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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.2012-11-15
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    Thank you. What if we replace the 1 with "all separator objects" of $\mathcal{C}$?2012-11-15
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    Please formulate exactly what you mean: as it stands your claim does not generalise.2012-11-15
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    What about (perhaps regular/split) *monomorphisms*? For what purpose you need it?2012-11-17
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    Sorry, now this questions sounds a bit pointless and confusing to me. I am now happy with the axioms for sub-object.2012-11-17

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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?2012-11-15
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    @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.2012-11-15