How does hocolim connect to Hom?

In a common group $\mathcal{C}$ one can draw the colim out of the Hom similar to this: $\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\colim}{colim}\DeclareMathOperator{\hocolim}{hocolim}\DeclareMathOperator{\Ho}{Ho}\DeclareMathOperator{\holim}{holim}$ $$\Hom\nolimits_\mathcal{C}(\colim A_i,B)=\lim \Hom\nolimits_\mathcal{C}(A_i,B)$$

I am seeking an equivalent declaration for hocolims - allows claim in simplicial collections, yet if there are extra basic declarations, that is also much better.

E.g. I can visualize. $$\Hom\nolimits_{\mathcal{C}}(\hocolim A_i,B)=\lim \Hom\nolimits_{\Ho(\mathcal{C})}(A_i,B)$$ - possibly one requires to have $B$ fibrant and also the A_i cofibrant below, i.e. that the Homs on the right are $\mathbb{R}Homs$.

Making use of the inner Hom in simplicial collections I can additionally visualize variations similar to this:. $$\Hom(\colim A_i,B)=\holim \Hom(A_i,B)$$. $$\Hom(\colim A_i,B)=\holim \mathbb{R}\!\Hom(A_i,B)$$

What is the appropriate declaration and also what is the area to discover this hocolim - yoga exercise?

Many thanks! N.B.

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2019-05-18 22:21:12
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Answers: 1

A formula like the one you are requesting for can be the following (Bousfield - Kan, "Homotopy restrictions, conclusions and also localizations", phase XII, suggestion 4.1):

$$ \mathrm{hom}_* (\mathrm{hocolim}\ \mathbf{A}, B) \cong \mathrm{holim}\ \mathrm{hom}_* (\mathbf{A}, B) \ . $$

Here $B$ is a sharp simplicial set, $\mathbf{A} : I \longrightarrow \Delta^{\mathrm{o}}\mathbf{Set}_*$ a functor from a tiny group $I$ to the group of sharp simplicial collections, and also for sharp simplicial collections $A, B$

$$ \mathrm{hom}_* (A,B) \in \Delta^{\mathrm{o}}\mathbf{Set}_* $$

is the sharp simplicial function room which $n$ - simplices are maps in $\Delta^{\mathrm{o}}\mathbf{Set}_* $

$$ \left( \Delta [n] \times A \right) / \left( \Delta [n] \times * \right) \longrightarrow B $$

(op.cit., phase VIII, 4.8).

I really did not experience the information, yet, if it frustrates you, it appears to me that you can go down the "sharp" point almost everywhere simply removing "sharp", $*$, and also $\Delta [n] \times * $ in what Bousfield - Kan claim.

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2019-05-21 06:15:34
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