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I am having some trouble understanding the difference between the $p$-completion and a $p$-local space.

If $X$ a simply connected space has all higher homotopy groups finitely generated, then the $p$-local approximation $X\rightarrow L_{p}X$ coincides with the $p$-completion $X_{p}^{\wedge}$. Is this correct?

Now if $X$ is not simply connected such that the fundamental group is abelian, do we have that $ L_{p}X \simeq X_{p}^{\wedge}$?

I think it is true if $X$ is a circle, more precisely, $S^{1\wedge}_{p}\sim L_{p}S^{1}\sim K(\mathbf{Z}_{p}^{\wedge},1)$. Am I wrong ?

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    $\begingroup$ On the first point, the p-localization localizes the homotopy groups whereas the p-completion completes the homotopy groups. So if all the higher homotopy groups are {\it finite} the completion and localization are same, but finitely generated doesn't suffice. For example take $K(Z,2)$. Localization gives $K(Z_{(p)},2)$ whereas the completion gives $K(Z_p^{\wedge},2)$. $\endgroup$
    – user43326
    Commented Aug 24, 2015 at 8:24
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    $\begingroup$ As to the second point, when $X$ is a circle, its $\pi _1$ acts nilpotently on all $\pi _i(X)$ since higher homotopy groups are trivial and the action on $pi _1$ is by identity, so localization and completion behave reasonably, but even in this case localization and completion differ. By localizing one gets $K(Z_{(p)},1)$, by completing one gets $K(Z_p^{\wedge },1)$. $\endgroup$
    – user43326
    Commented Aug 24, 2015 at 8:37

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If $L_{\mathrm{H}\mathbf{Z}_{(p)}}X$ denotes the $p$-localization of $X$, then $\pi_\ast L_{\mathrm{H}\mathbf{Z}_{(p)}}X\cong \pi_\ast X\otimes\mathbf{Z}_{(p)}$. On the other hand, if $L_{M(p)}X$ denotes the $p$-completion of $X$ (here $M(p)$ is the mod-$p$ Moore spectrum), then $\pi_\ast L_{M(p)}X\cong\pi_\ast X\otimes\mathbf{Z}_p$. Notice that $\pi_\ast X\otimes\mathbf{Z}_{(p)}$ differs from $\pi_\ast X\otimes\mathbf{Z}_p$ in general, so $L_{\mathrm{H}\mathbf{Z}_{(p)}}X$ differs from $L_{M(p)}X$ in general (unless, for eg. $\pi_\ast X$ is finite). For your question, recall that $S^1=K(\mathbf{Z},1)$. Completion at $p$ gives $K(\mathbf{Z}_p,1)$, while localizing at $p$ gives $K(\mathbf{Z}_{(p)},1)$, and these differ.

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  • $\begingroup$ I realized that this has been said in the comments as well. $\endgroup$
    – user62675
    Commented Sep 5, 2015 at 17:03
  • $\begingroup$ Does any functor also yield $p$-completion of $\pi_*(X)$? Because in general it differs from $\pi_*(X)\otimes{\mathbf Z}_p$, right? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 5, 2015 at 17:18
  • $\begingroup$ @მამუკაჯიბლაძე In general it does differ from $\pi_\ast X\otimes\mathbf{Z}_p$; if $\pi_\ast X$ is finitely generated, then then the two are isomorphic. I'm not sure I understand the first question; are you asking if there is a functor which takes a group to its $p$-completion? $\endgroup$
    – user62675
    Commented Sep 5, 2015 at 17:24
  • $\begingroup$ Yes exactly. More precisely, is there a functor $F$ on spaces such that $\pi_*(F(X))=\pi_*(X)^\wedge_p$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 5, 2015 at 17:35
  • $\begingroup$ @მამუკაჯიბლაძე There is a functor which takes a group to its $p$-completion. A map $X\to Y$ induces a map $L_{M(p)}X\to L_{M(p)}Y$, and you can take homotopy groups to get a map $\pi_\ast L_{M(p)}X\to \pi_\ast L_{M(p)}Y$. Now $\pi_\ast L_{M(p)}X\cong\pi_\ast X\otimes\mathbf{Z}_p$, so this is isomorphic to $\pi_\ast X^\wedge_p$ if $\pi_\ast X$ is finitely generated. Bousfield localization at $M(p)$ therefore gives you such a functor, but only on the subcategory of $\mathrm{Top}$ spanned by those spaces $X$ such that $\pi_\ast X$ is finitely generated. $\endgroup$
    – user62675
    Commented Sep 5, 2015 at 17:50

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