Uniqueness of preadditive structures

Proof. The morphism $\alpha : X \oplus Y \to X \times Y$ is defined by the equations $$p_1 \circ \alpha \circ i_1 = \id_X, \quad p_2 \circ \alpha \circ i_2 = \id_Y,$$ $$p_2 \circ \alpha \circ i_1 = 0,\quad p_1 \circ \alpha \circ i_2 = 0.$$ It does not depend on the choice of preadditive structure since zero morphisms are unique. It is an isomorphism: Define $$\beta \coloneqq i_1 \circ p_1 + i_2 \circ p_2 : X \times Y \to X \oplus Y.$$ Then $\alpha \circ \beta = \id_{X \times Y}$ because $$p_1 \circ \alpha \circ \beta = p_1 \circ \alpha \circ i_1 \circ p_1 + p_1 \circ \alpha \circ i_2 \circ p_2 = \id_X \circ p_1 + 0 \circ p_2 = p_1$$ and likewise $p_2 \circ \alpha \circ \beta = p_2$. We also have $\beta \circ \alpha = \id_{X \oplus Y}$ 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

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