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If $C$ is a simplicial model category, $F\colon I\longrightarrow J$ a functor between small categories, and $X\colon J\longrightarrow C$ a diagram, the canonical map $holim_JX\longrightarrow holim_IX\circ F$ is a weak equivalence if all the categories $F/j$ are contractible.

For $C=Top_\ast$ the category of (compactly generated Hausdorff) based spaces, is there a theorem about the connectivity of $holim_JX\longrightarrow holim_IX\circ F$? Thinking of holim as a hom object, the connectivity should depend on a notion of "connectivity for the diagram X", and of "relative dimension" for the map of $J$-shaped diagrams of spaces $BF/(-)\longrightarrow BJ/(-)$.

As a baby example, let $\mathcal{P}(n)$ be the poset of subsets of the set with $n+1$ elements, and $F\colon \mathcal{P}(n)\backslash\emptyset \longrightarrow \mathcal{P}(m)\backslash\emptyset$, for $n\leq m$, the inclusion that sends $S\subset n$ to $S\cup(m\backslash n)$. Let $X\colon \mathcal{P}(m)\backslash\emptyset\longrightarrow Top_\ast$ be the diagram with $X_m=Y$ for some fixed based space $Y$, and $X_S=\ast$ a point for all proper subsets $S\subset m$. Then the canonical map on homotopy limits is the restriction map $\Omega^{m}Y\longrightarrow \Omega^nY$ between loop spaces. This is as connected as the connectivity of $Y$ minus the dimension of $S^m/S^n$.

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  • $\begingroup$ Regarding the example, surely if $n < m$ then the restriction map is nullhomotopic? (It factors through a space of pointed maps from the $n$-disc to $Y$.) $\endgroup$ Jul 18, 2013 at 10:23
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    $\begingroup$ Yeah it is not a great example. The map is null homotopic means that it is an iso in $\pi_k$ for $k$ below the minimum of the connectivities of $\Omega^nY$ and $\Omega^mY$. This is the same as the connectivity of $Y$ minus $m=\dim S^m/S^n$ for $n<m$. $\endgroup$ Jul 18, 2013 at 10:41
  • $\begingroup$ @Emanuele: I am having some difficulties understanding your example. What is the value of $X$ on subsets of $m$ strictly containing $n$? Or did you perhaps mean $X_m = Y$ and $X_S =\ast$ for all other $S\neq m$? If so, then it seems that $X\circ F = \ast$ if $n<m$. Please let me know what I am misunderstanding. /// In any case, this is a very interesting question, and I am looking forward to learn about any answers. $\endgroup$ Jul 18, 2013 at 19:45
  • $\begingroup$ @Ricardo: You're right, The functor $F$ I want to consider is not the standard inclusion. The small category should contain $m$ as final object. I am going to edit the question. $\endgroup$ Jul 18, 2013 at 20:40

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