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Jul 2, 2015 at 2:05 answer added Noam D. Elkies timeline score: 3
Jul 2, 2015 at 1:59 comment added Kimball More generally than Todd's finite group example, you can consider finite groupoids. For example, the ideal classes of a quaternion algebra form a finite groupoid (the Brandt groupoid).
Jul 2, 2015 at 1:11 answer added Andreas Blass timeline score: 4
Jul 1, 2015 at 21:29 review Close votes
Jul 2, 2015 at 18:59
Jul 1, 2015 at 20:21 comment added Eric Wofsey For what it's worth, any finite category that has all finite products (or coproducts) is a preorder.
Jul 1, 2015 at 18:38 history made wiki Post Made Community Wiki by Todd Trimble
Jul 1, 2015 at 18:34 comment added Benjamin Steinberg If Q is a finite acyclic directed graph (called a quiver in this context) then the free category on Q (with vertex set the same as Q and arrows directed paths) is an interesting finite category. Functors from this category into the category of vector spaces is the same things as a quiver representation.
Jul 1, 2015 at 17:11 answer added Johannes Hahn timeline score: 11
Jul 1, 2015 at 17:07 comment added A Question Asker Sorry, I meant a category that is a finite preorder (essentially 5 objects 1 through 5 which are a preorder).
Jul 1, 2015 at 16:09 comment added Todd Trimble Well, the category of finite preorders is not a finite category, so I'm not sure what you're really after. The Monster group is an interesting finite category with one object and quite a few morphisms. The poset of truth values $\{0 \leq 1\}$ has a lot of nice properties (e.g., being cartesian closed) that are worth contemplating. Something that you might wish to contemplate is why the category of finite categories does not have coequalizers!
Jul 1, 2015 at 15:57 review First posts
Jul 1, 2015 at 16:00
Jul 1, 2015 at 15:57 history asked A Question Asker CC BY-SA 3.0