Binary Coproduct in Preadditive Category is Biproduct

Theorem

Let $a_1, a_2$ be objects of $A$.

Let $(a_1 \sqcup a_2, i_1, i_2)$ be their binary coproduct, assuming it exists.

Let $p_1 : a_1 \sqcup a_2 \to a_1$ be the unique morphism with:

$p_1 \circ i_1 = 1 : a_1 \to a_1$

$p_1 \circ i_2 = 0 : a_1 \to a_2$

Let $p_2 : a_1 \sqcup a_2 \to a_2$ be the unique morphism with:

$p_2 \circ i_1 = 0 : a_2 \to a_1$

$p_2 \circ i_2 = 1 : a_2 \to a_2$

where $1$ is the identity morphism and $0$ is the zero morphism.

Then $(a_1 \sqcup a_2, i_1, i_2, p_1, p_2)$ is the binary biproduct of $a_1$ and $a_2$.

By definition of binary biproduct, it remains to verify that:

$i_1\circ p_1 + i_2 \circ p_2 = 1 : a_1 \sqcup a_2 \to a_1 \sqcup a_2$.

By definition of binary coproduct and identity morphism, $1 : a_1 \sqcup a_2 \to a_1 \sqcup a_2$ is the unique morphism with:

$1 \circ i_1 = i_1$

$1 \circ i_2 = i_2$

Thus it remains to verify that $i_1\circ p_1 + i_2 \circ p_2$ satisfies the same condition:

$\ds (i_1\circ p_1 + i_2 \circ p_2) \circ i_1$

$=$

$\ds i_1 \circ (p_1 \circ i_1) + i_2 \circ (p_2 \circ i_1)$

$\ds $

$=$

$\ds i_1 \circ 1 + i_2 \circ 0$

$\ds $

$=$

$\ds i_1$

$\ds (i_1\circ p_1 + i_2 \circ p_2) \circ i_2$

$=$

$\ds i_1 \circ (p_1 \circ i_2) + i_2 \circ (p_2 \circ i_2)$

$\ds $

$=$

$\ds i_1 \circ 0 + i_2 \circ 1$

$\ds $

$=$

$\ds i_2$

$\blacksquare$