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 $\mathop{Mor}\nolimits _\mathcal {A}(x, x)$.

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 $\mathop{Mor}\nolimits (x,x)$ is a trivial abelian group containing $\text{id}_ x$, thus $\text{id}_ x = 0$ in $\mathop{Mor}\nolimits (x, x)$, which shows that each of (1) and (2) implies (3).

Now assume that $\text{id}_ x = 0$ in $\mathop{Mor}\nolimits (x,x)$. Let $y$ be an arbitrary object of $\mathcal{A}$ and let $f \in \mathop{Mor}\nolimits (x ,y)$. Denote $C : \mathop{Mor}\nolimits (x,x) \times \mathop{Mor}\nolimits (x,y) \to \mathop{Mor}\nolimits (x,y)$ 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 $f \in \mathop{Mor}\nolimits (y, x)$ 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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00ZZ. Beware of the difference between the letter 'O' and the digit '0'.