$p_Ai_B=0$ and $p_Bi_A=0$ in additive category Let $\cal A$ be an additive category. Then for any $A,B\in\textrm{Ob}(\cal A)$ the direct sum $A\oplus B$ is both their product and their coproduct. Let $i_A:A\rightarrow A\oplus B$ and $i_B:B\rightarrow A\oplus B$ be canonical embeddings and $p_A:A\oplus B\rightarrow A$ and $p_B:A\oplus B\rightarrow B$ be canonical projections. Then $p_Ai_A=1_A,\:p_Bi_B=1_B$ and $i_Ap_A+i_Bp_B=1_{A\oplus B}.$
It feels like these relations must imply $p_Ai_B=0$ and $p_Bi_A=0.$ I've tried to manipulate them but could not prove the desired. Could you please give me a hint?
 A: They are true, even if $\mathcal A$ is only semi-additive.
You have to think of how $i_A$ is defined: it is defined as the inclusion $A\to A\times B$ given by $id_A$ and $0$ on the $B$-coordinate. In particular by its very definition $p_Bi_A = 0$.
However, in the semi-additive case, this does not follow from the previous equations. In the additive case, PrudiiArca has explained how to deduce it, and it crucially uses minus - the point is that it needs to.
Here is a counterexample in the semi-additive case: consider $\mathcal A$ the category of $R$-modules in commutative monoids, where $R$ is the semi-ring $\{0,1\}$ where $1+1 = 1$.
Then we can choose $A=B= R$ and $p_A = p_B = i_A=i_B = id_R$, and we find $i_Ap_A +i_Bp_B = id_R + id_R = id_R$, and similarly $p_Ai_A = p_Bi_B= id_R$; while $i_Ap_B = id_R \neq 0$.
In fact, in an arbitrary semi-additive category, the equations you listed + $p_Ai_B = 0, p_Bi_A = 0$ completely determine $A\oplus B$.
A: Hint Having an additive category means that you can subtract morphisms. Rewrite the equation $i_Ap_A + i_Bp_B = 1$, postcompose with $p_A$ and use that $p_B$ is split epic.
