[proofplan] We prove the two universal properties directly from the biproduct identities. For the product property, a pair of morphisms $f:X \to A$ and $g:X \to B$ determines the candidate mediating morphism $i_A \circ f + i_B \circ g:X \to P$, and the projection equations verify existence. Uniqueness follows by decomposing any morphism $h:X \to P$ through the identity $\operatorname{id}_P=i_A \circ p_A+i_B \circ p_B$. The coproduct argument is the dual computation, using $r \circ p_A+s \circ p_B:P \to Y$ as the candidate mediating morphism. [/proofplan] [step:Construct the unique morphism into the biproduct with prescribed projections] Let $X \in \operatorname{Ob}(\mathcal C)$, and let \begin{align*} f &: X \to A, & g &: X \to B \end{align*} be morphisms in $\mathcal C$. Since $\mathcal C$ is preadditive, $\operatorname{Hom}_{\mathcal C}(X,P)$ is an abelian group and composition is bilinear. Define the morphism \begin{align*} u &: X \to P, & u &:= i_A \circ f + i_B \circ g. \end{align*} We verify that $u$ has the required projections. Using bilinearity of composition and the biproduct identities, \begin{align*} p_A \circ u &= p_A \circ (i_A \circ f + i_B \circ g) \\ &= (p_A \circ i_A) \circ f + (p_A \circ i_B) \circ g \\ &= \operatorname{id}_A \circ f + 0_{B,A} \circ g \\ &= f. \end{align*} Similarly, \begin{align*} p_B \circ u &= p_B \circ (i_A \circ f + i_B \circ g) \\ &= (p_B \circ i_A) \circ f + (p_B \circ i_B) \circ g \\ &= 0_{A,B} \circ f + \operatorname{id}_B \circ g \\ &= g. \end{align*} Thus $u:X \to P$ is a morphism with $p_A \circ u=f$ and $p_B \circ u=g$. Now let $h:X \to P$ be any morphism satisfying \begin{align*} p_A \circ h &= f, & p_B \circ h &= g. \end{align*} Using the biproduct identity $\operatorname{id}_P=i_A \circ p_A+i_B \circ p_B$ and bilinearity of composition, \begin{align*} h &= \operatorname{id}_P \circ h \\ &= (i_A \circ p_A+i_B \circ p_B)\circ h \\ &= i_A \circ (p_A \circ h)+i_B \circ (p_B \circ h) \\ &= i_A \circ f+i_B \circ g \\ &= u. \end{align*} Therefore $u$ is the unique morphism $X \to P$ with projections $f$ and $g$. Since $X$, $f$, and $g$ were arbitrary, $(P,p_A,p_B)$ is a product of $A$ and $B$. [/step] [step:Construct the unique morphism out of the biproduct with prescribed coprojections] Let $Y \in \operatorname{Ob}(\mathcal C)$, and let \begin{align*} r &: A \to Y, & s &: B \to Y \end{align*} be morphisms in $\mathcal C$. Since $\operatorname{Hom}_{\mathcal C}(P,Y)$ is an abelian group and composition is bilinear, define the morphism \begin{align*} v &: P \to Y, & v &:= r \circ p_A+s \circ p_B. \end{align*} We verify that $v$ has the required coprojection restrictions. Using bilinearity of composition and the biproduct identities, \begin{align*} v \circ i_A &= (r \circ p_A+s \circ p_B)\circ i_A \\ &= r \circ (p_A \circ i_A)+s \circ (p_B \circ i_A) \\ &= r \circ \operatorname{id}_A+s \circ 0_{A,B} \\ &= r. \end{align*} Similarly, \begin{align*} v \circ i_B &= (r \circ p_A+s \circ p_B)\circ i_B \\ &= r \circ (p_A \circ i_B)+s \circ (p_B \circ i_B) \\ &= r \circ 0_{B,A}+s \circ \operatorname{id}_B \\ &= s. \end{align*} Thus $v:P \to Y$ is a morphism with $v \circ i_A=r$ and $v \circ i_B=s$. Now let $k:P \to Y$ be any morphism satisfying \begin{align*} k \circ i_A &= r, & k \circ i_B &= s. \end{align*} Using the biproduct identity $\operatorname{id}_P=i_A \circ p_A+i_B \circ p_B$ and bilinearity of composition, \begin{align*} k &= k \circ \operatorname{id}_P \\ &= k \circ (i_A \circ p_A+i_B \circ p_B) \\ &= (k \circ i_A)\circ p_A+(k \circ i_B)\circ p_B \\ &= r \circ p_A+s \circ p_B \\ &= v. \end{align*} Therefore $v$ is the unique morphism $P \to Y$ with $v \circ i_A=r$ and $v \circ i_B=s$. Since $Y$, $r$, and $s$ were arbitrary, $(P,i_A,i_B)$ is a coproduct of $A$ and $B$. [/step]