Let $F:\mathcal{A}\longrightarrow\mathcal{B}$ be a functor, assume that both $\mathcal{A}$ and $\mathcal{B}$ are complete and cocomplete. I want to prove that the following two conditions are equivalent:
1) $F$ preserves binary products
2) $F$ preserves binary coproducts
My book says that this comes applying duality. But I would like to understand it in practice. Here are my thoughts:
Let's denote with $F^{\operatorname{op}}:\mathcal{A}^{\operatorname{op}}\longrightarrow\mathcal{B}^{\operatorname{op}}$ the functor defined by $$F^{\operatorname{op}}(A):=F(A)$$ $$F^{\operatorname{op}}(f):=F(f)$$
Assume that $F$ preserves binary products and try to prove that it preserves binary coproducts. Then take a binary coproduct in $\mathcal{A}$. It is a product in $\mathcal{A}^{\operatorname{op}}$ and since $F^{\operatorname{op}}$ is the same as $F$ and $F$ preserves products, it is mapped to a product in $\mathcal{B}^{\operatorname{op}}$, that is into a coproduct of $\mathcal{B}$.
Do you think I am correct? Are there smoother ways to proceed?
Your statement is false in general. For, take the forgetful functor $$U\colon\mathbf{Ab}\longrightarrow \mathbf{Set},$$ from the category of abelian groups to the category of sets. Note that both $\mathbf{Ab}$ and $\mathbf{Set}$ are complete and cocomplete, so the same is true of their opposite categories. Now, since the underlying set of the product of two abelian groups $A$, $B$ is the cartesian product of the underlying sets of $A$ and $B$, $U$ preserves (binary) products. However, the coproduct of $A$ and $B$ is their direct sum $A\oplus B$ (which is also their product), whose underlying set is not the coproduct of the underlying sets of $A$ and $B$, so $U$ does not preserve (binary) coproducts.
Note: the statement becomes true if you restrict yourself to additive categories and additive functors between them.