Let be a class of models (structures) of a given signature. Consider a (non-empty) family of structures in . If and is an embedding, we say that is a factor embedding. [1, 2] If additionally satisfies the condition that is the identity on , where is the th projection, then is said to be a strong factor embedding.  is said to be a factor embeddable class iff for every (non-empty) family of structures in and every there is a factor embedding . [1, 2]
The definition above does not require the product to be a member of , however many interesting examples of factor embeddable classes are in fact closed under products. Factor embeddable classes that are closed under finite products (or equivalently under binary products) have the joint embedding property. Factor embeddable classes closed under arbitrary products have the strong joint embedding property.
is factor embeddable.
For every pair of models there exists a homomorphism .
To see the above, suppose is factor embeddable and consider models . Then there exists a factor embedding from into the product . Composing this embedding with the projection onto gives a homomorphism . Conversely suppose such a homomorphism exists for all and consider a family in and . We can define a strong factor embedding by choosing homomorphisms for all with the identity map on , and then for all setting for each . 
The above proof shows that the factor embeddings guaranteed to exist for a factor embeddable class can always be chosen to be strong factor emebeddings. 
A corollory of the above is that if there exists a model which is a retract of every member of a class then, is factor embeddable - in particular if the members of have one element submodels, then is factor embeddable.  (A retract of a model is a submodel which is also a quotient model such that the quotient map composed with the submodel embedding is the identity map.)
The following are examples of factor embeddable classes:
The class of all Boolean algebras is an example of a class which is not factor embeddable - there is no way to embed the trivial Boolean algebra into a product of itself with any non-trivial Boolean algebras. (The trivial Boolean algebra satisfies the identity which is not satisfied by any Boolean algebra having more than one element.)
|Date of creation||2013-03-22 19:36:55|
|Last modified on||2013-03-22 19:36:55|
|Last modified by||Naturman (26369)|
|Defines||factor embeddable class|
|Defines||strong factor embedding|