TITLE: Right adjoint preserves products
QUESTION [7 upvotes]: Given two additive categories $\mathcal{C}$ and $\mathcal{D}$ and a functor $F: \mathcal{C} \rightarrow \mathcal{D}$, then:
Definition: $F$ is additive iff $F$ 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 sequence of natural isomorphisms:
$\begin{array}{ccc} Hom(A,G(B \times C)) & \simeq & Hom(F(A),B\times C) \\ & \simeq & Hom(F(A),B)\times Hom(F(A),C) \\ & \simeq & Hom(A,G(B))\times Hom(A,G(C)) \\ &\simeq&  Hom(A,G(B)\times G(C))\end{array};$
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?

REPLY [2 votes]: 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.