[proofplan] The proof uses only the defining identities of finite biproducts in an additive category. We first show that a prescribed family of component morphisms reconstructs a unique morphism $A\to B$ by summing injection-component-projection composites. Then we prove that every morphism is recovered from its components by inserting the biproduct resolutions of the identity on $A$ and $B$. Finally, the composition formula follows by inserting the identity decomposition of the middle biproduct $B$ between $g$ and $f$. [/proofplan] [step:Record the biproduct identities used throughout] For the biproduct $A=\bigoplus_{j=1}^{m}A_j$, the structure morphisms satisfy \begin{align*} \pi_j\iota_s &= \begin{cases} 1_{A_j}, & j=s,\\ 0_{A_s,A_j}, & j\neq s, \end{cases} & 1_A &= \sum_{j=1}^{m}\iota_j\pi_j. \end{align*} For the biproduct $B=\bigoplus_{i=1}^{n}B_i$, the structure morphisms satisfy \begin{align*} \rho_r\kappa_i &= \begin{cases} 1_{B_r}, & r=i,\\ 0_{B_i,B_r}, & r\neq i, \end{cases} & 1_B &= \sum_{i=1}^{n}\kappa_i\rho_i. \end{align*} Here $0_{X,Y}:X\to Y$ denotes the zero morphism in the abelian group $\mathcal C(X,Y)$. Since $\mathcal C$ is additive, composition is bilinear in the hom-groups, so finite sums may be composed termwise on either side. [guided] The only category-theoretic input is the finite biproduct identities. For $A=\bigoplus_{j=1}^{m}A_j$, the maps $\iota_j:A_j\to A$ are the injections and the maps $\pi_j:A\to A_j$ are the projections. Their defining relations are \begin{align*} \pi_j\iota_s &= \begin{cases} 1_{A_j}, & j=s,\\ 0_{A_s,A_j}, & j\neq s, \end{cases} & 1_A &= \sum_{j=1}^{m}\iota_j\pi_j. \end{align*} The first identity says that projecting the $s$-th summand onto the $j$-th summand is the identity if $j=s$ and the zero morphism otherwise. The second identity says that the identity map on the biproduct is the sum of the endomorphisms that project to one summand and reinclude it. For $B=\bigoplus_{i=1}^{n}B_i$, with injections $\kappa_i:B_i\to B$ and projections $\rho_i:B\to B_i$, the analogous identities are \begin{align*} \rho_r\kappa_i &= \begin{cases} 1_{B_r}, & r=i,\\ 0_{B_i,B_r}, & r\neq i, \end{cases} & 1_B &= \sum_{i=1}^{n}\kappa_i\rho_i. \end{align*} Because $\mathcal C$ is additive, every hom-set is an abelian group and composition distributes over addition. Thus expressions such as $\rho_r(\sum_i h_i)\iota_s=\sum_i\rho_r h_i\iota_s$ are valid. [/guided] [/step] [step:Reconstruct a morphism from a prescribed matrix of components] Let $(f_{ij})_{i,j}$ be a family of morphisms with $f_{ij}:A_j\to B_i$. Define the morphism \begin{align*} \widetilde f:A&\to B\\ A&\xrightarrow{\ \sum_{i=1}^{n}\sum_{j=1}^{m}\kappa_i f_{ij}\pi_j\ } B \end{align*} by \begin{align*} \widetilde f = \sum_{i=1}^{n}\sum_{j=1}^{m}\kappa_i f_{ij}\pi_j. \end{align*} For fixed indices $1\leq r\leq n$ and $1\leq s\leq m$, bilinearity of composition gives \begin{align*} \rho_r\widetilde f\iota_s &= \sum_{i=1}^{n}\sum_{j=1}^{m} \rho_r\kappa_i f_{ij}\pi_j\iota_s. \end{align*} By the biproduct identities, every summand is zero unless $i=r$ and $j=s$, and the remaining summand is \begin{align*} \rho_r\kappa_r f_{rs}\pi_s\iota_s = 1_{B_r} f_{rs} 1_{A_s} = f_{rs}. \end{align*} Therefore \begin{align*} \rho_r\widetilde f\iota_s=f_{rs} \end{align*} for all $r,s$. [guided] Suppose we are given the would-be matrix entries $f_{ij}:A_j\to B_i$. The only possible reconstruction formula is: project $A$ onto $A_j$, apply $f_{ij}$, include $B_i$ into $B$, and sum over all pairs of indices. Thus define \begin{align*} \widetilde f:A&\to B\\ A&\xrightarrow{\ \sum_{i=1}^{n}\sum_{j=1}^{m}\kappa_i f_{ij}\pi_j\ } B \end{align*} by \begin{align*} \widetilde f = \sum_{i=1}^{n}\sum_{j=1}^{m}\kappa_i f_{ij}\pi_j. \end{align*} This is a valid morphism $A\to B$: each composite $\kappa_i f_{ij}\pi_j$ has type \begin{align*} A\xrightarrow{\pi_j}A_j\xrightarrow{f_{ij}}B_i\xrightarrow{\kappa_i}B, \end{align*} and the hom-set $\mathcal C(A,B)$ is an abelian group, so the finite sum is defined. We now check that the reconstructed morphism has the prescribed components. Fix $r$ and $s$. Using bilinearity of composition in the additive category, \begin{align*} \rho_r\widetilde f\iota_s &= \rho_r \left( \sum_{i=1}^{n}\sum_{j=1}^{m}\kappa_i f_{ij}\pi_j \right) \iota_s\\ &= \sum_{i=1}^{n}\sum_{j=1}^{m} \rho_r\kappa_i f_{ij}\pi_j\iota_s. \end{align*} The biproduct identities now isolate exactly one summand. If $i\neq r$, then $\rho_r\kappa_i=0_{B_i,B_r}$, so the summand is zero. If $j\neq s$, then $\pi_j\iota_s=0_{A_s,A_j}$, so the summand is zero. The only nonzero contribution occurs when $i=r$ and $j=s$, giving \begin{align*} \rho_r\kappa_r f_{rs}\pi_s\iota_s = 1_{B_r} f_{rs} 1_{A_s} = f_{rs}. \end{align*} Hence $\rho_r\widetilde f\iota_s=f_{rs}$ for every pair $(r,s)$, exactly as required. [/guided] [/step] [step:Recover every morphism from its component matrix] Let $f:A\to B$ be a morphism. Define its component morphisms $f_{ij}:A_j\to B_i$ by \begin{align*} f_{ij}:=\rho_i f\iota_j. \end{align*} Using the identity decompositions of $A$ and $B$, we compute \begin{align*} f &= 1_B f 1_A\\ &= \left(\sum_{i=1}^{n}\kappa_i\rho_i\right) f \left(\sum_{j=1}^{m}\iota_j\pi_j\right)\\ &= \sum_{i=1}^{n}\sum_{j=1}^{m} \kappa_i\rho_i f\iota_j\pi_j\\ &= \sum_{i=1}^{n}\sum_{j=1}^{m} \kappa_i f_{ij}\pi_j. \end{align*} Thus applying the reconstruction formula to the component matrix of $f$ returns $f$ itself. Together with the previous step, this proves that the component map is a bijection with the stated inverse. [guided] Now begin with an actual morphism $f:A\to B$. Its matrix entry in row $i$ and column $j$ is defined by projecting the target to $B_i$ and restricting the source to $A_j$: \begin{align*} f_{ij}:=\rho_i f\iota_j:A_j\to B_i. \end{align*} We must prove that these entries determine $f$ uniquely. The biproduct identities give decompositions of the identity morphisms: \begin{align*} 1_A=\sum_{j=1}^{m}\iota_j\pi_j, \qquad 1_B=\sum_{i=1}^{n}\kappa_i\rho_i. \end{align*} Insert these identities on the right and left of $f$. Since composition is bilinear, the finite sums expand termwise: \begin{align*} f &= 1_B f 1_A\\ &= \left(\sum_{i=1}^{n}\kappa_i\rho_i\right) f \left(\sum_{j=1}^{m}\iota_j\pi_j\right)\\ &= \sum_{i=1}^{n}\sum_{j=1}^{m} \kappa_i\rho_i f\iota_j\pi_j\\ &= \sum_{i=1}^{n}\sum_{j=1}^{m} \kappa_i f_{ij}\pi_j. \end{align*} This is precisely the reconstruction formula applied to the component matrix $(f_{ij})_{i,j}$. Therefore no information is lost when passing from $f$ to its matrix of components. The previous step showed that every prescribed matrix is obtained from its reconstruction, so the two constructions are inverse to each other. [/guided] [/step] [step:Compute composition by inserting the middle biproduct identity] Let $C=\bigoplus_{\ell=1}^{r}C_\ell$ be a finite biproduct with injections $\lambda_\ell:C_\ell\to C$ and projections $\sigma_\ell:C\to C_\ell$. Let $f:A\to B$ and $g:B\to C$ be morphisms. Define \begin{align*} f_{ij}&:=\rho_i f\iota_j:A_j\to B_i,\\ g_{\ell i}&:=\sigma_\ell g\kappa_i:B_i\to C_\ell. \end{align*} For fixed $1\leq \ell\leq r$ and $1\leq j\leq m$, insert the identity decomposition of $B$ between $g$ and $f$: \begin{align*} \sigma_\ell g f\iota_j &= \sigma_\ell g 1_B f\iota_j\\ &= \sigma_\ell g \left(\sum_{i=1}^{n}\kappa_i\rho_i\right) f\iota_j\\ &= \sum_{i=1}^{n} \sigma_\ell g\kappa_i\rho_i f\iota_j\\ &= \sum_{i=1}^{n} g_{\ell i}f_{ij}. \end{align*} Thus the $(\ell,j)$-entry of the matrix of $gf$ is obtained by the usual row-by-column formula, with multiplication given by composition in $\mathcal C$ and addition given by the abelian group structure on hom-sets. This proves compatibility with matrix multiplication and completes the proof. [/step]