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Let $\mathcal{C}$ be a preadditive category, i.e. each hom-set admits an abelian group structure and composition with respect to these structures is bilinear. There is the well-known fact that if finite products exist, then so fo finite coproducts and they coincide. I am having some trouble proving the zero-ary case:

If $\mathcal{C}$ has an initial object $\varnothing$, then $\varnothing$ is terminal.

We have to show that for every object $X \in \mathcal{C}$ there exists a unique arrow $X \to \varnothing$. Existence is easy. Since each hom-set is a group, we simply let $0 : X \to \varnothing$ be the zero element of the group $\mathcal{C}(X,\varnothing)$. However, uniqueness is bothering me. I think there is not much choice: We suppose there exists another morphism $f : X \to \varnothing$. Since $\varnothing$ is initial, we find $h : \varnothing \to X$. Then since $\mathcal{C}(\varnothing,\varnothing)$ has only one element, we get that $$0 \circ h = f \circ h = \operatorname{id}_\varnothing$$ But then I am stucked...any hint or alternative way would be nice (I am aware of the proof found here, but somehow I would like to keep my phrasing).

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You only need the existence of zero morphisms for this. A category is said to have zero morphisms if between every pair of objects there are morphisms $0_{cd} : c \to d$ called zero morphisms which are absorbing in the sense that

  • if $f : d \to e$ is any morphism, then $f \circ 0_{cd} = 0_{ce}$, and
  • if $g : b \to c$ is any morphism, then $0_{cd} \circ g = 0_{bd}$.

This is implied by the existence of a zero object but does not require it, and is satisfied by any preadditive category. Zero morphisms are unique when they exist in the sense that if $0_{cd}$ is a collection of morphisms satisfying the above definition and $0_{cd}'$ is another collection of morphisms satisfying the above definition then $0_{cd} = 0_{cd}'$ (easy to verify by composing on either side by a zero endomorphism). Abstractly, a category with zero morphisms is a category enriched over pointed sets.

Now, suppose $0$ is an initial object in a category with zero morphisms. Then there is a unique morphism $0 \to 0$, which must be both the zero morphism and the identity. Hence if $f : X \to 0$ is any morphism (there is at least one, namely the zero morphism), we have

$$\text{id}_0 \circ f = f = 0 \circ f = 0$$

and so $f = 0$ is unique. So $0$ is terminal. Note that this argument only uses the fact that $\text{End}(0)$ has one element, so proves a stronger statement: in a category with zero morphisms, any object with only one endomorphism (equivalently, whose identity endomorphism is zero) is a zero object.

A nice follow-up exercise is to show that if a functor between two categories with zero morphisms preserves zero morphisms, then it preserves zero objects. Hence zero objects are absolute colimits for categories enriched over pointed sets.

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  • $\begingroup$ But I do not yet know that a zero object exists to get a zero morphism. This is exactly the thing I want to prove: If I have an initial object, then I do have a terminal object, hence a zero object. $\endgroup$ – TheGeekGreek Jan 13 '18 at 20:51
  • $\begingroup$ I'm calling it $0$ but I'm only using the fact that it's an initial object. $\endgroup$ – Qiaochu Yuan Jan 13 '18 at 20:52
  • $\begingroup$ Sorry...I am a bit confused. Where exactly do you use the fact that $0$ is initial? The problem I think is, that for defining a zero morphism we need a zero object... $\endgroup$ – TheGeekGreek Jan 13 '18 at 21:00
  • $\begingroup$ @TheGeekGreek: no you don't. A category has zero morphisms when it's enriched over pointed sets; this means that between every pair of objects there is a morphism called $0$ which is absorbing in that any composition of $0$ with another morphism produces another $0$. In particular, any preadditive category has zero morphisms given by the identity elements of each abelian group. The only place I use that $0$ is initial is to conclude that there is a unique morphism $0 \to 0$, which must be both the zero morphism and the identity. So in fact the conclusion is stronger: ... $\endgroup$ – Qiaochu Yuan Jan 13 '18 at 21:06
  • $\begingroup$ ...in a category with zero morphisms, any object with only one endomorphism is a zero object. I'll rewrite the argument with more details. $\endgroup$ – Qiaochu Yuan Jan 13 '18 at 21:06

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