Lemma 12.3.2. Let $\mathcal{A}$ be a preadditive category. Let $x$ be an object of $\mathcal{A}$. The following are equivalent

1. $x$ is an initial object,

2. $x$ is a final object, and

3. $\text{id}_ x = 0$ in $\mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(x, x)$.

Furthermore, if such an object $0$ exists, then a morphism $\alpha : x \to y$ factors through $0$ if and only if $\alpha = 0$.

Proof. First assume that $x$ is either (1) initial or (2) final. In both cases, it follows that $\mathop{\mathrm{Mor}}\nolimits (x,x)$ is a trivial abelian group containing $\text{id}_ x$, thus $\text{id}_ x = 0$ in $\mathop{\mathrm{Mor}}\nolimits (x, x)$, which shows that each of (1) and (2) implies (3).

Now assume that $\text{id}_ x = 0$ in $\mathop{\mathrm{Mor}}\nolimits (x,x)$. Let $y$ be an arbitrary object of $\mathcal{A}$ and let $f \in \mathop{\mathrm{Mor}}\nolimits (x ,y)$. Denote $C : \mathop{\mathrm{Mor}}\nolimits (x,x) \times \mathop{\mathrm{Mor}}\nolimits (x,y) \to \mathop{\mathrm{Mor}}\nolimits (x,y)$ the composition map. Then $f = C(0, f)$ and since $C$ is bilinear we have $C(0, f) = 0$. Thus $f = 0$. Hence $x$ is initial in $\mathcal{A}$. A similar argument for $f \in \mathop{\mathrm{Mor}}\nolimits (y, x)$ can be used to show that $x$ is also final. Thus (3) implies both (1) and (2). $\square$

There are also:

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).