4
$\begingroup$

Let $S$ be an object of category $C$. We say $S$ is a separator object of $C$ if whenever $Y \stackrel{f_1} \longleftarrow X \stackrel{f_2}{\longrightarrow} Y $ $(\forall x[S\stackrel{x}\longrightarrow X \Rightarrow f_1x=f_2x]) \Rightarrow f_1 =f_2$

This reminds of the definition of an epimorphism. I am wondering if we could define separator object as an object which all of its morphisms are epic.

  • 2
    The point of this definition is that $\text{Hom}(S, -)$ is a faithful functor.2012-07-16

1 Answers 1

4

The definition of a separator object appears to enshrine "collective" right-cancellation, but not the individual right-cancellation that is mandated by epicness. IOW, an object is a separator object when $f_1x=f_2x$ for specific $f_1,f_2$ can be right-cancelled if it holds for all $x$, whereas a specific $x$ is epic if $f_1x=f_2x$ can be right-cancelled for any $f_1,f_2$. These do not mean the same thing (which arrows are being universally quantified vs. individually specified is distinct) and aren't necessarily compatible, so we can't define a separator object as one with all outbound morphisms epic.

Consider the following situation: given $S$, there exist arrows from $S$ to $X$, $S\xrightarrow{x}X$ and $S\xrightarrow{y}X$ , and arrows from $X$ to $Y$, $X\xrightarrow{f}Y$ and $X\xrightarrow{g}Y$, such that $fx=gx$ but $fy\ne gy$ and $f\ne g$. This is not in contradiction with the definition of a separator object (so $S$ could still be a separator), while it does preclude $x$ from being an epimorphism. We can even devise a category with precisely these objects and arrows (plus $1_S,1_X,1_Y$), and $S$ will be a separator because the definition is vacuously fulfilled.

On the other hand, if an object $S$ has the property that all outbound morphisms are epic, then it will also be a separator object, so they are related.

  • 2
    I probably should have made this explicit... Let $G$ be a generator and let $X$ be an arbitrary object. Assuming that arbitrary coproducts exist, Exercise 0 about generators is that the natural map $\coprod_{f \in \operatorname{Hom}(G,X)} G \to X$ coming from the universal property of the coproduct is an epimorphism.2012-07-15