0, and so the matrix (ε_{π,φ}) is triangular.

In order to find the diagonal entries of the matrix, let π = φ and assume that the condition W_{φ} φ_{A} = i holds. Then

(8.3)  f = ⋁_{\begin{matrix}y∈X \\ f≥φ(y) \end{matrix}} φ_{A}(y) = ( ⋁_{\begin{matrix}y∈X \\ f≥φ(y) \\ φ(y)∉A \end{matrix}} φ(y)) ∨ ( ⋁_{\begin{matrix}z∈X \\ f≥φ(z) \\ φ(z)∈A \end{matrix}} s_{T} (φ(z))) ,

because φ_{A}(z) = s_{T} (φ(z)) when φ(z) ∈ A. Suppose that A ≠ ∅ and let f ∈ A. Since φ ∈ M_{X}, we have E ⊆ φ(X), so there exists z_{0} ∈ X such that f = φ(z_{0}). The equation (8.3) yields in particular f ≥ s_{T} (φ(z_{0})), hence

f ≥ s_{T} (φ(z_{0})) > φ(z_{0}) = f ,

a contradiction. Therefore A must be empty, φ_{∅} = φ, and we get

ε_{φ,φ} = ∑_{\begin{matrix}A⊆E \\ W_{φ} φ_{A}=i \end{matrix}}(−1)^{|A|} = \{ \begin{matrix}1 & if W_{φ} φ = i , \\ 0 & otherwise . \end{matrix}

But since E ⊆ φ(X), for any f ∈ E, there exists y_{f} ∈ X such that f = φ(y_{f}). Therefore

W_{φ} φ(f) = ⋁_{\begin{matrix}y∈X \\ f≥φ(y) \end{matrix}} φ(y) = φ(y_{f}) = f = i(f) .

This shows that the condition W_{φ} φ = i holds and so the diagonal coefficient ε_{φ,φ} is equal to 1. Therefore, the matrix is unitriangular, hence invertible. By inverting the matrix, we deduce that the column vector (λ_{φ}) is equal to zero. Therefore ω = ∑_{φ∈M_{X}} λ_{φ}φ̂ = 0. This completes the proof of Theorem 8.1.

8.4. Corollary. The map

μ_{X} : S_{T} (X) ⟶ ∏_{W ∈C(E,X)} S_{T} (E) , ω ↦ (W ω)_{W ∈C(E,X)} .

is injective.

Proof : Since S_{T} (E) = Q_{T} (E) by Lemma 7.1, the map μ_{X} is the restriction to S_{T} (X) of the map μ_{X} of Theorem 8.1.

8.5. Remark. We can interpret Theorem 8.1 in a more functorial way. Evaluation at E is a functor from the category of correspondence functors to the category of (left) kR_{E}-modules. This has a left adjoint, mapping a kR_{E}-module V to the correspondence functor

L_{E,V} := kC(−, E) ⊗_{kR(E)} V ,

as well as a right adjoint, mapping V to the correspondence functor

L_{E,V}^{0} := Hom_{kR(E)} (kC(E, −), V ) .

For any correspondence functor F, the unit of the latter adjunction yields a morphism of correspondence functors

F ⟶ L_{E,F (E)}^{0} .

Taking F = Q_{T} and evaluating at a finite set X, we get a map

Q_{T} (X) ⟶ L_{E,Q_{T} (E)}^{0}(X) = Hom_{kR(E)} (kC(E, X), Q_{T} (E)) ⊆ ∏_{W ∈C(E,X)} Q_{T} (E)

which is equal to the injective map μ_{X} of Theorem 8.1. Therefore, this theorem asserts in fact that the correspondence functor Q_{T} is isomorphic to a subfunctor of L_{E,Q_{T} (E)}^{0}, which is a quite explicit functor in view of the description of Q_{T} (E) in Corollary 3.16.

9. Fundamental functors

Fundamental functors play an important role in our work and are in particular essential in [BT4]. In this section, our main purpose is to prove that the functor S_{T} := S_{T}^{E} of the present paper is isomorphic to the fundamental functor S_{E,M} where M = M_{E,R^{op}} is the fundamental module defined in Section 7. Here E = Mirr(T) as in Section 8.

We need to define the notion of fundamental functor. Recall from Section 2 in [BT3] that, for any left kR_{E}-module V, there is an associated correspondence functor

L_{E,V} := kC(−, E) ⊗_{kR(E)} V ,

(see Remark 8.5 above), a subfunctor J_{E,V} defined by

J_{E,V} (X) := { ∑_{i} α_{i} ⊗ w_{i} ∈ L_{E,V} (X) | ∀β ∈ kC(E, X), ∑_{i}(βα_{i}) · w_{i} = 0}

and a quotient

S_{E,V} := L_{E,V} /J_{E,V} .

When E is equal to Mirr(T) (for some finite lattice T) and M = M_{E,R^{op}} is the fundamental kR_{E}-module, we define

𝕊_{E,R} := S_{E,M}

and call it the fundamental correspondence functor associated to the finite poset (E, R). Here, and for the rest of this paper, we emphasize that a finite poset is a pair (E, R) where E is a finite set and R is a partial order relation on the set E.

9.1. Theorem. Let T be a finite lattice and M = M_{E,R^{op}} be the fundamental kR_{E}-module, where E = Mirr(T) and R^{op} is the opposite order of the natural partial order on E. Then S_{T} is isomorphic to the fundamental correspondence functor 𝕊_{E,R}.

Proof : The isomorphism M ≅ S_{T} (E) of Theorem 7.4 extends to a morphism of functors

η : L_{E,M} ⟶ S_{T} .

This is surjective because S_{T} is generated by î, which belongs to S_{T} (E). The definition of the subfunctor J_{E,M} shows that the image of J_{E,M} under η is equal to the subfunctor K of S_{T} defined by

K(X) = {ω ∈ S_{T} (X) | W ω = 0 for all W ∈ C(E, X)} .

By Corollary 8.4, we have K(X) = {0}. Therefore K = 0 and J_{E,M} is the kernel of η. It follows that

𝕊_{E,R} = S_{E,M} = L_{E,M}/J_{E,M} ≅ S_{T} ,

as required.

By using Theorem 6.3, we recover in particular a k-basis for 𝕊_{E,R}(X) and a formula for its rank, providing a new proof of Theorem 6.6 in [BT4].

10. Simple functors

Simple correspondence functors are classified and studied in [BT2, BT3, BT4]. Our purpose in this section is to show that they can be constructed by using in a direct fashion the functor S_{T} of the present paper, independently of our previous work. In order to describe this construction, we first show that the group A = Aut(E, R) acts on the right on the functor S_{T}. As usual, T is a finite lattice, Mirr(T) = (E, R) and we let A = Aut(E, R).

For any finite set X and any correspondence U ∈ C(X, E), we define α_{X}(U) = U î and we extend this by k-linearity to a k-linear map

α_{X} : kC(X, E) ⟶ S_{T} (X) ,

which is surjective by the definition of S_{T}. This clearly defines a morphism of correspondence functors α : kC(−, E) ⟶ S_{T}. The group algebra kA has a right permutation action on kC(X, E) defined by U · γ = UΔ_{γ} for any U ∈ C(X, E) and any γ ∈ A. As before, we write Δ_{τ} = {(τ(e), e) | e ∈ E} for any permutation τ of E. We intend to show that this action passes to the quotient kC(X, E)/ Ker(α_{X}) ≅ S_{T} (X). Explicitly, we will show that Ker(α_{X}) is a kA-submodule of kC(X, E), so that the right kA-module structure can be transported from kC(X, E) to S_{T} (X) via the map α_{X}. We first prove this in the case X = E.

10.1. Lemma. Ker(α_{E}) is a kA-submodule of kC(E, E). Consequently, the action

(U î) · γ := (UΔ_{γ}) î, ∀ U ∈ C(E, E) , ∀ γ ∈ A ,

is a well-defined right kA-module structure on S_{T} (E).

Proof : Let ω = ∑_{U∈C(E,E)} λ_{U}U ∈ kC(E, E), where λ_{U} ∈ k, and suppose that

ω ∈ Ker(α_{E}), that is,

∑_{U∈C(E,E)} λ_{U}U î = 0 .

This belongs to S_{T} (E) ≅ M_{E,R^{op}} (Theorem 7.4) and the definition (7.2) yields

U î = \{ \begin{matrix}Δ_{τ} î & if ∃ τ ∈ Σ_{E} such that Δ ⊆ Δ_{τ^{−1}}U ⊆ R^{op} , \\ 0 & otherwise, \end{matrix}

with τ unique if it exists. This uniqueness allows us to sum over all permutations τ, under the condition Δ ⊆ Δ_{τ^{−1}}U ⊆ R^{op}. Explicitly, we get

U î = ∑_{\begin{matrix}τ∈Σ_{E} \\ Δ⊆Δ_{τ^{−1}} U⊆R \end{matrix}} Δ_{τ} î,

where all terms of the sum are zero, except at most one. It follows that

0 = ∑_{U∈C(E,E)} λ_{U}U î = ∑_{U∈C(E,E)} λ_{U} ∑_{\begin{matrix}τ∈Σ_{E} \\ Δ⊆Δ_{τ^{−1}} U⊆R^{op} \end{matrix}} Δ_{τ} î

= ∑_{τ∈Σ_{E}} ( ∑_{\begin{matrix}U∈C(E,E) \\ Δ⊆Δ_{τ^{−1}} U⊆R^{op} \end{matrix}} λ_{U})Δ_{τ} î.

Since {Δ_{τ} î | τ ∈ Σ_{E}} is a k-basis of S_{T} (E) (Lemma 7.1), the coefficients must vanish. Thus, for any τ ∈ Σ_{E}, we get

(10.2)  ∑_{\begin{matrix}U∈C(E,E) \\ Δ⊆Δ_{τ^{−1}} U⊆R^{op} \end{matrix}} λ_{U} = 0 .

Now we have to prove that ωΔ_{γ} belongs to Ker(α_{E}) for any γ ∈ A. We have ωΔ_{γ} = ∑_{U∈C(E,E)} λ_{U}UΔ_{γ} and the definition (7.2) yields

UΔ_{γ} î = \{ \begin{matrix}Δ_{τγ} î & if ∃ τ ∈ Σ_{E} such that Δ ⊆ Δ_{τ^{−1}}U ⊆ ^{γ}R^{op} , \\ 0 & otherwise. \end{matrix}

with τ unique if it exists. Here ^{γ}R^{op} = Δ_{γ}R^{op}Δ_{γ^{.1}} and it easy to check that

(10.3)  γ ∈ Aut(E, R) = Aut(E, R^{op}) ⟺ ^{γ}R^{op} = R^{op} .

Therefore the condition above is simply Δ ⊆ Δ_{τ^{−1}}U ⊆ R^{op}, as before. We then compute

ωΔ_{γ} î = ∑_{U∈C(E,E)} λ_{U}UΔ_{γ} î = ∑_{U∈C(E,E)} λ_{U} ∑_{\begin{matrix}τ∈Σ_{E} \\ Δ⊆Δ_{τ^{−1}} U⊆R^{op} \end{matrix}} Δ_{τγ} î

= ∑_{τ∈Σ_{E}} ( ∑_{\begin{matrix}U∈C(E,E) \\ Δ⊆Δ_{τ^{−1}} U⊆R^{op} \end{matrix}} λ_{U})Δ_{τγ} î.

The inner sum is zero, thanks to the equation (10.2) above. Therefore

α_{E}(ωΔ_{γ}) = ωΔ_{γ} î = 0 ,

as was to be shown.

Now we move from E to an arbitrary finite set X. Recall that G = G^{Mirr(T)} = ⋁E ⊔ G̃, as defined in (4.4).

10.4. Lemma.

(a) Ker(α_{X}) is a kA-submodule of kC(X, E).

(b) The action

(U î) · γ := (UΔ_{γ}) î, ∀ U ∈ C(X, E) , ∀ γ ∈ A ,

is a well-defined right kA-module structure on S_{T} (X).

(c) Right multiplication by γ ∈ A is a morphism of correspondence functors S_{T} → S_{T}.

(d) Every γ ∈ A extends to an automorphism of the full subposet G of T and we have

φ̂ · γ = { \atop γ^{−1}φ} ∀ φ ∈ G^{X} such that E ⊆ φ(X) .

Proof : (a) Let ω ∈ kC(X, E) such that α_{X}(ω) = 0. We apply the (left) action of a correspondence W ∈ C(E, X). Since W ω î belongs to S_{T} (E), we can use the action of γ ∈ A, which is well-defined in S_{T} (E) by Lemma 10.1. Thus we have

0 = (W α_{X}(ω)) · γ = (W ω î) · γ = W ωΔ_{γ} î = α_{E}(W ωΔ_{γ}) = W α_{X}(ωΔ_{γ}) ,

using the fact that α : kC(−, E) ⟶ S_{T} is a morphism of functors. The equality W α_{X}(ωΔ_{γ}) = 0 holds for any correspondence W ∈ C(E, X), so by Corollary 8.4, we get α_{X}(ωΔ_{γ}) = 0. In other words, ωΔ_{γ} ∈ Ker(α_{X}), as was to be shown.

(b) As mentioned at the beginning of this section, the right kA-module structure can be transported from kC(X, E) to S_{T} (X) via the map α_{X}.

(c) Since γ ∈ A acts on the right while correspondences act on the left, it is clear that these actions commute.

(d) Recall that G = ⋁E ⊔ G̃. If t ∈ ⋁E, we define

γ(t) = ⋁_{\begin{matrix}e∈E \\ e≤t \end{matrix}} γ(e) .

If h ∈ G̃, then Lemma 4.5 shows that h = s_{T} (b_{1}) where b_{1} = ρ(h) and b_{1} ∈ E. We define γ(h) = s_{T} (γ(b_{1})). Actually, γ maps the whole chain below h in Lemma 4.5 to a chain below γ(h) with the same properties. It follows that Aut(E, R) = Aut(G), as already observed in Corollary 2.31 of [Bo2]. Recall from Theorem 6.3 that

φ̂ = ∑_{C⊆P}(−1)^{|C|} (V_{φ}U_{C}) î,

where V_{φ} = {(x, φ(x)) | x ∈ X} ⊆ X × G and where P and U_{C} are defined by (5.1) and (5.3). We claim that U_{C}Δ_{γ} = Δ_{γ}U_{γ^{−1}(C)} and V_{φ}Δ_{γ} = V_{γ^{−1}φ}. It follows from these claims that

φ̂ · γ = ∑_{C⊆P}(−1)^{|C|} (V_{φ}U_{C}) î· γ = ∑_{C⊆P}(−1)^{|C|} (V_{φ}U_{C}Δ_{γ}) î
= ∑_{C⊆P}(−1)^{|C|} (V_{φ}Δ_{γ}U_{γ^{−1}(C)}) î = ∑_{C⊆P}(−1)^{|C|} (V_{γ^{−1}φ}U_{γ^{−1}(C)}) î
= ∑_{D⊆P}(−1)^{|D|} (V_{γ^{−1}φ}U_{D}) î = { \atop γ^{−1}φ} ,

as required.

We are left with the proof of the claims. Let g ∈ G and e ∈ E.

(g, e) ∈ U_{C}Δ_{γ} ⟺ (g, γ(e)) ∈ U_{C} ⟺ \{ \begin{matrix}γ(e) < g if g ∈ (G−E) ⊔ C , \\ γ(e) ≤ g if g ∈ E−C. \end{matrix}
⟺ \{ \begin{matrix}e < γ^{−1}(g) if γ^{−1}(g) ∈ (G−E) ⊔ γ^{−1}(C) , \\ e ≤ γ^{−1}(g) if g ∈ (E− γ^{−1}(C)). \end{matrix}
⟺ (γ^{−1}(g), e) ∈ U_{γ^{−1}(C)} ⟺ (g, e) ∈ Δ_{γ}U_{γ^{−1}(C)} .

proving the first claim. Now let x ∈ X and g ∈ G.

(x, g) ∈ V_{φ}Δ_{γ} ⟺ (φ(x), g) ∈ Δ_{γ} ⟺ g = γ^{−1}φ(x) ⟺ (x, g) ∈ V_{γ^{−1}φ} ,

proving the second claim.

10.5. Lemma. S_{T} (X) is a free right kA-module.

Proof : By Theorem 6.3, S_{T} (X) has a k-basis B̂_{X} where

B_{X} = {φ ∈ T^{X} | E ⊆ φ(X) ⊆ G} = {φ ∈ G^{X} | E ⊆ φ(X)} .

Any γ ∈ A permutes B_{X} via left composition by γ^{−1}. If γ^{−1}φ = φ, then γ^{−1}φ(x) = φ(x) for all x ∈ X, so in particular γ^{−1}(e) = e for all e ∈ E because E ⊆ φ(X). Since γ is an automorphism of E, this shows that γ = id. Therefore the permutation action of A on B_{X} is free.

By Corollary 3.11, the map φ ↦ φ̂ is a bijection B_{X} → B̂_{X}. Moreover, part (d) of Lemma 10.4 shows that this map also preserves the right action of A. It follows that the permutation action of A on B̂_{X} is free as well. In other words, the kA-module S_{T} (X) is free.

Let A = Aut(E, R). For any left kA-module V, we define a correspondence functor S_{T,V} := S_{T} ⊗_{kA} V, or more precisely

S_{T,V} (X) := S_{T} (X) ⊗_{kA} V ,

using the right action of kA on S_{T} (X) defined in Lemma 10.4 above. Of course, the (left) action of a correspondence U ∈ C(Y, X) on S_{T} (X) ⊗_{kA} V is given by U(ω ⊗ v) = (Uω) ⊗ v, for any ω ∈ S_{T} (X) and any v ∈ V.

In order to treat simple modules and simple functors, we can assume that the base ring k is a field.

10.6. Theorem. Let k be a field. Let (E, R) be a partial order, A = Aut(E, R), and T a finite lattice with Mirr(T) ≅ (E, R).

(a) If V is a simple left kA-module, S_{T,V} is a simple correspondence functor.

(b) If S is a simple correspondence functor, then S ≅ S_{T,V} for some finite lattice T with Mirr(T) ≅ (E, R) and some simple kA-module V.

(c) Any simple correspondence functor S_{T,V} is isomorphic to a quotient of a fundamental functor S_{T}.

(d) The dimension of S_{T,V} (X) is given by the formula

dim_{k} S_{T,V} (X) = \frac{dim_{k}(V)}{| Aut(E, R)|} ∑_{i=0}^{|E|}(−1)^{i}\binom{|E|}{i}(|G| − i)^{|X|} .

Proof : (a) By Corollary 8.4, there is an injective map

μ_{X} : S_{T} (X) ⟶ ∏_{W ∈C(E,X)} S_{T} (E) , ω ↦ (W ω)_{W ∈C(E,X)} .

This is obviously right kA-linear. By Lemma 10.5, S_{T} (X) is a free right kA-module and since the group algebra of a finite group is self-injective, S_{T} (X) is an injective right kA-module. Therefore the kA-linear injection μ_{X} splits. Since tensoring with V preserves split injections, the map

μ_{X} ⊗ id_{V} : S_{T} (X) ⊗_{kA} V ⟶ ∏_{W ∈C(E,X)} S_{T} (E) ⊗_{kA} V

is still injective.

Let J be a nonzero subfunctor of S_{T,V} and let ω ∈ J(X) be nonzero for some finite set X. By injectivity of μ_{X} ⊗ id_{V}, there exists W ∈ C(E, X) such that u := W ω ≠ 0. This is an element of J(E) ⊆ S_{T,V} (E) = S_{T} (E) ⊗_{kA} V. Writing M = S_{T} (E) and m_{σ} = Δ_{σ}î for simplicity, recall from Theorem 7.4 that M has a k-basis {m_{σ} | σ ∈ Σ_{E}} and that it is isomorphic to the fundamental kR_{E}-module M_{E,R^{op}}, where R_{E} = C(E, E). Moreover, the right action of γ ∈ A permutes freely the basis of M and satisfies

m_{σ} · γ = (Δ_{σ} î) · γ = Δ_{σ}Δ_{γ} î = Δ_{σγ} î = m_{σγ} .

Letting [Σ_{E}/A] denote a set of representatives of cosets σA, we obtain

M ⊗_{kA} V = ⨁_{σ∈[Σ_{E}/A]} m_{σ} ⊗ V ,

and in particular u = ∑_{σ∈[Σ_{E}/A]} m_{σ} ⊗ v_{σ} for some v_{σ} ∈ V. Since u ≠ 0, we have v_{π} ≠ 0 for some π ∈ [Σ_{E}/A].

The action of relations on the fundamental kR_{E}-module is described in (7.2). In particular the action of ^{π}R^{op} is given by

πR^{op} m_{σ} = \{ \begin{matrix}m_{τσ} & if ∃ τ ∈ Σ_{E} such that Δ ⊆ Δ_{τ^{−1}} ^{π}R^{op} ⊆ ^{σ}R^{op} , \\ 0 & otherwise. \end{matrix}

The first containment relation implies Δ_{τ} ⊆ ^{π}R^{op}. This is impossible if τ ≠ id because a nontrivial permutation cannot be contained in a partial order relation. So τ = id and we are left with ^{π}R^{op} ⊆ ^{σ}R^{op}, which implies ^{π}R^{op} = ^{σ}R^{op} because the cardinalities are equal. Thus σ^{−1}π stabilizes R^{op}, hence σ^{−1}π ∈ A by (10.3). Since σ and π are representatives in [Σ_{E}/A], we deduce that π = σ. In other words

πR^{op}m_{σ} = \{ \begin{matrix}m_{σ} & if σ = π , \\ 0 & otherwise. \end{matrix}

Going back to u = ∑_{σ} m_{σ} ⊗v_{σ}, we get ^{π}R^{op}u = m_{π} ⊗v_{π}. This is a nonzero element of J(E) because u ∈ J(E).

Take now γ ∈ A and τ = πγπ^{−1} ∈ Σ_{E}. Then

Δ_{τ} m_{π} ⊗ v_{π} = m_{τπ} ⊗ v_{π} = m_{πγ} ⊗ v_{π} = m_{π} · γ ⊗ v_{π} = m_{π} ⊗ γ · v_{π} .

Therefore m_{π} ⊗ γ · v_{π} ∈ J(E) for every γ ∈ A. Since the kA-module V is simple, it is generated by its nonzero element v_{π} and we deduce that m_{π} ⊗ V ⊆ J(E). Now for any σ ∈ [Σ_{E}/A], we obtain m_{σ} ⊗ V = Δ_{σπ^{−1}}m_{π} ⊗ V ⊆ J(E), hence M ⊗_{kA} V ⊆ J(E). In other words J(E) = S_{T} (E) ⊗_{kA} V = S_{T,V} (E).

In particular î⊗ V ⊆ J(E) and therefore, for any correspondence U ∈ C(X, E), we get

U î⊗ V = U(î⊗ V) ⊆ U J(E) ⊆ J(X) .

It follows that S_{T} (X) ⊗_{kA} V ⊆ J(X), so that J(X) = S_{T,V} (X), hence J = S_{T,V}. This completes the proof of the simplicity of S_{T,V}.

(b) Let S be a simple correspondence functor and let E be a set of minimal cardinality such that S(E) ≠ {0}. Then S(E) is a kR_{E}-module, where R_{E} = C(E, E) is the set of all relations on E. Let I = ∑_{|X|<|E|} kC(E, X)C(X, E) be the ideal generated by all relations which factorize through a set of cardinality < |E|. Since S(X) = {0} whenever |X| < |E|, S(E) is a kR_{E}/I-module, where kR_{E}/I is the algebra of so-called essential relations on E. By Theorem 8.1 in [BT1], S(E) ≅ M ⊗_{kA} V, where V is a simple left kA-module and M is a suitable (kR_{E}/I, kA)-bimodule. More precisely, the left action of kR_{E}/I on M is described in Proposition 8.5 of [BT1], which asserts that M is isomorphic to the fundamental kR_{E}-module of Section 7.

Take now any finite lattice T with Mirr(T) ≅ (E, R) (e.g. the lattice of all down-sets in E, see Section 11). Then we know that S_{T} (E) ≅ M by Theorem 7.4, so there are isomorphisms of kR_{E}-modules

S(E) ≅ M ⊗_{kA} V ≅ S_{T} (E) ⊗_{kA} V ≅ S_{T,V} (E) .

But if two simple correspondence functors have isomorphic nonzero evaluations at some set E, they are isomorphic functors. This is a well-known general fact of the representation theory of small categories, which goes back to [Bo1] and is explicit in Proposition 2.7 of [BT2] or in Proposition 3.2 of [We]. Therefore we obtain S ≅ S_{T,V}, as was to be shown.

(c) A simple kA-module V is generated by a single nonzero element v_{0} ∈ V. It follows that any element of S_{T} (X)⊗_{kA}V can be written φ⊗v_{0} for some φ ∈ S_{T} (X). Therefore we get a surjective morphism

S_{T} ⟶ S_{T,V} = S_{T} ⊗_{kA} V , φ ↦ φ ⊗ v_{0} .

This shows that S_{T,V} is isomorphic to a quotient of S_{T}.

(d) Since S_{T} (X) is a free right k Aut(E, R)-module (Lemma 10.5), we get

dim_{k} S_{T,V} (X) = \frac{dim_{k}(V)}{| Aut(E, R)|} dim_{k} S_{T} (X)

and the result follows from the formula in Theorem 6.3.

Note that, by the formula in (d), we recover the main result for simple correspondence functors obtained in Theorem 7.10 of [BT4].

It should be noted that we may have two nonisomorphic lattices T and T′ such that S_{T,V} ≅ S_{T′,V}, but T and T′ must have isomorphic subposets Mirr(T) ≅ (E, R) ≅ Mirr(T′). More precisely, simple correspondence functors are parametrized by triples (E, R, V), independently of the choice of T with Mirr(T) ≅ (E, R). We discuss this question in the next Section 11.

11. Choosing the lattice

As mentioned above, the choice of a lattice T is irrelevant for the parametrization of simple functors. The purpose of this section is to clarify this question. We prove that several constructions of the present paper do not depend on the choice of a finite lattice T, but only on the poset Mirr(T) ≅ (E, R).

We fix a finite poset (E, R). Recall that a down-set in E is a subset F of E such that, for any f ∈ F and any e ∈ E with e ≤ f, we have e ∈ F. Let D be the set of all down-sets in E. This is a lattice with respect to union and intersection of subsets (and it is actually a distributive lattice). For any e ∈ E, define

\underline{e} = \{f ∈ E | f ≤ e\} ,  \underline{E} = \{\underline{e} | e ∈ E\} ,
e^{∗} = {f ∈ E | f ≱ e} ,  E^{∗} = {e^{∗} | e ∈ E} .

It is elementary to check that \underline{E} = Jirr(D), because any down-set X is a union X = ⋃_{e∈X} \underline{e} and each \underline{e} is join-irreducible. Similarly, any up-set X is a union X = ⋃_{e∈X} E_{≥e} and by passing to the complement we get E−X = ⋂_{e∈X} e^{∗}, because E−E_{≥e} = {f ∈ E | f ≱ e} = e^{∗}. Therefore E^{∗} = Mirr(D). Moreover, for all e, f ∈ E,

e ≤ f ⟺ \underline{e} ⊆ \underline{f} ⟺ e^{∗} ⊆ f^{∗} .

In other words, the maps e ↦ \underline{e} and e ↦ e^{∗} are poset isomorphisms E ≅ \underline{E} ≅ E^{∗}.

Throughout this paper, we used meet-irreducible elements and here again it is E^{∗} = Mirr(D) which comes into play.

11.1. Lemma. Let T be a finite lattice such that Mirr(T) = (E, R) and let D be the lattice of all down-sets in E. The map

α : T ⟶ D , α(t) = {f ∈ E | f ≱ t}

is a full join-morphism whose restriction to E is the isomorphism E ≅ E^{∗}. In particular α is injective.

Proof : If t_{1}, t_{2} ∈ T, we have f ≥ (t_{1} ∨ t_{2}) if and only if f ≥ t_{1} and f ≥ t_{2}. Therefore

α(t_{1} ∨ t_{2}) = {f ∈ E | f ≱ (t_{1} ∨ t_{2})}
= {f ∈ E | f ≱ t_{1}} ∪ {f ∈ E | f ≱ t_{2}}
= α(t_{1}) ∪ α(t_{2}) .

Moreover, α(ˆ0_{T}) = {f ∈ E | f ≱ ˆ0_{T} } = ∅ = ˆ0_{D}, so α is a join-morphism. To prove that α is full, suppose that α(t_{1}) ⊇ α(t_{2}). The negation of the conditions yields {f ∈ E | f ≥ t_{1}} ⊆ {f ∈ E | f ≥ t_{2}}, hence

⋀_{\begin{matrix}f∈E \\ f≥t_{1} \end{matrix}} f ≥ ⋀_{\begin{matrix}f∈E \\ f≥t_{2} \end{matrix}} f .

By (2.3), any t ∈ T is the meet of all meet-irreducible elements larger than t. Therefore we obtain t_{1} ≥ t_{2}, proving the α is full. In particular, α is injective.

The definition of α yields α(e) = {f ∈ E | f ≱ e} = e^{∗}, for any e ∈ E, so the restriction of α to E is the isomorphism E ≅ E^{∗}.

For any finite lattice L such that Mirr(L) = (E, R), let s_{L} : L → L be the map defined in (2.2). Also, write ρ_{L} for the map ρ : L → L defined in (4.3). Let G_{L} be the subset G of L defined in (4.4), namely

(11.2)  G_{L} = ⋁E ⊔ G̃_{L} , where G̃_{L} = {h ∈ L | s_{L}^{∞}ρ_{L}^{∞}(h) = h, h ∉ ⋁E} .

We view G_{L} as a full subposet of L.

11.3. Lemma. Let T be a finite lattice such that Mirr(T) = (E, R) and let α : T → D be the join-morphism of Lemma 11.1.

(a) α s_{T} = s_{D} α.

(b) α ρ_{T} = ρ_{D} α.

(c) α(G_{T}) = G_{D}. In particular, the posets G_{T} and G_{D} are isomorphic, so that G_{T} does not depend on T, but only on (E, R) up to isomorphism.

Proof : (a) Let t ∈ T. Recall that s_{T} (t) = t if and only if t ∉ Mirr(T), while, if t ∈ Mirr(T), s_{T} (t) is the unique minimal element in T_{>t} and the interval ]t, s_{T} (t)[ is empty.

If t ∉ E. Then α(t) ∉ E^{∗} by injectivity of α. Therefore s_{T} (t) = t and s_{D}(α(t)) = α(t), so that α s_{T} (t) = s_{D} α(t).

If e ∈ E, then E_{≥e} = E_{>e} ⊔ {e} = E_{≥s_{T} (e)} ⊔ {e}. Therefore E−E_{≥s_{T} (e)} = (E−E_{≥e}) ⊔ {e}. In other words, α(s_{T} (e)) = α(e) ⊔ {e} = e^{∗} ⊔ {e}. On the other hand s_{D}(α(e)) = s_{D}(e^{∗}) = e^{∗} ⊔ {e} because if f^{∗} ∈ E^{∗} satisfies f^{∗} ⊃ e^{∗}, then f > e and e ∈ f^{∗}, hence f^{∗} ⊇ e^{∗} ⊔ {e}, showing that e^{∗} ⊔ {e} is the minimal element of D_{⊃e^{∗}}. It follows that α(s_{T} (e)) = s_{D}(α(e)).

(b) Let f ∈ E. If f ≤ t, then f^{∗} ⊆ α(t). Conversely, if f^{∗} ⊆ α(t), then (E−E_{≥f}) ⊆ (E−E_{≥t}), hence E_{≥f} ⊇ E_{≥t}. Since t is the meet of all meet-irreducible elements larger than t, by (2.3), we get

f = ⋀_{e≥f} e ≤ ⋀_{e≥t} e = t ,

so that f ≤ t. This shows that f < t if and only if f^{∗} ⊂ α(t). Therefore

α(ρ_{T} (t)) = α( ⋁_{f<t} f) = ⋃_{f<t} α(f) = ⋃_{f^{∗}⊂α(t)} f^{∗} = ρ_{D}(α(t)) ,

as was to be shown.

(c) Since α(E) = E^{∗}, we get α(⋁E) = ⋁E^{∗}. Now applying (a) and (b) to the characterization of G̃_{T} recalled in (11.2) above, we see immediately that α(G̃_{T}) = G̃_{D} and therefore α(G_{T}) = G_{D}. Since G_{D} is a full subposet of D and α is full (by Lemma 11.1), we see that α_{|G_{T}} : G_{T} → G_{D} is an isomorphism of posets. Now D only depends on (E, R), by its very definition, so it is clear that G_{T} ≅ G_{D} only depends on (E, R).

The fact that the poset G_{T} does not depend on T, but only on (E, R) up to isomorphism, was left without proof in Corollary 6.7 of [BT4]. This was settled in Theorem 4.2 of [Bo2], as a by-product of a far reaching generalization of the construction of G. The proof above shows that the result on its own is in fact rather elementary.

Since the map α of Lemma 11.1 is a join-morphism, it induces a morphism of correspondence functors α : F_{T} → F_{D} mapping φ ∈ T^{X} to αφ ∈ D^{X}.

11.4. Theorem. Let T be a finite lattice such that Mirr(T) = (E, R) and let α : T → D be the join-morphism of Lemma 11.1.

(a) α : F_{T} → F_{D} restricts to an isomorphism of functors S_{T} → S_{D}.

(b) For any finite set X, the isomorphism F_{T} (X) → F_{D}(X) is also an isomorphism of right k Aut(E, R)-modules.

(c) α induces an isomorphism of simple correspondence functors S_{T,V} → S_{D,V}, for any simple left k Aut(E, R)-module V. In particular, this simple functor does not depend on T, but only on (E, R) up to isomorphism.

Proof : (a) By Theorem 6.3, S_{T} (X) has a k-basis B̂_{X} = {φ̂ | φ ∈ B_{X}}, where B_{X} = {φ ∈ T^{X} | E ⊆ φ(X) ⊆ G_{T} }. Since α maps φ ∈ B_{X} to the composition αφ and since α(E) = E^{∗} (Lemma 11.1) and α(G_{T}) = G_{D} (Lemma 11.3), we see that α maps B_{X} to

B_{X}^{∗} := {αφ ∈ D^{X} | E^{∗} ⊆ αφ(X) ⊆ G_{D}} = {ψ ∈ D^{X} | E^{∗} ⊆ ψ(X) ⊆ G_{D}} .

Now we claim that α̂φ = αφ̂. Postponing the proof of the claim, it follows that

α(B̂_{X}) = {αφ̂ | φ ∈ B_{X}} = {α̂φ | φ ∈ B_{X}} = {α̂φ | αφ ∈ B_{X}^{∗}}
= {ψ̂ | ψ ∈ B_{X}^{∗}} = B̂_{X}^{∗} .

Theorem 6.3 asserts that B̂_{X}^{∗} is a basis of S_{D}(X), so we see that α maps bijectively a basis of S_{T} (X) to a basis of S_{D}(X). Therefore the restriction of α is an isomorphism S_{T} → S_{D}.

To prove the claim, we first show that αφ_{A} = (αφ)_{A^{∗}} where A^{∗} = α(A) for any A ⊆ E. This is because

αφ_{A}(x) = \{ \begin{matrix}αφ(x) & if φ(x) ∉ A , \\ αs_{T} φ(x) & if φ(x) ∈ A , \end{matrix}

while

(αφ)_{A^{∗}}(x) = \{ \begin{matrix}αφ(x) & if αφ(x) ∉ A^{∗} , \\ s_{D}αφ(x) & if αφ(x) ∈ A^{∗} , \end{matrix}

so that αφ_{A} = (αφ)_{A^{∗}} since αs_{T} = s_{D}α by Lemma 11.3. It follows that

αφ̂ = ∑_{A⊆E}(−1)^{|A|}αφ_{A} = ∑_{A^{∗}⊆E^{∗}}(−1)^{|A^{∗}|}(αφ)_{A^{∗}} = α̂φ ,

proving the claim.

(b) The isomorphism α : S_{T} (E) → S_{D}(E) maps î to α î, which is equal to the generator α̂ i of S_{D}(E), by the claim above. It follows that the isomorphism S_{T} (X) → S_{D}(X) maps U î to Uα̂ i. By the definition of the right action of γ ∈ Aut(E, R) (see Lemma 10.4), we obtain

α((U î) · γ) = α(UΔ_{γ} î) = UΔ_{γ} α î = UΔ_{γ} α̂ i = (U α̂ i) · γ = α(U î) · γ ,

as required.

(c) Since S_{T,V} = S_{T} ⊗_{kA} V and S_{D,V} = S_{D} ⊗_{kA} V, where A = Aut(E, R), the isomorphism S_{T,V} → S_{D,V} follows from (a) and (b).

11.5. Remark. In the proof of (b), we have used S_{D}(E), whereas, since Mirr(D) = E^{∗}, the functor S_{D} is naturally generated by S_{D}(E^{∗}) instead of S_{D}(E). But one can pass from S_{D}(E) to S_{D}(E^{∗}) by left composition with the correspondence J = {(e^{∗}, e) | e ∈ E}, which realizes the isomorphism α_{|E} : E → E^{∗}. Explicitly, the computation gives, for any φ ∈ D^{E},

(Jφ)(e^{∗}) = φ(e) = φ α_{|E}^{−1}(e^{∗}) ,

and we see that we have pre-composed by the inverse of this isomorphism. As a result, S_{D} is naturally generated by î^{∗} := α îα_{|E}^{−1} ∈ S_{D}(E^{∗}).

Theorem 10.6 and Theorem 11.4 imply that we recover the parametrization of simple functors obtained in [BT2], but with a rather different approach.

11.6. Corollary. The simple correspondence functors are parametrized by isomorphism classes of triples (E, R, V), where (E, R) is a finite poset and V is a simple k Aut(E, R)-module.

12. Functors associated to a join-morphism

Associated to any join-morphism of finite lattices, there is a duality studied in [BT5] giving rise to more correspondence functors. The purpose of this section is to establish a link between this construction and the approach of the present paper.

The starting point in [BT5] is a join-morphism α : T → T′^{op}, where T and T′ are finite lattices, together with the subset

Φ_{α} = {f ∈ Jirr(T) | α(f) ∈ Jirr(T′), α^{op}α(f) = f} .

Note that the definition of α uses T′^{op} whereas Φ_{α} uses T′. In view of the approach of the present paper, it is convenient to set L = T′^{op}, so that α is simply a joinmorphism from T to L. As a result, one gets Jirr(T′) = Jirr(L^{op}) = Mirr(L), and therefore

Φ_{α} = {f ∈ Jirr(T) | α(f) ∈ Mirr(L), α^{op}α(f) = f} ,
Φ′_{α} = {f′ ∈ Mirr(L) | α^{op}(f′) ∈ Jirr(T), αα^{op}(f′) = f′} .

This is the notation used in [BT5]. For simplicity, we shall define E := Φ_{α} and E′ := Φ′_{α}.

12.1. Lemma. Let α : T → L be a join-morphism of finite lattices. Let E, viewed as a full subposet of T, and E′, viewed as a full subposet of L, be defined as above. The restriction of α to E yields an isomorphism of posets E → E′. Its inverse is induced by the restriction of α^{op} to E′.

Proof : This is Lemma 3.6 in [BT5], observing that α : T → L is order-preserving, whereas in [BT5], α was viewed as an order-reversing map α : T → T′.

We also extend the notation to elements of E and E′ and to subsets of E and E′, as follows:

if e ∈ E, then e′ = α(e) ∈ E′, and if e′ ∈ E′, then e = α^{op}(e′) ∈ E .

Similarly,

if B ⊆ E, then B′ = α(B) ⊆ E′, and if B′ ⊆ E′, then B = α^{op}(B′) ⊆ E .

If α(f) ∈ Jirr(T′) = Mirr(L), the unique maximal element of the interval [0_{T′}, α(f)[ in T′ is r_{T′}(α(f)), which is of course equal to s_{L}(α(f)), the unique minimal element of the interval ]α(f), 1_{L}] in L. Working with Mirr(L) and the operator s_{L} is precisely the point of view used in the present paper.

Associated with a given join-morphism α : T → L, there is a correspondence functor 𝕊_{α} obtained by means of a certain duality which is a main tool in [BT5]. But it is shown in Corollary 3.15 of [BT5] that 𝕊_{α} can also be described in a different way (up to isomorphism). Using this alternative description, we define

𝕊_{α} = ᾰ(F_{T}) ,

the image of the morphism ᾰ : F_{T} → F_{L} induced by the following linear combination of join-morphisms from T to L :

ᾰ = ∑_{A⊆E}(−1)^{|A|}α_{A} ,

where α_{A} is defined by

α_{A}(t) = \{ \begin{matrix}s_{L}α(t) & if t ∈ A , \\ α(t) & if t ∈ T −A . \end{matrix}

This sums resembles the definition of φ^{E} in (3.8) and the purpose of this section is to explore this resemblance. More precisely, for any φ ∈ T^{X}, we have αφ ∈ L^{X} and α̂φ^{E′} resembles ᾰφ, but we shall see that it is not always equal to ᾰφ. As before,

for any map ψ : X → L, we write for simplicity ψ̂^{E′} = ψ̂.

First define, for all t ∈ T,

\overline{t} = \{ \begin{matrix}α^{op}α(t) & if α(t) ∈ E′ , \\ t & if α(t) ∈ L−E′ . \end{matrix}

For any φ ∈ T^{X}, one defines also \overline{φ} ∈ T^{X} by \overline{φ}(x) = \overline{φ(x)} for all x ∈ X. Notice that α\overline{φ} = αφ because αα^{op}α = α.

12.2. Lemma. Let φ ∈ T^{X}.

(a) If A ⊆ E and A′ = α(A) ⊆ E′, we have an equality (αφ)_{A′} = α_{A}\overline{φ}.

(b) α̂φ = ᾰ\overline{φ}.

(c) If φ(x) ∈ T −A but αφ(x) ∈ A′, one has (αφ)_{A′}(x) ≠ α_{A}φ(x).

Proof : (a) Let x ∈ X. By Definition 3.7, one gets

(αφ)_{A′}(x) = \{ \begin{matrix}s_{L}αφ(x) & if αφ(x) ∈ A′ , \\ αφ(x) & if αφ(x) ∈ L−A′ . \end{matrix}

On the other hand, the definition of α_{A} given above and the fact that α\overline{φ} = αφ imply

α_{A}\overline{φ}(x) = \{ \begin{matrix}s_{L}α\overline{φ}(x) = s_{L}αφ(x) & if \overline{φ}(x) ∈ A , \\ α\overline{φ}(x) = αφ(x) & if \overline{φ}(x) ∈ T −A . \end{matrix}

The condition \overline{φ}(x) ∈ A yields either α^{op}αφ(x) ∈ A or φ(x) ∈ A, according to the definition of \overline{φ}. Applying α and using αα^{op}α = α, we see that αφ(x) ∈ A′ in both cases. Conversely, if αφ(x) ∈ A′, then αφ(x) ∈ E′ and the first case of the definition of \overline{φ}(x) yields \overline{φ}(x) = α^{op}αφ(x) ∈ A. We see that the conditions correspond to each other, proving the equality (αφ)_{A′} = α_{A}\overline{φ}.

(b) By (a), one obtains

α̂φ = ∑_{A′⊆E′}(−1)^{|A′|}(αφ)_{A′} = ∑_{A⊆E}(−1)^{|A|}α_{A}\overline{φ} = ᾰ\overline{φ} .

(c) When φ(x) ∈ T −A , one has α_{A}φ(x) = αφ(x), while when αφ(x) ∈ A′, one gets (αφ)_{A′}(x) = s_{L}αφ(x) ≠ αφ(x).

Note that case (c) in Lemma 12.2 arises for instance when t := φ(x) ∉ E but α(t) ∈ E′. In the notation of Lemma 4.7 and Proposition 4.9 in [BT5], one gets \overline{t} = βα(t) and t < βα(t) ∈ E, where β = α^{op}. The distinction between t and \overline{t} is the main ingredient in Lemma 12.2.

Actually, if we had \overline{t} = t for all t ∈ T, then ᾰφ would be equal to α̂φ, which belongs to Q_{L}^{E′} by Lemma 3.9. In that case, it would follow that 𝕊_{α} = ᾰ(F_{T}) is a subfunctor of F_{L} contained in Q_{L}^{E′}. We now show that this containment in fact always holds.

12.3. Theorem. The subfunctor 𝕊_{α} = ᾰ(F_{T}) of F_{L} has the following properties :

(a) 𝕊_{α} contains S_{L}^{E′}.

(b) 𝕊_{α} is contained in Q_{L}^{E′}. Explicitly, ˆ ᾰφ = ᾰφ for any function φ ∈ T^{X}.

Proof : (a) Let j : E′ → L be the inclusion map. By definition, S_{L}^{E′} is generated by ĵ ∈ F_{L}(E′). Since α induces a bijection δ : E → E′ by Lemma 12.1, S_{L}^{E′} is also generated by ĵδ ∈ F_{L}(E). But we have clearly ĵδ = ĵδ (see Lemma 6.1). If i : E → T denotes the inclusion map, we have jδ = αi, because the evaluation at e ∈ E gives α(e) in both cases (since δ(e) = α(e) by definition). By Lemma 12.2, we obtain

ĵδ = α̂i = ᾰ\overline{i} .

But \overline{i} = i because \overline{i}(e) = α^{op}αi(e) = δ^{−1}δ(e) = e = i(e) for all e ∈ E. It follows that

ĵδ = ĵδ = ᾰ\overline{i} = ᾰi

and therefore the generator ĵδ of S_{L}^{E′} belongs to ᾰ(F_{T})(E) = 𝕊_{α}(E).

(b) We have to prove that ᾰφ belongs to Q_{L}^{E′}(X) for any φ ∈ T^{X}. By Proposition 3.17, we need to show that ˆ ᾰφ = ᾰφ. But ᾰ = ∑_{B⊆E}(−1)^{|B|}α_{B}, so that

ˆᾰφ = ∑_{B⊆E}(−1)^{|B|}{ \atop α_{B} φ} = ∑_{B⊆E}(−1)^{|B|} ∑_{A′⊆E′}(−1)^{|A′|}(α_{B} φ)_{A′}
= ∑_{B⊆E}(−1)^{|B|}α_{B} φ + ∑_{B⊆E, A′≠∅}(−1)^{|B|+|A′|}(α_{B} φ)_{A′}
= ᾰφ + ∑_{B⊆E, A′≠∅}(−1)^{|B|+|A′|}(α_{B} φ)_{A′} .

We are going to prove that the latter sum is zero. We fix a nonempty subset A′ of E′ and we intend to show that the sum ∑_{B⊆E}(−1)^{|B|}(α_{B} φ)_{A′} is zero. We define

M′ = {e′ ∈ A′ | s_{L}(e′) ∉ A′}

and we note that all maximal elements of A′ belong to M′, so that M′ is nonempty since A′ is nonempty. Using again the bijection of Lemma 12.1, recall that A = α^{op}A′ and M = α^{op}M′.

Any subset B ⊆ E is a disjoint union B = C ∪ D with C ⊆ E−M and D ⊆ M. Therefore

∑_{B⊆E}(−1)^{|B|}(α_{B} φ)_{A′} = ∑_{C⊆E−M}(−1)^{|C|}( ∑_{D⊆M}(−1)^{|D|}(α_{C∪D} φ)_{A′})

and it suffices to prove that, for any fixed subset C ⊆ E−M, the inner sum ∑_{D⊆M}(−1)^{|D|}(α_{C∪D} φ)_{A′} is zero. By Lemma 12.4 below,

∑_{D⊆M}(−1)^{|D|}(α_{C∪D} φ)_{A′} = ∑_{D⊆M}(−1)^{|D|}(α_{C} φ)_{A′} .

But ∑_{D⊆M}(−1)^{|D|} = 0, because M is nonempty. Therefore

∑_{D⊆M}(−1)^{|D|}(α_{C∪D} φ)_{A′} = 0

as required. This will complete the proof of Theorem 12.3, provided we prove Lemma 12.4 below.

12.4. Lemma. Let M be as above and let C be a subset of E−M. For any subset D of M, the functions (α_{C∪D} φ)_{A′} and (α_{C} φ)_{A′} are equal.

Proof : Let x ∈ X. We first compute (α_{C} φ)_{A′}(x) by discussing the values of φ(x). If φ(x) ∈ C, we have α_{C} φ(x) = s_{L}αφ(x). But

φ(x) ∈ C ⇒ φ(x) ∉ M ⇒ αφ(x) ∉ M′ ⇒ s_{L}αφ(x) ∈ A′

by the definition of M′. Thus α_{C} φ(x) ∈ A′ and we obtain :

(12.5) If φ(x) ∈ C , α_{C} φ(x) ∈ A′ and (α_{C} φ)_{A′}(x) = s_{L}α_{C} φ(x) = s_{L}s_{L}αφ(x) .

If now φ(x) ∉ C, we get α_{C} φ(x) = αφ(x), hence :

(12.6)  If φ(x) ∉ C , (α_{C} φ)_{A′}(x) = \{ \begin{matrix}αφ(x) & if αφ(x) ∉ A′ , \\ s_{L}αφ(x) & if αφ(x) ∈ A′ . \end{matrix}

Now we compute (α_{C∪D} φ)_{A′}(x) by discussing again the values of φ(x).

If φ(x) ∈ C, we have α_{C∪D} φ(x) = s_{L}αφ(x) = α_{C} φ(x) and this belongs to A′, as noticed in (12.5). It follows that :

(12.7)  If φ(x) ∈ C , (α_{C∪D} φ)_{A′}(x) = s_{L}α_{C∪D} φ(x) = s_{L}s_{L}αφ(x) .

If φ(x) ∈ D, we have α_{C∪D} φ(x) = s_{L}αφ(x). But

φ(x) ∈ D ⇒ φ(x) ∈ M ⇒ αφ(x) ∈ M′ ⇒ αφ(x) ∈ A′ and s_{L}αφ(x) ∉ A′

by the definition of M′. Thus we obtain :

(12.8) If φ(x) ∈ D , αφ(x) ∈ A′ and (α_{C∪D} φ)_{A′}(x) = (s_{L}αφ)_{A′}(x) = s_{L}αφ(x) .

Finally, if φ(x) ∉ C ∪ D, we get α_{C∪D} φ(x) = αφ(x), hence :

(12.9)  If φ(x) ∉ C ∪ D , (α_{C∪D} φ)_{A′}(x) = \{ \begin{matrix}αφ(x) & if αφ(x) ∉ A′ , \\ s_{L}αφ(x) & if αφ(x) ∈ A′ . \end{matrix}

The comparison between (12.5) and (12.7) in case φ(x) ∈ C, respectively between (12.6), (12.8) and (12.9) in case φ(x) ∉ C, shows that (α_{C} φ)_{A′}(x) = (α_{C∪D} φ)_{A′}(x). This completes the proof of Lemma 12.4 and so Theorem 12.3 is now fully established.

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Serge Bouc, CNRS-LAMFA, Université de Picardie - Jules Verne, 33, rue St Leu, F-80039 Amiens Cedex 1, France.

serge.bouc@u-picardie.fr

Jacques Thévenaz, Institut de mathématiques, EPFL, Station 8, CH-1015 Lausanne, Switzerland.

jacques.thevenaz@epfl.ch