Hom set in Additive Category is $0$ I know this is a really easy question but for some reason I'm having trouble with it.
If $M$ is an object in an additive category $\mathcal C$, and $\text{Hom}_{\mathcal C}(M,M) = 0$, then $M = 0$.
I know that this implies $\text{id}_M =0$ but I'm having trouble showing that $M$ satisfies the condition to be the zero object, or showing that the map from $0\rightarrow M$ is necessarily invertible.
We have that this composite $M \rightarrow 0 \rightarrow M$ is the $0$ map and the identity simultaneously but I'm stumped.
I think I'm thinking about it too much.
 A: Given an object $A$ there will be a morphism from $A$ to $M$, the
zero morphism. Composing an arbitrary morphism $f:A\to M$ with the
unique map from $M$ to $M$ you get $f$, as the unique map from $M$ to $M$
is the identity, and also the zero morphism, as it factors through the
zero object. There is therefore one morphism from $A$ to $M$: $M$ is a terminal object. Terminal objects are all isomorphic. So $M$ is isomorphic
to the zero object.
A: You always have that $0 \to M \to 0$ equals $\mathrm{id}_0$, and you just showed that $M\to 0 \to M$ equals $\mathrm{id}_M$. Hence $0\to M$ is an iso (with inverse $M \to 0$).
Remark: the additive structure is not really used here, this is rather a property of pointed categories.
A: More generally, let $\mathcal C$ be a category. Let $x, y ∈ \operatorname{Ob} \mathcal C$ be objects of $\mathcal C$ whose only endomorphisms are their respective identities $\mathrm{id}_x$, $\mathrm{id}_y$. Then
$$x \cong y \Longleftrightarrow \text{there are arrows $x → y$ and $y → x$}.$$
This is because whenever one can form paths $x → y → x$ and $y → x → y$, they both have to compose to the only endomorphisms of $x$ and $y$ – the identities (as already pointed out by others).
