# Zero object in additive category

I am new in category theory and I got stuck with this following problem - An additive category $$\mathcal{C}$$ is a category with $$\mathcal{C}(A, B)$$ an abelian group for every pair $$A, B$$ and composition of morphisms being a group homomorphism, an object $$0$$ such that $$\mathcal{C}(0,0)=\{*\}$$ the trivial group with one element also denoted $$\mathcal{C}(0,0)=0$$, and finite sums and products always exist.

Given this information, how to show the object $$0$$ is an initial and terminal object? That is, there are unique maps $$0\to A$$ and $$A\to 0$$ for every $$A\in Ob \mathcal{C}$$.

I can see there must be at least one morphism $$0\to A$$ and $$A\to 0$$ namely one can choose the identity element of the groups $$\mathcal{C}(0, A)$$ and $$\mathcal{C}(A, 0)$$, but then I am unable to show how that is the only possible morphism. I suppose we have to look at the group homomorphism $$\mathcal{C}(0, A)\times \mathcal{C}(A, 0)\to \mathcal{C}(0,0)$$ but not sure how to use that fact. Thanks in advance.

• Use bilinearity. The point is that in an additive category there are two canonical endomorphisms of every object $c$, the identity $\text{id}_c$ and zero $0_c$, and the zero object is the unique object such that these two agree. Now multiply any morphism into or out of the zero object by this unique morphism. Commented Feb 14, 2023 at 8:37
• I get the intuition for what you're saying, but what is $0_c$ in general? It makes sense when it's the unique morphism corresponding to the zero object, but here by definition this object isn't initial or terminal, but only some object which has no nontrivial endomorphisms. With this definition if $f:c\to 0$ some morphism then composing with the $0$ map only gives me back $f$. What am I missing here? Commented Feb 14, 2023 at 9:02
• $0_c$ is the additive identity of $\text{Hom}(c, c)$; more generally there is a zero morphism $0_{c, d} : c \to d$ which is the additive identity of $\text{Hom}(c, d)$. By bilinearity it has the property that any composition involving a zero morphism is another zero morphism. Now apply this fact here. Commented Feb 14, 2023 at 9:19

As Qiaochu Yuan says there, and also as you state in your question, there is an identity element for each abelian group $$C(A,B)$$, and in particular for $$C(A,0)$$ and $$C(0,A)$$. The usual notation for the operation between morphisms is "+", and therefore we will call these elements $$0_{AB}$$ ($$0_A$$ in the case of $$C(A,A)$$). You also note that, for the zero object $$0$$, the group $$C(0,0)$$ is trivial and its only element is $$0_0$$. This element must be equal to $$Id_0$$, the neutral element for the composition that exists for every object in a category.
Now, we want to show that $$C(A,0)$$ has exactly one element. Let us take therefore $$f:A\rightarrow 0$$. On the one hand, you already noticed that
$$Id_0\circ f=f.$$
But on the other hand, since $$Id_0=0_0$$, and applying first that $$0_0$$ is neutral for addition, and second distribution of composition over addition, we have
$$f=Id_0\circ f=0_0\circ f=(0_0+0_0)\circ f= 0_0\circ f+0_0\circ f= Id_0\circ f +Id_0\circ f= f+f.$$
Therefore $$f=0_{A0}$$, and $$C(A,0)$$ has exactly one element for every object $$A$$ in the category. This means that $$0$$ is terminal. Similarly, one can prove that it is also initial.