Uniqueness of preadditive structures

Claim

Let

\mathcal{C}

be a preadditive category (or more generally, a category enriched in commutative monoids) with finite products and finite coproducts. Then for all objects

X,Y

the canonical morphism

$\alpha : X \oplus Y \to X \times Y$

is an isomorphism. Moreover, the preadditive structure is unique: If

f,g : A \rightrightarrows B

are morphisms, their sum

$f+g : A \to B$

is the composite of

(f,g) : A \to B \times B

, the inverse

\alpha^{-1} : B \oplus B \to B \times B

, and the codiagonal

\nabla : B \oplus B \to B

Proof

The morphism

\alpha : X \oplus Y \to X \times Y

is defined by the equations

$p_1 \circ \alpha \circ i_1 = \mathrm{id}_X, \quad p_2 \circ \alpha \circ i_2 = \mathrm{id}_Y$

$p_2 \circ \alpha \circ i_1 = 0,\quad p_1 \circ \alpha \circ i_2 = 0$ .

It does not depend on the choice of preadditive structure since zero morphisms are unique. It is an isomorphism: Define

$\beta := i_1 \circ p_1 + i_2 \circ p_2 : X \times Y \to X \oplus Y$ .

Then

\alpha \circ \beta = \mathrm{id}_{X \times Y}

because

$p_1 \circ \alpha \circ \beta = p_1 \circ \alpha \circ i_1 \circ p_1 + p_1 \circ \alpha \circ i_2 \circ p_2 = \mathrm{id}_1 \circ p_1 + 0 \circ p_2 = p_1$

and likewise

p_2 \circ \alpha \circ \beta = p_2

. We also have

\beta \circ \alpha = \mathrm{id}_{X \oplus Y}

with a very similar calculation that shows

\beta \circ \alpha \circ i_1 = i_1

and

\beta \circ \alpha \circ i_2 = i_2

Therefore, for morphisms $f,g : A \rightrightarrows B$ the composite $A \to B$ in the claim is equal to

\begin{aligned} \nabla \circ \beta \circ (f,g) & = \nabla \circ (i_1 \circ p_1 + i_2 \circ p_2) \circ (f,g) \\ & = \nabla \circ i_1 \circ p_1 \circ (f,g) + \nabla \circ i_2 \circ p_2 \circ (f,g) \\ & = p_1 \circ (f,g) + p_2 \circ (f,g) \\ & = f + g. \end{aligned}

Usage

This lemma is referenced in the following categories: