# Identity making each $\operatorname{Hom}_\mathcal{A}(A,B)$ an abelian group in an additive category the Zero map?

In an additive category, we have a zero object and every Hom-set is given the structure of an abelian group such that composition distributes over addition. That is, given $f \in \operatorname{Hom}_\mathcal{A}(A,B) \quad g,g' \in \operatorname{Hom}_\mathcal{A}(B,C) \;$ we have: $$f(g+g') = fg + fg'$$

$\textbf{1.}$ Is the identity in each Hom-set the unique zero map? I can't find any axiom saying that it is and no reason as of yet that it should be.

$\textbf{2.}$ When defining an $exact$ functor (or left/right exact functor) we say that $F$ is an additive functor taking an exact sequence $$0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0$$
is taken to the exact sequence $$0 \rightarrow F(A) \xrightarrow{F(f)} F(B) \xrightarrow{F(g)} F(C) \rightarrow 0$$

Is this to say that $F(0)=0?$ I suppose that if the answer to question $\textbf{1.}$ is yes then since $F$ is additive then $F(0:0\rightarrow A) = 0:F(0) \rightarrow F(A)$ so that we can replace these $F(0)$s with $0$ anyway?

• Actually, I've just seen this question, but im not sure that it actually answers the question. math.stackexchange.com/questions/335824/… Is it claiming there that composition is an abelian group too? – SEWillB May 9 '18 at 9:58

To answer 1., first notice that in any group (not necessarily abelian) the only idempotent is the identity, i.e. $x^2 = x$ if and only if $x = 1_G$, or in additive notation, $x + x = x$ if and only if $x = 0_G$. Thus, we'd like to show that $0_{AB}+0_{AB} = 0_{AB}$ for the zero morphism $0_{AB}\colon A\to B$.
First consider what exactly $0_{AB}$ is. It is given as the unique morphism that factors through zero object, i.e. it is the composition $A\to 0\to B$. So, we can write $0_{AB} = 0_{0B}\circ 0_{A0}$ and therefore we have $$0_{AB}+0_{AB} = 0_{0B}\circ 0_{A0} + 0_{0B}\circ 0_{A0} = (0_{0B}+0_{0B})\circ 0_{A0} = 0_{0B}\circ 0_{A0} = 0_{AB}$$
where $0_{0B}+0_{0B}=0_{0B}$ because $\operatorname{Hom}(0,B)$ is trivial group by the definition of zero object.
For 2., notice that $X$ is zero object if and only if $0_X = \operatorname{id}_X$. One direction is obvious. To prove that $0_X = \operatorname{id}_X$ implies that $X$ is initial (and analogously it is terminal), consider $f,g\colon X\to A$ for any object $A$. By 1. we know that $$f = f\circ\operatorname{id}_X = f\circ 0_X = g\circ 0_X = g\circ \operatorname{id}_X = g,$$ and we are done. To conclude that $F0 \cong 0$, observe that additive $F$ preserves both zero and identity morphism.
• @SEWillB, just a comment on your last sentence. That argument doesn't work since not all objects of $F$'s codomain are necessarily of the form $FA$ or isomorphic to it. And even if so, that still doesn't mean that there are no other maps between $F0$ and $FA$ besides zero map. – Ennar May 9 '18 at 14:15