[proofplan] We prove the two assertions directly from the definition of a preadditive category. First we check that postcomposition by a morphism gives a homomorphism of abelian groups and that associativity and identities make this assignment functorial. Then we verify additivity on morphisms using bilinearity of composition. The contravariant case is the same argument with precomposition in place of postcomposition, with the direction encoded by passing to $\mathcal C^{\mathrm{op}}$. [/proofplan] [step:Define the covariant Hom assignment and verify functoriality] Fix an object $A \in \operatorname{Ob}(\mathcal C)$. Define a map on objects \begin{align*} F_A: \operatorname{Ob}(\mathcal C) &\to \operatorname{Ob}(\mathrm{Ab}) \\ B &\mapsto \mathcal C(A,B). \end{align*} This is well-defined because $\mathcal C$ is preadditive, so every Hom set $\mathcal C(A,B)$ is an abelian group. For a morphism $f \in \mathcal C(B,C)$, define \begin{align*} F_A(f): \mathcal C(A,B) &\to \mathcal C(A,C) \\ h &\mapsto f \circ h. \end{align*} This map is a homomorphism of abelian groups: for $h_1,h_2 \in \mathcal C(A,B)$, bilinearity of composition in the second variable gives \begin{align*} F_A(f)(h_1+h_2) &= f \circ (h_1+h_2) \\ &= f \circ h_1 + f \circ h_2 \\ &= F_A(f)(h_1)+F_A(f)(h_2). \end{align*} Also $F_A(f)(0_{\mathcal C(A,B)})=0_{\mathcal C(A,C)}$ by the same bilinearity. It remains to check functoriality. For each object $B$, the identity morphism $1_B \in \mathcal C(B,B)$ satisfies \begin{align*} F_A(1_B)(h)=1_B\circ h=h \end{align*} for every $h \in \mathcal C(A,B)$, so $F_A(1_B)=1_{\mathcal C(A,B)}$. If $f \in \mathcal C(B,C)$ and $g \in \mathcal C(C,D)$, then for every $h \in \mathcal C(A,B)$, associativity of composition gives \begin{align*} F_A(g\circ f)(h) &= (g\circ f)\circ h \\ &= g\circ(f\circ h) \\ &= F_A(g)(F_A(f)(h)). \end{align*} Hence $F_A(g\circ f)=F_A(g)\circ F_A(f)$, and $F_A=\mathcal C(A,-):\mathcal C\to \mathrm{Ab}$ is a functor. [/step] [step:Show the covariant Hom functor preserves addition on morphisms] Let $B,C \in \operatorname{Ob}(\mathcal C)$. We prove that the morphism map \begin{align*} \mathcal C(B,C) &\to \operatorname{Hom}_{\mathrm{Ab}}(\mathcal C(A,B),\mathcal C(A,C)) \\ f &\mapsto F_A(f) \end{align*} is a group homomorphism. Let $f,g \in \mathcal C(B,C)$. For every $h \in \mathcal C(A,B)$, bilinearity of composition in the first variable gives \begin{align*} F_A(f+g)(h) &= (f+g)\circ h \\ &= f\circ h + g\circ h \\ &= F_A(f)(h)+F_A(g)(h). \end{align*} The addition on $\operatorname{Hom}_{\mathrm{Ab}}(\mathcal C(A,B),\mathcal C(A,C))$ is pointwise addition, so this equality for all $h$ gives \begin{align*} F_A(f+g)=F_A(f)+F_A(g). \end{align*} Likewise, for every $h \in \mathcal C(A,B)$, \begin{align*} F_A(0_{\mathcal C(B,C)})(h) &= 0_{\mathcal C(B,C)}\circ h \\ &= 0_{\mathcal C(A,C)}, \end{align*} so $F_A(0_{\mathcal C(B,C)})=0_{\operatorname{Hom}_{\mathrm{Ab}}(\mathcal C(A,B),\mathcal C(A,C))}$. Therefore $\mathcal C(A,-)$ is additive. [/step] [step:Define the contravariant Hom assignment and verify functoriality] Fix again $A \in \operatorname{Ob}(\mathcal C)$. Define a map on objects \begin{align*} G_A: \operatorname{Ob}(\mathcal C^{\mathrm{op}}) &\to \operatorname{Ob}(\mathrm{Ab}) \\ B &\mapsto \mathcal C(B,A). \end{align*} For a morphism $f^{\mathrm{op}}: C \to B$ in $\mathcal C^{\mathrm{op}}$, corresponding to a morphism $f:B\to C$ in $\mathcal C$, define \begin{align*} G_A(f^{\mathrm{op}}): \mathcal C(C,A) &\to \mathcal C(B,A) \\ k &\mapsto k\circ f. \end{align*} This map is a homomorphism of abelian groups: for $k_1,k_2 \in \mathcal C(C,A)$, bilinearity of composition in the first variable gives \begin{align*} G_A(f^{\mathrm{op}})(k_1+k_2) &= (k_1+k_2)\circ f \\ &= k_1\circ f + k_2\circ f \\ &= G_A(f^{\mathrm{op}})(k_1)+G_A(f^{\mathrm{op}})(k_2). \end{align*} For each object $B$, the identity morphism $1_B^{\mathrm{op}}$ in $\mathcal C^{\mathrm{op}}$ corresponds to $1_B$ in $\mathcal C$, and \begin{align*} G_A(1_B^{\mathrm{op}})(k)=k\circ 1_B=k \end{align*} for every $k \in \mathcal C(B,A)$. Thus $G_A(1_B^{\mathrm{op}})=1_{\mathcal C(B,A)}$. Now let $f^{\mathrm{op}}: C\to B$ and $g^{\mathrm{op}}: D\to C$ be morphisms in $\mathcal C^{\mathrm{op}}$, corresponding to $f:B\to C$ and $g:C\to D$ in $\mathcal C$. Their composite in $\mathcal C^{\mathrm{op}}$ is $(g\circ f)^{\mathrm{op}}: D\to B$. For every $k \in \mathcal C(D,A)$, associativity of composition in $\mathcal C$ gives \begin{align*} G_A((g\circ f)^{\mathrm{op}})(k) &= k\circ(g\circ f) \\ &= (k\circ g)\circ f \\ &= G_A(f^{\mathrm{op}})(G_A(g^{\mathrm{op}})(k)). \end{align*} Hence $G_A((g\circ f)^{\mathrm{op}})=G_A(f^{\mathrm{op}})\circ G_A(g^{\mathrm{op}})$, which is exactly functoriality for $G_A:\mathcal C^{\mathrm{op}}\to\mathrm{Ab}$. [/step] [step:Show the contravariant Hom functor preserves addition on morphisms] Let $B,C \in \operatorname{Ob}(\mathcal C)$. We prove that the map \begin{align*} \mathcal C(B,C) &\to \operatorname{Hom}_{\mathrm{Ab}}(\mathcal C(C,A),\mathcal C(B,A)) \\ f &\mapsto \bigl(k\mapsto k\circ f\bigr) \end{align*} is a group homomorphism. Let $f,g \in \mathcal C(B,C)$. For every $k \in \mathcal C(C,A)$, bilinearity of composition in the second variable gives \begin{align*} G_A(f^{\mathrm{op}}+g^{\mathrm{op}})(k) &= k\circ(f+g) \\ &= k\circ f + k\circ g \\ &= G_A(f^{\mathrm{op}})(k)+G_A(g^{\mathrm{op}})(k). \end{align*} Therefore \begin{align*} G_A(f^{\mathrm{op}}+g^{\mathrm{op}}) = G_A(f^{\mathrm{op}})+G_A(g^{\mathrm{op}}). \end{align*} Similarly, for every $k \in \mathcal C(C,A)$, \begin{align*} G_A(0_{\mathcal C(B,C)}^{\mathrm{op}})(k) &= k\circ 0_{\mathcal C(B,C)} \\ &= 0_{\mathcal C(B,A)}, \end{align*} so zero morphisms are preserved. Thus $\mathcal C(-,A):\mathcal C^{\mathrm{op}}\to\mathrm{Ab}$ is additive. [/step] [step:Conclude additivity in both variables] The first two steps prove that for each object $A$, postcomposition defines an additive functor $\mathcal C(A,-):\mathcal C\to\mathrm{Ab}$. The last two steps prove that for each object $A$, precomposition defines an additive functor $\mathcal C(-,A):\mathcal C^{\mathrm{op}}\to\mathrm{Ab}$. Hence the Hom construction is additive in each variable separately. [/step]