The set products satisfy the coproduct property in the Abelian category.
The set product, where elements are pairs, is capable of acting as a
co-product object in the category . In other words, the following
diagram commutes in where is uniquely determined
by the rest of the diagram.
Commutative diagram
and , the inclusion functions
The function and are special inclusion functions given by:
They place the input in either the first or second entry of a pair, leaving the other entry
to be the identity element of the corresponding group, or . The fixed identity entry allows the variable entry to inherit the group behaviour of the input. This insures that
the inclusion functions satisfy the requirements to be group homomorphisms.
The morphism
It is not a matter of searching for , for it's form was anticipated from the beginning.
can be expressed precisely in terms of and :
The definition of follows (is forced from) from the constraints.
What is interesting is that is actually possible! And only just.
Can you remember why works in
but not in ?
as a co-product
is a valid group in both and . It's group operation is the parallel application of the group operations from and . In this sense, is not particularly interesting.
What is interesting is that satisfies the co-product property, even though it wasn't constructed for this purpose. The caveat here is that we have been careful to specify the behaviour of the inclusion functions to be used, and . These inclusion functions are not standard inclusion functions, as the domain is not a subset of the co-domain.
as a group homomorphism
To be a group homomorphism, must satisfy the following equality,
for any :
Consider first the LHS:
Now the RHS:
For these two expressions to be equal, commutativity must be present.
vs.
being a group homomorphism is a critical requirement for to act as co-product.
Even if both and are commutative, we can find in the category a target group which is not commutative. With not being Abelian, we can't find a homomorphism from to in . The universal property is only satisfied if there exists a
unique morphism for all , so any failure to find
a morphism will prevent the universal property from being satisfied. This
explains why cannot act as a co-product in ,
even if and are commutative groups.
Example
The morphism
As an example, consider . The inclusion functions send elements of to the highlighted column on the left and send elements of to the highlighted row on the top. For the diagram above, the non-highlighted center section do not play a role in the actually set-function commutativity, as the elements are never mapped to by the inclusion functions; however, the center section does affects whether:
forms a group
forms a morphism in the category
both of which are requirements for the commutativity diagram to be valid diagram of morphisms between objects.