[proofplan] The product and coproduct hypotheses already provide the universal properties required of a binary biproduct. The four displayed identities give all biproduct equations except the decomposition identity on $P$. We prove that missing identity by comparing the two endomorphisms $i_A \circ p_A + i_B \circ p_B$ and $1_P$ after postcomposition with the product projections $p_A$ and $p_B$; the product universal property then forces the two endomorphisms to be equal. [/proofplan] [step:Use the product projections to test equality of endomorphisms of $P$] Since $(P,p_A,p_B)$ is a product of $A$ and $B$, the following uniqueness property holds: for every object $X$ of $\mathcal C$ and every pair of morphisms \begin{align*} f_A &: X \to A, & f_B &: X \to B, \end{align*} there exists a unique morphism $f:X \to P$ such that \begin{align*} p_A \circ f &= f_A, & p_B \circ f &= f_B. \end{align*} Applying this with $X=P$, two morphisms $u,v:P \to P$ are equal whenever \begin{align*} p_A \circ u &= p_A \circ v, & p_B \circ u &= p_B \circ v. \end{align*} Indeed, both $u$ and $v$ are then morphisms from $P$ to the product object $P$ with the same two component morphisms to $A$ and $B$, so the uniqueness clause in the product universal property gives $u=v$. [/step] [step:Compute the projections of the candidate decomposition map] Define the endomorphism $e:P \to P$ by \begin{align*} e := i_A \circ p_A + i_B \circ p_B. \end{align*} This sum is defined because $\mathcal C$ is preadditive, so $\operatorname{Hom}_{\mathcal C}(P,P)$ is an abelian group and composition is bilinear. We compute its composite with $p_A:P \to A$. By bilinearity of composition in a preadditive category, \begin{align*} p_A \circ e &= p_A \circ (i_A \circ p_A + i_B \circ p_B) \\ &= p_A \circ i_A \circ p_A + p_A \circ i_B \circ p_B \\ &= 1_A \circ p_A + 0_{B,A} \circ p_B \\ &= p_A + 0_{P,A} \\ &= p_A. \end{align*} Similarly, using $p_B \circ i_A = 0_{A,B}$ and $p_B \circ i_B = 1_B$, \begin{align*} p_B \circ e &= p_B \circ (i_A \circ p_A + i_B \circ p_B) \\ &= p_B \circ i_A \circ p_A + p_B \circ i_B \circ p_B \\ &= 0_{A,B} \circ p_A + 1_B \circ p_B \\ &= 0_{P,B} + p_B \\ &= p_B. \end{align*} [/step] [step:Compare with the identity map using the product universal property] The identity morphism $1_P:P \to P$ has projection composites \begin{align*} p_A \circ 1_P &= p_A, & p_B \circ 1_P &= p_B. \end{align*} The preceding step showed that $e:P \to P$ has the same two projection composites: \begin{align*} p_A \circ e &= p_A, & p_B \circ e &= p_B. \end{align*} By the product uniqueness property established above, applied to the two morphisms $e,1_P:P \to P$, we obtain \begin{align*} i_A \circ p_A + i_B \circ p_B = e = 1_P. \end{align*} [/step] [step:Assemble the biproduct axioms] By hypothesis, $(P,p_A,p_B)$ is a product of $A$ and $B$, and $(P,i_A,i_B)$ is a coproduct of $A$ and $B$. The four compatibility identities \begin{align*} p_A \circ i_A &= 1_A, & p_B \circ i_B &= 1_B, & p_A \circ i_B &= 0_{B,A}, & p_B \circ i_A &= 0_{A,B} \end{align*} are assumed, and the previous step proves the remaining identity \begin{align*} i_A \circ p_A + i_B \circ p_B = 1_P. \end{align*} Therefore the object $P$, with injections $i_A,i_B$ and projections $p_A,p_B$, satisfies the defining axioms of a binary biproduct of $A$ and $B$ in the preadditive category $\mathcal C$. [/step]