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Does$$ \forall x \left(1 \stackrel{x}\longrightarrow X\right) \Rightarrow 1 \stackrel{x} \longrightarrow A $$ means $A \subset B $? Is there any better way to express this with arrows?

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In the category of sets, yes, provided you label morphisms $1 \to X$ by their values. Otherwise false in general. –  Zhen Lin Nov 15 '12 at 17:17
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Thank you. What if we replace the 1 with "all separator objects" of $\mathcal{C}$? –  Hooman Nov 15 '12 at 19:11
    
Please formulate exactly what you mean: as it stands your claim does not generalise. –  Zhen Lin Nov 15 '12 at 19:52
    
What about (perhaps regular/split) monomorphisms? For what purpose you need it? –  Berci Nov 17 '12 at 14:23
    
Sorry, now this questions sounds a bit pointless and confusing to me. I am now happy with the axioms for sub-object. –  Hooman Nov 17 '12 at 14:39

1 Answer 1

There is a category whose objects are sets and morphism are inclusions.

So if $A$, $B$ are sets then $A \longrightarrow B$ means we have an inclusion $A \subset B$. We have identity morphism since $A \subset A$ and you can compose $A \subset B$ with $B \subset C$ by transitivity to get $A \subset C$.

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That is a subcategory of Set, morphisms are inclusions. What is inclusion? Is it possible to express that with Slice category? –  Hooman Nov 15 '12 at 15:19
    
@Hooman, nice point about it being a subcat, I hadn't thought of that. I don't think we can get it as a slice or any other purely categorical way - since category theory generally only classifies things up to isomorphism, we'd have to actually look at the elements inside a set similar to how you were doing. –  sperners lemma Nov 15 '12 at 15:34

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