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 $K(P, 1)$….


  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.

1 Answer 1


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 $\pi_1(X)$.

  • $\begingroup$ Could you explain why $T(\widetilde{X})$ admits a free action of $\pi_1(X)$? $\endgroup$ Feb 18, 2022 at 15:13
  • $\begingroup$ 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. $\endgroup$
    – IJL
    Feb 18, 2022 at 21:54

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