Right adjoint preserves products Given two additive categories $\mathcal{C}$ and $\mathcal{D}$ and a functor $F \colon \mathcal{C} \rightarrow \mathcal{D}$, then:
Definition: $F$ is additive if and only if $F$ turns product diagrams into product diagrams.
I saw an argument using the Yoneda Lemma that if I have an adjunction $F \dashv G$ then I have the following sequence of natural isomorphisms:
\begin{align*}
  \mathrm{Hom}(A, G(B \times C))
  &\simeq \mathrm{Hom}(F(A), B \times C) \\
  &\simeq \mathrm{Hom}(F(A), B) \times \mathrm{Hom}(F(A), C) \\
  &\simeq \mathrm{Hom}(A, G(B)) \times \mathrm{Hom}(A, G(C)) \\
  &\simeq \mathrm{Hom}(A, G(B) \times G(C)) \,.
\end{align*}
This implies that $G(B \times C) \simeq G(B) \times G(C)$.
But, here I'm just saying that the objects are isomorphic not that $G$ preserves product diagrams. What am I missing here?
 A: Given objects $B^{\prime}, C^{\prime}\in{\mathcal C}$, endowing an object $X\in{\mathcal C}$ with the structure of a product of $B^{\prime}$ and $C^{\prime}$ is the same as providing an isomorphism $\text{Hom}_{\mathcal C}(-,X)\cong\text{Hom}_{\mathcal C}(-,B^{\prime})\times\text{Hom}_{\mathcal C}(-,C^{\prime})$ of functors ${\mathcal C}^{\text{op}}\to\text{Set}$. In your situation, $B^{\prime} = G(B)$, $C^{\prime} = G(C)$ and $X = G(B\times C)$, and at the fourth stage of your chain of isomorphisms you are done.
Note that you do not have to assume that ${\mathcal C}$ has products for that.
A: I think you are missing that the product is determined only up to isomorphism, moreover if $P$ is a product of objects $B$ and $C$ (i.e. the following diagram is universal),
$$\begin{matrix}
P&\overset{\pi_B}\longrightarrow\  B \\
 & \searrow{\scriptstyle{\pi_C} } \\
& \ \ \ C
\end{matrix}$$
 and $f:P\to Q$ is an isomorphism, then composing both legs of the product diagram with $P$ by $f^{-1}$ we get again a product diagram, where $Q$ plays the role of the product object.
Put it in other words, assume that products in $\mathcal C$ are not yet 'prepared', then still the first 3 isomorphisms of hom sets make sense and are valid, so
$$\hom(A,G(P))\ \simeq\ \hom(A,G(B))\times\hom(A,G(C)) $$
which proves that $G(P)$ is a product object. If you put $A=G(P)$ and take element $1_{G(P)}$ from left side, check that the above isomorphisms will naturally evaluate $(G(\pi_B),G(\pi_C))$ as the corresponding element on the right side.
