Timeline for Homotopy colimits over a certain subset category.
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2013 at 16:00 | comment | added | Dan Ramras | By saying $J$ is cofinal in $I$, I just meant that for each $i$ in $I$, there is an element $j$ in $J$ such that $i$ is contained in $j$. Here I'm thinking of $I$ and $J$ as posets. | |
| Mar 29, 2013 at 10:48 | comment | added | Dedalus | Karol: Thank you! That was very helpful. I am working with simplicial sets with an action of a profinite group, so I think it should work in that case. I will look into the article you gave more explicitly. Dan: Thank you too! With cofinal - you mean homotopy cofinal , right? | |
| Mar 28, 2013 at 22:43 | comment | added | Karol Szumiło |
The category $I$ is not directed (unless $S$ is finite, but that's a boring case). However, it is direct and for such categories there is a very explicit procedure for constructing homotopy colimits (of nice enough i.e. Reedy cofibrant diagrams). It proceeds by inductively attaching objects lying over $I_n$ and the taking the mapping telescope of the resulting sequence. It is described in detail in Theorem 9.3.5 of arxiv.org/abs/math/0610009v4.
|
|
| Mar 28, 2013 at 18:27 | comment | added | Dan Ramras | When $S$ is finite then $S$ itself is a terminal element of $I$ and you can just take $J$ to be the trivial category with this one element. If $S$ is countable, then we can assume $S =\mathbb{N}$ and let $J$ be the subcategory consisting of the sets $\{1, \ldots, n\}$. Then $J$ is cofinal (and directed), so the homotopy colimit restricted to $J$ is the same as the full homotopy colimit (up to homotopy); this is proven somewhere in Bousfield and Kan. | |
| Mar 28, 2013 at 18:27 | comment | added | Dan Ramras | The morphisms in your category $I$ are simply inclusions? By directed, do you mean isomorphic to the poset of natural numbers? Since the union of two finite sets is finite, your category $I$ is always a directed poset in the usual sense (for any two objects, there is a third that is greater than both) but if you want to talk about mapping telescopes I guess you need a poset isomorphic to the natural numbers under the usual ordering. | |
| Mar 28, 2013 at 18:19 | comment | added | Eric Wofsey | I don't understand what you're asking for. You could take $J=I$, since $I$ itself is directed. | |
| Mar 28, 2013 at 15:10 | history | asked | Dedalus | CC BY-SA 3.0 |