In an additive category show that we have products iff we have coproducts. Show the equivalence between the definition of (co)product in terms of representable functors and the usual one in terms of universal properties. An additive category is a special case of a category enriched in pointed sets. A remarkable property is that every two objects $X$ and $Y$ are connected by some morphism $f: X \rightarrow Y$. If we have a terminal $1$ and initial object $0$ then there is a morphism $1 \rightarrow 0$. Note that $hom(0,0)$ and $hom(1,1)$ are singletons. Thus the mere existence of $1 \rightarrow 0$ entails that $0\backsimeq 1$.
Let $\mathcal{C}$ be a triangulated category and $\mathcal{N}$ a null system and $Q: \mathcal{C} \rightarrow \mathcal{C}/\mathcal{N}$ the canonical functor. Suppose that $X \in \mathcal{N}$. Then $Q(X) \backsimeq 0$. To see this take $X \rightarrow 0$. By the triangulated category axioms this can be extended to a triangle $X \rightarrow 0 \rightarrow Z \rightarrow TX$. By the definition of null system we have that $Z \in \mathcal{N}$ since by definition also $0 \in \mathcal{N}$. Hence by definition we have that $X \rightarrow 0$ is in $S(\mathcal{N})$ so that it is an isomorphism in $\mathcal{C}/\mathcal{N}$.
