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Assume that we have a connective spectrum $X$, and denote the $p$-completion of this spectrum in the sense of Bousfield by $X^{\wedge}_p$ (which is given by the function spectrum $F(S^{-1}\mathbb{Z}/p^{\infty},X)$). Let $(\Omega^{\infty}_0X)^{\wedge}_p$ be the profinite $p$-completion of the zeroth component of the infinite loop space of $X$. Is that true that $\Omega^{\infty}_0X^{\wedge}_p\simeq (\Omega^{\infty}_0X)^{\wedge}_p$, meaning that they are weakly equivalent as topological spaces ?

Remark. I was thinking about two specific spectra $X$ and $Y$ and a map between them which I spare you with their descriptions that I know the map induces equivalence between $p$-completions $X^{\wedge}_p\simeq Y^{\wedge}_p$. I'd like to show $H_*(\Omega^{\infty}_0X,\mathbb{Z}/p)\cong H_*(\Omega^{\infty}_0Y,\mathbb{Z}/p)$. This made me to ask the previous question.

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    $\begingroup$ Under mild assumptions, both $p$-completions complete the homotopy groups at $p$. Since the homotopy groups of the infinite loop spaces is those of the spectrum, you get the equivalence that you want (under those assumptions). $\endgroup$ – user43326 Apr 1 '14 at 17:28
  • $\begingroup$ What are those assumptions? Do you know any possible references? $\endgroup$ – Sam Nariman Apr 2 '14 at 19:28
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There is a standard natural short exact sequence for the calculation of $\pi_n(X^{\wedge}_p)$ from $\pi_n(X)$, and it applies whether $X$ is a spectrum or a space. (I'm assuming you mean Bousfield completion at $p$ for both). For spaces that is in Bousfield-Kan, and a more recent treatment is in More Concise Algebraic Topology [MP], by Kate Ponto and myself. See [Theorem 11.1.2, MP]. The homotopy groups of $\Omega^{\infty}_0 (X^{\wedge}_p)$ are the positive degree homotopy groups of $X^{\wedge}_p$ and are $p$-complete in the sense defined in [MP]. Therefore $\Omega^{\infty}_0 (X^{\wedge}_p)$ is $p$-complete by [Theorem 11.1.1, MP]. By the universal property of completion at $p$, there is a map $(\Omega^{\infty}_0 X)^{\wedge}_p \to \Omega^{\infty}_0 (X^{\wedge}_p)$ under $\Omega^{\infty}_0X$ . It induces isomorphisms on homotopy groups, as you want.

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Suppose $X$ is of finite type (i.e., $\pi _n(X)$ finitely generated for all $n$) and $0$-connected. Then one can use Milnor-type exact sequence to show that

$\pi _*(X_p^{\wedge})\cong \pi _*(X)_p^{\wedge}$

On the other hand, since $X$ is of finite type, so is $\Omega ^{\infty }X$, and according to Bousfield, Kan, {\it Homotopy Limits, Completions, and Localizations} (Splinger Lecture Notes in Mathematics 304}, Part I, Chapter VI example 5.2, we have

$\pi _*((\Omega ^{\infty }X)_p^{\wedge})\cong (\pi _*(\Omega ^{\infty }X))_p^{\wedge}$

(recall that we have finite type hypothesis, so tensoring with $p$-adics amounts to completing at $p$) where $(\Omega ^{\infty }X)_p^{\wedge}$ denotes the Bousfield-Kan $p$-completion of the space $\Omega ^{\infty }X$. It is also known (loc.cit) that under this assumption, Bousfield-Kan completion agrees with QUillen's or Sullivan's profinite $p$-completion.

Combining all these and the isomorphism

$\pi _*(Y)\cong \pi _*(\Omega ^{\infty}Y)\mbox{ for $*\geq 0$)}$ for any spectra $Y$, we see that $(\Omega ^{\infty }X)_p^{\wedge}$ and $\Omega ^{\infty }(X_p^{\wedge})$ are weakly equivalent.

(if you prefer, one can say that the natural map $\Omega ^{\infty }X \rightarrow \Omega ^{\infty }(X_p^{\wedge})$ factors through $(\Omega ^{\infty }X)_p^{\wedge}$ and the resulting map is a weak equivalence.

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    $\begingroup$ I think there is $\pi_0$ issue, even for $X=H\mathbb{Z}$ we don't have $\Omega^{\infty}X^{\wedge}_p\simeq (\Omega^{\infty}X)^{\wedge}_p$ because they don't have the same $\pi_0$. That is why I asked for $\Omega^{\infty}_0X^{\wedge}_p\simeq (\Omega^{\infty}_0X)^{\wedge}_p$. The thing is I don't want to impose finite type conditions because I am working with Thom spectrum over $BG^{\delta}$ where $G^{\delta}$ is some lie group with discrete topology. It is connective spectrum but not of finite type. $\endgroup$ – Sam Nariman Apr 6 '14 at 0:03
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    $\begingroup$ You are right, I added $0$-connected hypothesis to avoid the $\pi _0$ issue. As to the non-finiteness, I don't know if there is any good theory to deal with this, but anyway we still have the Milnor exact sequence, so we can still compute the homotopy groups of $\Omega ^{\infty}(X_p^{\wedge})$. $\endgroup$ – user43326 Apr 6 '14 at 16:23

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