∀*n*∈ℕ, 〈1* _{n}*〉

∀*u* ∈*M*^{B},
∃*f*∈Mor_{L}(*K*,*M*),
*f*_{|B}=*u*

∃*f*∈Mor_{L}(*K*,*M*^{MB}),∀*b*
∈*B*, *f*(*b*)=π_{b}

In the category of ∃

The axioms of module of abstract clones, say that this algebra satisfies all

The satisfaction condition is equivalent to : for any equality between terms found true in

Indeed, the morphism

Conversely, any formula made of universal quantifiers over an equality of terms

Except if it is a statement of equality between 2 variables; to avoid such exceptions we may need to modify our above definition by seeing

Thus, these *L*-algebras are also *K*-algebras where *K*
is seen as a set of symbols with a common set *B* of arguments
(thus a common arity). The interpretation of *K* in an *L*-algebra
*M* satisfying (*K*,*B*) is characterized (we may say
"recursively defined") by the condition of the above *f* ,
namely (using similar notations to those of group action which are a
particular case): ∀*u*∈*M ^{B}*,

- ∀
*b*∈*B*,*b*⋅*u*=*u*_{b}

- ∀
*s*∈*L*, ∀*x*∈*K*,^{ns}*s*(_{K}*x*) ⋅*u*=*s*((_{M}*x*⋅_{j}*u*)_{j<ns})

(

⇒ (

⇒

Conversely, any

Namely, any

If

If

So, in this category, the set

The class of all

If the family (

- Each
*K*was only an_{i}*L*-algebra, not necessarily satisfying (*K*,_{j}*B*). Can we replace (_{j}*K*,_{i}*B*) by some (_{i}*K'*,_{i}*B'*) where_{i}*K'*satisfies all (_{i}*K*,_{j}*B*), but that remains equivalent as a predicate on the class of algebras satisfying all (_{j}*K*,_{j}*B*) for_{j}*j*≠*i*? - Can two conditions (
*K*,*B*) and (*K*',*B'*) with the same arity (*B*and*B'*are in bijection) be fused into one ?

These questions will be positively answered below.

We can easily verify that:

Any subalgebra of an *L*-algebra satisfying (*K*,*B*)
also satisfies (*K*,*B*) (because *K* is generated
by *B*).

Any product of *L*-algebras satisfying (*K*,*B*)
also satisfies (*K*,*B*).

Now comes the hard stuff:

**Theorem.** For any *L*-algebra *A* and any
variety *V* of *L*-algebras, the category of all (*M*,*f*)
where *M* is an *L*-algebra in *V* and *f*
∈Mor_{L}(*A*,*M*), has an initial object (*A*',φ), called the quotient of *A* by the
family of conditions defining the variety.

Proof.

LetNow we can answer the previous questions.Hbe the set of all equivalence relationshonAsuch thatA/his anL-algebra inV, with the canonical projectionp_{h}∈Mor_{L}(A,A/h).

Let φ =∏_{h}_{∈}_{H}p_{h}andA' = Im φ ⊂P=∏_{h}_{∈}_{H}A/h.

A' satisfies all (K,_{i}B)_{i}_{i∈I}because it is a subalgebra of a productPofL-algebras that do.

To verify that (A',φ) is an initial object in this category, let (M,f) be another object. We need to show that there is a unique morphismgfrom (A',φ) to (M,f), i.e.g∈Mor_{L}(A',M) such thatg০φ=f.

In the work on quotients (2.9) we saw the existence of a uniquegsuch thatg০φ=f, writteng=f/φ with domain Im φ =A', provided that ~_{φ}⊂~_{f}. The condition of uniqueness here is Im φ =A'. The existence is obtained asg=j০π_{h}, whereh= ~_{f}, π_{h}∈Mor_{L}(A',A/h) is the restriction of the canonical projection ofPon its factorA/h, andj= (f/h) ∈ Mor_{L}(A/h,M). Indeed,০φ =

gj০π_{h}০φ =j০p_{h}=f. ∎

First, to replace (

If the restriction of this quotient φ to

If there is anNow assuming it injective, the arity is preserved; the verification of equivalence of the conditions (L-algebraMwith elementsx≠x' and satisfying all conditions then for allb≠b' inB, ∃u∈M,^{Bi}u(b)=x≠u(b')=x' and ∃f∈Mor_{L}(K,_{i}M),f_{|B}=uthus ∃g∈Mor_{L}(K',_{i}M),g০φ=f, and φ(b)≠φ(b') becausef(b)≠f(b').

Now for fusing several conditions (

Any

In fact, when

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