Category where morphism sets are Abelian groups

Let $$\mathcal{C}$$ be a category, and suppose for all objects $$X, Y$$ of $$\mathcal{C}$$, Mor$$_{\mathcal{C}}(X,Y)$$ is equipped with the structure of an Abelian group, such that the composition of morphisms is bilinear.

Now it can easily be shown that Mor$$_{\mathcal{C}}(X,X) =:$$ End$$_{\mathcal{C}}(X)$$ is a ring (under composition as multiplication). (In the absence of any knowledge about the Abelian group structure, distributivity still arises from bilinearity in this case.)

The next thing to be proved is that $$X$$ is a "zero object" if and only if End$$_{\mathcal{C}}(X)$$ is the zero ring. Now I see that this "zero object" takes the form of a trivial group in group theory, zero ring in ring theory, the space $$\{0\}$$ as a vector space, etc., but I haven't come across a clear general form of this concept, which makes it hard to prove the statement for a general object $$X$$ of this rather general category $$\mathcal{C}$$.

• What do you mean by a "clear general form", exactly? At least as it was introduced to me, a "zero object" was simply an object which is both a terminal and initial object in a category. That meaning, to each object, it has a unique morphism, and each object has a unique morphism to it, simultaneously. I feel like you probably know this since you seem to understand that, e.g., the trivial group is a zero object in the category of groups (though the zero ring is only terminal in the category of rings - the integers are initial there), and this notion is only a step away. [cont] Commented Mar 20, 2019 at 21:58
• And this notion is basically one step away from the notion of terminal/initial, so I feel like clarification could be helpful. Commented Mar 20, 2019 at 21:58

A zero object $$X$$ is defined to be both a terminal and an initial object, i.e. there is exactly one arrow $$A\to X$$ and exactly one arrow $$X\to A$$.
Since each $$\hom(A,B)$$ is an Abelian group, it always has a zero element, call it $$0_{AB}$$.
Note that bilinearity implies $$0_{AB}\circ f=0_{XB}$$ for any $$f:X\to A$$.
Now, if $$X$$ is a zero object, also $$\hom(X,X)$$ has only one element, hence $$1_X=0_{XX}$$, so it must be the zero ring.
Conversely, if $$\hom(X,X)$$ is the zero ring, we have $$1_X=0_{XX}$$ and thus, for any $$f:A\to X$$, we have $$f=1_X\circ f=0_{XX}\circ f=0_{AX}$$ So there's a unique arrow $$A\to X$$.
And similarly, $$0_{XA}$$ is the unique arrow $$X\to A$$.
• $A$ is an arbitrary object? Commented Mar 20, 2019 at 22:38
• Yup. Specifically this needs to hold for all objects $A$ of the category. Commented Mar 20, 2019 at 22:54