Binary Coproduct in Preadditive Category is Biproduct

Theorem
Let $A$ be a preadditive category.

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 $a_2$.

Proof
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:

Also see

 * Binary Product in Preadditive Category is Biproduct