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The classical localization of topological spaces at a given set of primes $\mathcal{P}$ is a functor $\mathcal{T}\xrightarrow{(-)_{(\mathcal{P)}}}\mathcal{T}$ from a suitable category of topological spaces to itself that has seen many equivalent definition. It factors $\mathcal{T}\xrightarrow{(-)_{(\mathcal{P)}}}\mathcal{T}_{\mathcal{(P)}}\hookrightarrow\mathcal{T}$ through the full subcategory of $\mathcal{P}$-local members of $\mathcal{T}$ and is left adjoint to the inclusion. Thus is preserves all colimits. It also preserves some limits. It preserves homotopy pullbacks and in particular fibre sequences. Which brings me to my question.

What are the conditions for a homotopy limit to be preserved by localization? More importantly, how does one compute the localization of infinite homotopy limits?

As an illustration I wish to calculate the rational homotopy type of the component of the based mapping space $Map_*^{B\iota}(BU(1),BU(2))$ where $\iota:U(1)\hookrightarrow U(2)$ is the canonical inclusion. I write $BU(1)=\mathbb{C}P^{\infty}$ as the homotopy colimit over $\mathbb{C}P^n$. Then I have

$Map_*^{B\iota}(BU(1),BU(2))_\mathbb{Q}=\left(Map_*^{B\iota}(hocolim_n\mathbb{C}P^n,BU(2))\right)_\mathbb{Q}=\left(holim_n Map_*^{B\iota|}(\mathbb{C}P^n,BU(2))\right)_\mathbb{Q}$.

On the inside I have

$Map_*^{B\iota|}(\mathbb{C}P^n,BU(2))_\mathbb{Q}=Map_*^{B\iota|\mathbb{Q}}(\mathbb{C}P^n,BU(2)_\mathbb{Q})=Map_*^{B\iota|_\mathbb{Q}}(\mathbb{C}P^n,K(\mathbb{Q},2)\times K(\mathbb{Q},4))=K(\mathbb{Q},2)$

and naively commuting localization with the homotopy inverse limit i get

$Map_*^{B\iota}(BU(1),BU(2))_\mathbb{Q}=holim K(\mathbb{Q},2)=K(\mathbb{Q},2)$

What is the obstruction to performing this naive commutation?

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    $\begingroup$ Localization (of abelian groups) does not commute with infinite products. Since localization (of simple spaces) localizes homotopy groups, I think one cannot expect a simple general answer. The question has to be first answered on the group level. $\endgroup$
    – Peter May
    Apr 14, 2016 at 1:55
  • $\begingroup$ Thank you. I wasn't aware that it was such an open problem. I;m not a group theorist and was only partially aware of the difficulties there. I expected at least a lim1-style measure of the obstruction - at least for sufficiently nice spaces/groups. $\endgroup$
    – Tyrone
    Apr 15, 2016 at 16:23

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