Biproduct

In category theory and its applications to mathematics, a biproduct of a finite collection of objects, in a category with zero objects, is both a product and a coproduct. In a preadditive category the notions of product and coproduct coincide for finite collections of objects. The biproduct is a generalization of finite direct sums of modules.

Definition

Let C be a category with zero morphisms. Given a finite (possibly empty) collection of objects ''A''₁, ..., ''A''_''n'' in C, their ''biproduct'' is an object $A_1 \oplus \dots \oplus A_n$ in C together with morphisms

*$p_k \!: A_1 \oplus \dots \oplus A_n \to A_k$ in C (the ''projection morphisms'')
*$i_k \!: A_k \to A_1 \oplus \dots \oplus A_n$ (the ''embedding morphisms'')
satisfying
*$p_k \circ i_k = 1_$, the identity morphism of $A_k,$ and
*$p_l \circ i_k = 0$, the zero morphism $A_k \to A_l,$ for $k \neq l,$
and such that
*$\left( A_1 \oplus \dots \oplus A_n, p_k \right)$ is a product for the $A_k,$ and
*$\left( A_1 \oplus \dots \oplus A_n, i_k \right)$ is a coproduct for the $A_k.$

If C is preadditive and the first two conditions hold, then each of the last two conditions is equivalent to $i_1 \circ p_1 + \dots + i_n\circ p_n = 1_$ when ''n'' > 0. An empty, or nullary, product is always a terminal object in the category, and the empty coproduct is always an initial object in the category. Thus an empty, or nullary, biproduct is always a zero object.

Examples

In the category of abelian groups, biproducts always exist and are given by the direct sum. The zero object is the trivial group.

Similarly, biproducts exist in the category of vector spaces over a field. The biproduct is again the direct sum, and the zero object is the trivial vector space.

More generally, biproducts exist in the category of modules over a ring.

On the other hand, biproducts do not exist in the category of groups. Here, the product is the direct product, but the coproduct is the free product.

Also, biproducts do not exist in the category of sets. For, the product is given by the Cartesian product, whereas the coproduct is given by the disjoint union. This category does not have a zero object.

Block matrix algebra relies upon biproducts in categories of matrices.H.D. Macedo, J.N. Oliveira
Typing linear algebra: A biproduct-oriented approach
Science of Computer Programming, Volume 78, Issue 11, 1 November 2013, Pages 2160-2191, , .

Properties

If the biproduct $A \oplus B$ exists for all pairs of objects ''A'' and ''B'' in the category C, and C has a zero object, then all finite biproducts exist, making C both a Cartesian monoidal category and a co-Cartesian monoidal category.

If the product $A_1 \times A_2$ and coproduct $A_1 \coprod A_2$ both exist for some pair of objects ''A''₁, ''A''₂ then there is a unique morphism $f: A_1 \coprod A_2 \to A_1 \times A_2$ such that

*$p_k \circ f \circ i_k = 1_,\ (k = 1, 2)$
*$p_l \circ f \circ i_k = 0$ for $k \neq l.$

It follows that the biproduct $A_1 \oplus A_2$ exists if and only if ''f'' is an isomorphism.

If C is a preadditive category, then every finite product is a biproduct, and every finite coproduct is a biproduct. For example, if $A_1 \times A_2$ exists, then there are unique morphisms $i_k: A_k \to A_1 \times A_2$ such that

*$p_k \circ i_k = 1_,\ (k = 1, 2)$
*$p_l \circ i_k = 0$ for $k \neq l.$

To see that $A_1 \times A_2$ is now also a coproduct, and hence a biproduct, suppose we have morphisms $f_k: A_k \to X,\ k=1,2$ for some object $X$. Define $f := f_1 \circ p_1 + f_2 \circ p_2.$ Then $f$ is a morphism from $A_1 \times A_2$ to $X$, and $f \circ i_k = f_k$ for $k = 1, 2$.

In this case we always have
*$i_1 \circ p_1 + i_2 \circ p_2 = 1_.$

An additive category is a preadditive category in which all finite biproducts exist. In particular, biproducts always exist in abelian categories.

References

*{{rp, at=Section 1.2

Additive categories
Limits (category theory)