0
$\begingroup$

Hi all:

I was reading a proof that localization of categories preserves additivity. In the proof the author uses a statement:

If $X$ is an object in a preadditive category $C$ and ${End}_C(X) = \{0_X\}$ then $X$ is the zero object in $C$.

I know that the statement is true when $C$ is an additive category, but I'm not quite sure if it's true for preadditive category.

To be precise, I traced back to the very definition of a zero object, that is, both initial and terminal. However, the condition in the statement only suggests the uniqueness of morphism $X\rightarrow Y$ for any object $Y\in \text{ob}C$, I don't see why such a morphism should exist.

Any suggestions or reference? Thanks.

Improve this question
$\endgroup$
1
  • $\begingroup$ For those who wonder what ends have to do with anything here, $End_C(X)$ means "endomorphisms on $X$". $\endgroup$
    – Andrej Bauer
    Commented Jan 23, 2013 at 9:16

1 Answer 1

Reset to default
6
$\begingroup$

In principle $\textrm{End}(X)$ could have two elements, $\textrm{id}_X$ and $0_X$. If it has only one element, then $\textrm{id}_X = 0_X$, so for all $f : X \to Y$, $f = f \circ \textrm{id}_X = f \circ 0_X = 0$. Similarly, for all $g : Z \to X$, $g = \textrm{id}_X \circ g = 0_X \circ g = 0$. Thus $X$ is both initial and terminal. This argument works in any category with a reasonable notion of zero morphism, not just preadditive categories.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Yes, if there's a morphism $X\rightarrow Y$ (or $Z\rightarrow X$), it must be unique. But I was wondering how do we know such a morphism exist? $\endgroup$
    – CJ L
    Commented Jan 23, 2013 at 9:03
  • 3
    $\begingroup$ Ah I just figured out. It exists since $\text{Hom}_C(X,Y)$ is an abelian group and hence non-empty. $\endgroup$
    – CJ L
    Commented Jan 23, 2013 at 9:11
  • 4
    $\begingroup$ This argument works for any category enriched over pointed sets (which is another way to say "has zero morphisms"). Note that this is a property of a category and not a structure. The upshot is that $0 = \text{id}$ is a purely "algebraic" condition and is therefore preserved by arbitrary enriched functors. See qchu.wordpress.com/2012/09/14/… for a discussion. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 23, 2013 at 9:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .