Let $\mathcal{C}$ and $\mathcal{D}$ two additive categories. Supose $T: \mathcal{C} \rightarrow \mathcal{D}$ a functor that preserves finite product, i.e. , for all $A,B$ objects $T(A \times B) \simeq T(A) \times T(B)$. I want to prove that $T(0) = 0$ for the zero morphism. I know that such a condition imply that $T$ is additive and the result is obvious, but i want a direct proof because i'm trying to use this to show that if i have a functor that preserves product then it preserves coproduct in this conditions. Make sense this question?
Thanks.
If $T$ preserves finite products, it should also preserve the empty product (the empty set is finite!) i.e. the terminal (here: zero) object, and the result follows. But you want that only binary products are preserved, right?
Let $0 \in C$ be a zero object. We want to prove that $T(0)$ is a zero object (this suffices). The natural morphism $T(0 \times 0) \to T(0) \times T(0)$ is an isomorphism (by assumption). There is a unique isomorphism $0 \times 0 \cong 0$. Hence, the natural diagonal morphism $T(0) \to T(0) \times T(0)$ is an isomorphism.
Lemma. Let $C$ be an additive category and let $A \in C$ be an object such that the diagonal morphism $A \to A \times A$ is an isomorphism. Then, $A$ is a zero object.
Proof. If $B$ is any object, then any two morphisms $f : B \to A$, $g : B \to A$ are equal, since $(f,g) : B \to A \times A$ factors through the diagonal. Applying this to $\mathrm{id}_A$ and the zero morphism $0 : A \to A$, we see that $A=0$.