# Kan–Thurston theorem and R-completion

A corollary of the Kan and Thurston theorem states that the space $$X$$ (path connected) can be recovered, up to homotopy, by applying the fiber-wise $$\mathbb{Z}$$-completion functor of Bousfield–Kan (notation: $$\dot{\mathbb{Z}}_\infty$$) to the fibration $$K(P, 1)\to TX = K(G, 1)\to K(G/P, 1)$$ where $$TX$$ is a Kan–Thurston construction of $$X$$ (see references), $$G/P = \pi_1(X)$$, $$P$$ is a perfect group (i. e. $$H_1(P; \mathbb{Z}) = 0$$, trivial coefficients) and groups $$P$$, $$G$$ and $$G/P$$ form the exact sequence $$1\to P \to G \to G/P \to 1.$$

For understanding why $$\dot{\mathbb{Z}}_\infty K(G, 1)\cong X$$, I want to use the Whitehead theorem:

Let $$f: X\to Y$$ be a map of pointed connected $$CW$$-complexes such that $$f$$ induces an isomorphism on fundamental groups and on homology with any local coefficient system. Then $$f$$ is a homotopy equivalence.

As $$P$$ is perfect, the space $$\mathbb{Z}_\infty K(P, 1)$$ is simply-connected. Hence, from the long exact sequence of fibration $$\mathbb{Z}_\infty K(P, 1)\to \dot{\mathbb{Z}}_\infty K(G, 1)\to K(G/P, 1)$$ we have that the Kan–Thurston map $$t: TX=K(G, 1)\to X$$ induces an isomorphism $$\pi_1(\dot{\mathbb{Z}}_\infty K(G, 1))\cong \pi_1(X)$$. Moreover, by the main Kan–Thurston result, $$t$$ induces isomorphisms $$H_\ast(TX; \mathcal{A}) \cong H_\ast(X; \mathcal{A})$$.

Now consider a commutative diagram $$\require{AMScd}$$ $$\begin{CD} K(P, 1) @>>> K(G, 1) @>>> K(G/P, 1) \\ @VVV @VVV @| \\ \mathbb{Z}_\infty K(P, 1) @>>> \dot{\mathbb{Z}}_\infty K(G, 1) @>>> K(G/P, 1). \end{CD}$$

The morphism of fibrations in the diagram induces a morphism of spectral sequences. And by Zeeman's comparison theorem it is sufficient to prove isomorphism $$H_\ast(\mathbb{Z}_\infty K(P, 1); \mathcal{A})\cong H_\ast(K(P, 1); \mathcal{A})$$ for any local coefficient system $$\mathcal{A}$$.

It is known that the Bousfield–Kan $$R$$-completion ($$R\subset \mathbb{Q}$$ or $$R = \mathbb{Z}/p$$) of a space $$Y$$ with $$R$$-perfect fundamental group ($$R\otimes H_1(\pi_1; \mathbb{Z}) = 0$$) is $$R$$-good, that is, $$H_\ast(R_\infty Y; R)\cong H_\ast(Y; R)$$. It implies the desired isomorphism in the case of $$\mathcal{A} = \mathbb{Z}$$. But I don't know why $$H_\ast(\mathbb{Z}_\infty K(P, 1); \mathcal{A})\cong H_\ast(P; \mathcal{A})$$ for any local coefficient system $$\mathcal{A}$$. Moreover, it looks weird that $$\mathcal{A}$$ on $$\mathbb{Z}_\infty K(P, 1)$$ is trivial (thanks to $$\pi_1(\mathbb{Z}_\infty K(P, 1)) = 0$$) but we can't say the same about the local system on $$K(P, 1)$$….

References

1. D. Kan and W. Thurston, Every connected space has the homology of a $$K(\pi, 1)$$, Topology Vol. 15. pp. 253–258, 1976.
2. A. K. Bousfield and D. M. Kan, Homotopy limits, completions and localizations, Lecture Notes in Math. 304, Springer, 1972.

The Kan-Thurston construction depends not just on the homotopy type of $$X$$, but it depends very heavily on the exact choice of cell structure for $$X$$. However, it does have some naturality properties. In particular, if $$\widetilde X$$ is the universal covering space of $$X$$, then $$T\widetilde X$$ admits a free action of $$G/P=\pi_1(X)$$ such that the quotient is isomorphic to $$TX$$. In particular, if you pull back your first fibration along the universal covering map $$\widetilde X\rightarrow X$$, you get a fibration $$K(P,1)\rightarrow T(\widetilde X)\rightarrow E(G/P)$$, where the base space is contractible. I would use this fibration to deduce that $$\mathbb{Z}_\infty K(P,1)$$ is homotopy equivalent to $$\widetilde X$$.
Incidentally, any local coefficient system on $$K(P,1)$$ that is induced by pulling back a local coefficient system on $$X$$ is trivial, because $$P$$ is in the kernel of the map to $$\pi_1(X)$$.
• Could you explain why $T(\widetilde{X})$ admits a free action of $\pi_1(X)$? Feb 18 at 15:13
• The construction $Y\mapsto T(Y)$ can be designed in such a way that it is natural for maps of simplicial complexes that do not collapse any simplices; in particular for simplicial automorphisms. The free action of $\pi_1(X)$ on $\widetilde{X}$ thus induces an action of $\pi_1(X)$ on $T(\widetilde{X})$ too.