# Homotopy colimits over a certain subset category.

Hi! Let $I$ denote all the finite subsets of some set (infinite or finite) S. For each n, let $I_n$ be the subset consisting of objects of cardinality n, so that there are no morphisms between the objects in $I_n$. Suppose that we have a diagram $X:I \rightarrow M$ where $M$ denotes some model category (if it is neccessary, assume some nice properties on it too). My question is the following:

Could we present $hocolim X$ as the homotopy colimit of some subdiagram $J \subset I$? Ideally, I would like to be able to find a $J$ that is directed, so that the homotopy colimit will be the mapping telescope. If so, why? Is there any additional properties on X I need for this to be true?

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I don't understand what you're asking for. You could take $J=I$, since $I$ itself is directed. – Eric Wofsey Mar 28 '13 at 18:19
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. – Dan Ramras Mar 28 '13 at 18:27
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. – Dan Ramras Mar 28 '13 at 18:27
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. – Karol SzumiĆo Mar 28 '13 at 22:43
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. – Dan Ramras Mar 29 '13 at 16:00