3
$\begingroup$

Suppose $D$ is a small category and $F\colon D\to sSets$ a functor. I want to calculate the simplicial set $$\operatorname{hocolim}F$$ via the bar construction: $\operatorname{hocolim}F$ is the diagonal of the bisimplicial set $$ \ldots\begin{array}{c}\to \\ \vdots\\\to\end{array}\coprod_{d_{k}\to ...\to d_0}F(d_{k})\begin{array}{c}\to \\ \vdots\\\to\end{array}...\begin{array}{c}\to \\ \to\\\to\end{array}\coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0). $$ which is called the bar resolution $BF\colon \Delta^{op}\to sSets$. We have also $\operatorname{hocolim}F = \operatorname{hocolim}BF$ by Bousfield-Kan.

In some cases, it seems to be sufficient to consider only the $k$-truncation of $BF\colon \Delta^{op}\to sSets$. For example, if $D=\mathbb{N}$ we have $$ \operatorname{hocolim}F= \operatorname{hocolim}\left( \coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0)\right). $$ As explained in in the answer to this question on MO, one may consider the $k$-truncation of $BF$, if the nerve $ND$ of $D$ is $k$-skeletal. However, the nerve $N\mathbb{N}$ is not $1$-skeletal and it suffices nevertheless to consider the $1$-truncation of the bar resolution. Apparently it does not suffice to observe that $N\mathbb{N}$ is contractible.

Why does this suffice for $D=\mathbb{N}$ and are there some easy to formulate conditions on $D$ such that one may take the $k$-truncated bar resolution for calculating $\operatorname{hocolim}F$?

$\endgroup$
  • $\begingroup$ You have to understand the formula correctly. The subtlety is in the indexing of the first coproduct: it's not a coproduct over all arrows in $\mathbb{N}$, just the generators. $\endgroup$ – Zhen Lin Aug 22 '14 at 12:57
  • $\begingroup$ Dear Zhen Lin, sorry but I don't understand what you mean. In $(BF)_1$, it is a coproduct indexed by all morphisms of $D$, right? Do you mean, that the indexing of the truncated bar resolution I wrote below is different? $\endgroup$ – user8463524 Aug 22 '14 at 13:07
  • 1
    $\begingroup$ The second formula. The first formula is the definition. $\endgroup$ – Zhen Lin Aug 22 '14 at 13:35
  • 1
    $\begingroup$ Because $\mathbb{N}$ is a free category. $\endgroup$ – Zhen Lin Aug 22 '14 at 13:37
  • 1
    $\begingroup$ Yes, I expect it should work, but I have never seen it written out explicitly. $\endgroup$ – Zhen Lin Aug 22 '14 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.