Talk:Filtered category

Meaning of definition
The third condition states something similar to the existence of coequalizers (without uniqueness). Ok, but what does the second condition mean? What's the point of two objects & arrows if no extra condition is imposed on them? That's just as if we state that there's an arrow between any two objects. Am I wrong? --Anton (talk) 02:22, 27 January 2009 (UTC)

-- Think about the case of posets. It is saying that for any two elements in a poset there is an element above them. 130.54.16.83 (talk) 09:51, 27 January 2009 (UTC)


 * Okay, but what changes if we require simply an arrow from any object to any other object? If a category has more than two objects, we can always choose an arbitrary extra one and get the same axiom. It works both ways. Am I wrong? --Anton (talk) 16:31, 28 January 2009 (UTC)


 * Anton, there are two examples to bear in mind. First, the category of U of open covers of a topological space X. So, an object of U is an open cover of X. A morphism A->B of covers is a refinement of A. That is, B is a refinement of A. Now, condition 2 of the definition holds in this case. Namely, we can have two covers A,B such that neither is a refinement of the other. So, there is no arrow between them. But, we can find a common refinement A->C, B->C. Second, consider the category whose objects are positive integers, and n->m if and only if m|n (m divides n). You can check that this category is filtered. —Preceding unsigned comment added by Benjamin Antieau (talk • contribs) 19:03, 5 February 2009 (UTC)


 * Oops, both of the examples I just gave are cofiltered categories. One usually uses the first in the context of presheaves of sets, and so the limits one thinks about are filtered in that case, since you move to the opposite category. --Benjamin Antieau (talk) 05:56, 10 February 2009 (UTC)


 * If for any two objects there is an arrow between then, then yes, the second condition is vacuously satisfied, but again, think about posets, not every two objects in a poset has an arrow between them. For example, the set of subsets of a set with inclusion as the arrows. 219.117.195.84 (talk) 11:43, 16 June 2009 (UTC)
 * Thank you, I see now. --Anton (talk) 14:58, 20 June 2009 (UTC)


 * I still insist that an intuitive justification for the third condition is desirable. In every preorder category, because it is thin, the third condition is satisfied, take $$k:=j, w:=id(j)$$, then $$u=v, w\circ u = w\circ v$$. The third condition becomes useful only if we go from preorders to categories, it is specific to categories. --Beroal (talk) 13:28, 20 September 2011 (UTC)


 * Also I am interested in examples of filtered categories which are not thin. --Beroal (talk) 13:43, 20 September 2011 (UTC)
 * I still don't understand why the third condition becomes useful, since it applies to $J$ given to be a category. Since you have two arrows between given $$i$$ and $$j$$, you may take $$k:=j$$, and choose $$w:=id(j)$$ as you said. And by definition of identity morphisms in a category, you still trivially have $$u=v \implies id(j)u = id(j)v \implies wu = wv$$ which strictly satisfies the condition as stated. 132.169.176.150 (talk) 09:43, 25 May 2022 (UTC)