Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Yesterday I was wandering for the $n$-lab and I've found the definition of $n$-poset. Following this post it seems that a $n$-poset should be a $(n,n+1)$-category. Now an $(n,r)$-category should be a category such that every $k$-morphism is an equivalence for $k\geq r$ and every pair of parallel $k$-morphisms with $k \geq n$ are equivalent. Now here're my problems: I suppose that this objects should generalize in any some way the notion of poset to higher categorical structure, i.e. it should be a categorification of the notion of poset, but I don't get why this should be the case

could anyone explain to me how $n$-posets generalize the notion of poset?

share|improve this question
    
Have you read the section "Special cases" at the nlab article? For me this directly answers your question. –  Martin Brandenburg Oct 12 '11 at 20:23
    
Technicality: an $n$-poset is an $(n-1,n)$-category rather than an $(n,n+1)$-category. –  Toby Bartels Oct 12 '11 at 20:35

1 Answer 1

up vote 3 down vote accepted

It seems to me that you are effectively asking why a $(0,1)$-category is a poset. Because if that is so, it makes sense to define an $n$-poset to be an $(n-1,n)$-category.

To see why a $(0,1)$-category is a poset, just unwind the definition: it contains possibly non-invertible 1-morphisms, but any two of them that have the same source and target are equivalent (and the space of choices of equivalences between them is contractible). By the characterization of posets as categories that means it is a poset.

share|improve this answer
1  
Note also that all of this "is" is up to equivalence (of, in general, $\infty$-categories). On the $n$Lab, that's the default anyway; there might be two different definitions of the same term, but as long as they're equivalent in this weak sense, then they're simply different ways to look at the same thing. –  Toby Bartels Oct 12 '11 at 20:37
1  
That's why saying "the space of choices of equivalences between them is contractible" amounts to the same as "there is just one equivalence between them". –  Toby Bartels Oct 12 '11 at 20:38
    
I think now I get it: the problem was that I didn't recognize the $(0,1)$-categories as poset because I was used to see poset in strict form: that is categories in which every pair of equivalent morphisms are equal. With this limitation $(n-1,n)$-categories seemed more like a generalization of preorder than poset, but now I see, $n$-poset are a weak higher dimensional version of poset. Thanks a lot. –  Giorgio Mossa Oct 13 '11 at 12:48
    
Yes, arguably the term ought to be "n-preorder"! –  Mike Shulman Oct 19 '11 at 0:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.