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I’m trying to reconcile a few things I’ve heard about the $n$-categorical ladder. First, $(n+1)$-categories should be something like categories enriched in $n$-categories. Second, $0$-categories are sets, and $(-1)$-categories are truth values. And third, every $n$-category can be considered as a discrete $(n+1)$-category.
So if I interpret ‘truth values’ to mean the poset of truth values $\text{False} \rightarrow \text{True}$ with monoidal structure from and, $\wedge$, then I should expect $0$-categories to be posets. But this would throw off the whole ladder, since then $1$-categories would be categories enriched in posets, rather than in sets. This is mentioned in the $(-1)$-category article on the lab. We could perhaps fix this situation by saying that sets really should be $0$-groupoids, which would be posets that are groupoids and so equivalence relations, which seems fine. But the upper levels of the ladder are still messed up, unless we said that an $(n+1)$-category is a category enriched in $n$-groupoids, in which case we could get back our usual $1$-categories. But then we’d be describing the $(n, 1)$-categorical ladder, rather than the $n$-categorical ladder. I suppose that the $(n, r)$-categorical formalism might help me here, but I haven’t been able to grok it yet.
On the other hand, I could interpret ‘truth values’ to mean set of of truth values $\{\text{False}, \text{True}\}$ with monoidal structure from $\wedge$ again. Now $0$-categories are enriched in this category, but since it is discrete, $[a, b] \wedge [b, c] = [a, c]$, and since $[a,a] = 1$, $$[a. b] \wedge [b, a] = [a, a] = 1$$
so that $[a, b] = 1$ for all $a$ and $b$. This is a set, but it is codiscrete, rather than discrete (as a $1$-category). So this interpretation of $(-1)$-categories gives me the right $0$-categories, but it gives me the ‘wrong’ kind of inclusion of $n$-categories into $(n+1)$-categories.
Is there a reason to prefer the discrete inclusion of $n$-categories into $(n+1)$-categories versus the codiscrete one? Is there a better way to think about $(-1)$-categories that disentangles this?
I should expect 0-categories to be posets. But this would throw off the whole ladder, since then 1-categories would be categories enriched in posets, rather than in sets.
There’s a distinction (I can’t remember where it’s discussed) between the $n$-groupoid ladder and the $n$-category ladder, of which this is an echo. The groupoid of truth values (say $\{\empty,1\}$) gives sets on enriching in it. Note that a category enriched in $n$-groupoids in an $(n,1)$-category, and a groupoid enriched in an $n$-groupoid is an $(n+1)$-groupoid.
One place is section 5.1 of arXiv:math/0608420.
David> Wouldn’t the groupoid of truth values also give codiscrete sets? I guess my main question is why do we prefer the discrete inclusion to the codiscrete inclusion…
Mike> Hmm, thanks! It seems on first glance that you have a bunch of ladders for the different levels of groupoidyness vs. posetyness. The groupoid ladder is diagonal though, and so a very different thing…
It strikes me that a major problem with considering the codiscrete inclusion of $n$-categories into $(n+1)$-categories is that it won’t reflect equivalence. In light of my brief skim of the relevant section of your notes (I just began to read them this afternoon, unrelated!), it would also mess with the idea that $(-2)$-categories should appear as contractible fibers; if all $0$-categories were contractible as $1$-categories this might pose an issue.
I don’t think there is much sense to the “codiscrete” inclusion.
Never mind
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