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Dec
5
comment Why “thin groupoids” are not ubiquitous?
Setoid is a set with an equivalence relation on it. (And I know it long ago before I've read your answer.) I understand this. I downvote because switching from a groupoid to a setoid leads to information loss, and this makes it not an answer to my question.
Dec
5
comment Why “thin groupoids” are not ubiquitous?
I need to describe a category. A category contains not only objects but also morphisms. As my category happens to be thin, there is a (not necessarily entire defined) function from pairs of objects into a morphism. Having a setoid we cannot define such a function. But I need this function. I need to be able to get the morphisms whenever a pair of objects is specified. Having only a setoid, I cannot do this.
Dec
5
comment Why “thin groupoids” are not ubiquitous?
But having two elements of setoids, I cannot restore particular morphism (such as $f\mapsto f\cap\Gamma$). It is not what I need
Dec
5
comment Why “thin groupoids” are not ubiquitous?
I yet don't understand you and don't see how to express this with setoids. When we switch from a thin groupoid to a setoid, the information about particular morphisms is lost (they are just replaced with a pair of objects), but the whole thing I need is information about what are particular morphisms, depicted as arrows in my diagrams.
Dec
5
comment Why “thin groupoids” are not ubiquitous?
I don't get how this is related with setoids. How to express my diagrams using setoids? I need particular isomorphisms not just the fact that two objects are isomorphic.
Dec
5
revised Why “thin groupoids” are not ubiquitous?
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Dec
5
asked Why “thin groupoids” are not ubiquitous?
Dec
2
comment Another way to express certain filter
See also counter-examples in this thread: groups.google.com/forum/#!topic/sci.math/Plru0S8ePzs
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answered Another way to express certain filter
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comment Another way to express certain filter
The other direction is surprisingly difficult
Dec
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comment Another way to express certain filter
In one direction: Let $X \in [\mathcal{A}]_{\mathfrak{B}}$. Then $\exists Y' \in \mathcal{A} : X \geq Y'$. Thus if $Y \geq X$ for $Y \in \mathfrak{A}$ then $Y \geq Y'$ and thus $Y \in \mathcal{A}$. So $\forall Y \in \mathfrak{A}: (Y \geq X \Rightarrow Y \in \mathcal{A})$.
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