# If we have finite coproducts in Ab-enriched category, does it follow that the zero object exists?

I am trying to understand what exactly are the axioms for an additive category and I got a bit lost. If we are given a category that is Ab-enriched and admits finite coproducts, can we derive it that the initial object coincides with the terminal one?

If we had it that products/coproducts are in fact the same (biproducts), then there would be no question (as we would get the zero object as a biproduct for the empty diagram), but I managed to prove the existence of biproducts only using the zero object already.

Thank you.

I think that an initial object in an Ab-enriched category is also terminal, thus a zero-object. Indeed, an initial object $I$ will have $\mathrm{End}(I)=0$ as it is initial. Therefore, $\mathrm{id}_I=0\mathrm{id}_I$. Now, if $f:X\rightarrow I$ is any morphism, then $f=\mathrm{id}_I\circ f=(0\mathrm{id}_I)\circ f=0(\mathrm{id}_I\circ f)=0$, because composition is linear. Thus $f=0$ and there is at most one morphism from $X$ to $I$. There is at least one, because $\mathrm{Hom}(X,I)$ is an abelian group, so has at least one object. Finally, for any $X$, there is a unique morphism $X\rightarrow I$.
I don't have enough reputation to comment on Mariano's answer. I just want to say that such a category cannot be Ab-enriched if it as at least two objects as $\mathrm{Hom}(X,Y)$ cannot be empty, it is an abelian group.
• You are using the notation of a zero morphism (and its property $f\circ0=0\circ f=0$) - but is it possible to do so if a zero morphism is defined through a zero object, the existence of which we are trying to prove? – Polydarya Dec 9 '13 at 10:39
• I used the fact that, if $I$ is an initial object, then its identity is the zero element in $\mathrm{End}(I)$ (because there is no one else). Then, I just used the composition rule : $f=\mathrm{id}_I\circ f$, which is the definition of an identity, and the linearity of $\circ$ which is an assumption in Ab-enriched categories. – Roland Dec 9 '13 at 11:04
• @user113969 Thank you very much, I think I simply failed to understand that when zero morphism is defined as the zero of the corresponding abelian group, it still holds that $0 \circ f = 0$ - which is kind of obvious. I am sorry. :) – Polydarya Dec 9 '13 at 21:06
• @martin-brandenburg Thanks for writing the definition, I was using the one which would work if the category didn't have the Ab-enrichment but did have the zero object ($A \to 0 \to B$). Now when I have the zero object it's easy to see that the two definitions are equivalent. – Polydarya Dec 9 '13 at 21:09