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I'm interested in something built upon the construction laid out in nlab article on Snaith's theorem

Let $(E, \mu, \iota)$ be a ring spectrum.

For $\beta \in \pi_n(E)$ an element of the $n$th stable homotopy group, then multiplication by $\beta$ is a homomorphism $$ \beta_\ast \;\colon\; E \to \Sigma^{-n} E \,. $$ The localization of $E$ at $\beta$ is the homotopy colimit over the iterated multiplication with $\beta$ $$ E[\beta^{-1}]= \underset{\to}{\lim} \left[ E \stackrel{\beta_\ast}{\to} \Sigma^{-n}E \stackrel{\Sigma^{-n} \beta_\ast}{\to} \Sigma^{-2n} E \to \cdots \right] $$ which has the universal property that $\mu_\beta$ becomes an equivalence on $E[\beta^{-1}]$.

Since $E[\beta^{-1}]$ is a spectrum, we can compute the homotopy groups. I want to know why

$$\pi_*(E[\beta^{-1}]) \cong \pi_*(E)[\beta^{-1}]$$ holds. I wrote down an "argument" (using the reduced telescope of the sequence) for why that is true, but after not having looked at it for some time I recognize it was circular.

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    $\begingroup$ Do you know that $\pi_*$ commutes with filtered (homotopy) colimits ? $\endgroup$ Commented Mar 9, 2023 at 20:16
  • $\begingroup$ Not as of yet. I will try to familiarize myself, would there be more steps to complete the proof? $\endgroup$
    – Excalibur
    Commented Mar 9, 2023 at 21:01
  • $\begingroup$ This would really be the main step. The second one would be understanding why $A[x^{-1}]$ can be written as a colimit for an ordinary (commutative) ring $A$ $\endgroup$ Commented Mar 9, 2023 at 21:03
  • $\begingroup$ Thanks you @MaximeRamzi for your comments. Unless someone comes up with a more elementary approach, I will look into it. $\endgroup$
    – Excalibur
    Commented Mar 11, 2023 at 15:20
  • $\begingroup$ It's a lot of busy writing but the key is: $S^n$ is an ANR. $\endgroup$
    – Wlod AA
    Commented Oct 12, 2023 at 14:10

1 Answer 1

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Giving details along the hint given, note that given a (discrete) commutative ring $A$ and an element $x\in A$, the colimit of the sequence $$ A \xrightarrow{x} A \xrightarrow{x} A \xrightarrow{x} A \xrightarrow{} \cdots $$ (in the category of $\mathbb{Z}$-modules) satisfies the universal property of $A[x^{-1}]$. Denote $A'$ this colimit and $f_n:A \to A', n\geq 0$ the canonical maps into the colimit. The $\mathbb{Z}$-module $A'$ inherits a ring structure by $[a_n]\cdot[a_m]:=[x^{m-n}a_nb_m]$ whose unit element is $f_0(1)$, and the map $f_0:A\to A'$ becomes a ring homomorphism. The element $f_1(1)\in A'$ provides an inverse to $x$ since $xf_1(1)=f_0(1)\in A'$. Moreover, if $g:A\to B$ is a ring homomorphism which sends $x$ to a unit, there is a unique extension of $g$ along $f_0:A\to A'$ given by the compatible family of maps $g_n:A \to A'$ defined by $g_n(a)=g(x)^{-n}g(a)$. This proves that $A'\cong A[x^{-1}]$.

Now note that the functor $\pi_* : Sp \to Ab^\mathbb{Z}$ which takes homotopy groups preserves filtered colimits since $\pi_n=[\Sigma^n\mathbb{S},-]$ and the (suspended) sphere, being a finite spectrum, is a compact object in spectra.

This explains why $\pi_*(E[\beta^{-1}])\cong \pi_*(E)[\beta^{-1}]$.

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