Homology equivalence and isomorphism on $\pi_1$ not enough for homotopy equivalence?

Question

It is a standard consequence of Hurewicz's theorem that a homology eqivalence between simply connected spaces is a weak equivalence (and hence a homotopy equivalence, if the spaces are CW-complexes).

What is more, it is even enough to assume that the map is a homology equivalence with local coefficients and an iso on $\pi_1$, see e.g. this question.

(On the other hand, a map which is only a homology equivalence does not need to be an isomorphism on $\pi_1$ and hence not a weak equivalence.)

What is a concrete example of a map that is an isomorphism on homology with $\mathbb Z$ coefficients, and also an isomorphism on $\pi_1$, but not a weak equivalence?

Answer 1

Let $X$ be the CW complex obtained from $S^1 \vee S^n$, $n>1$, by attaching an $(n+1)$-cell via a map $S^n\to S^1\vee S^n$ representing the element $2t-1$ in $\pi_n(S^1\vee S^n) \cong {\mathbb Z}[t,t^{-1}]$, so $\pi_n(X)\cong{\mathbb Z}[t,t^{-1}]/(2t-1)\cong {\mathbb Z}[1/2]\subset{\mathbb Q}$. The inclusion map $S^1 \to X$ then induces an isomorphism on homology and on $\pi_i$ for $i<n$ but not on $\pi_n$.

This example relies heavily on the nontriviality of the action of $\pi_1$ on $\pi_n$, but this is necessary since Whitehead's theorem that homology isomorphisms between simply-connected CW complexes are homotopy equivalences holds more generally for CW complexes with trivial action of $\pi_1$ on all homotopy groups including $\pi_1$. (This is Proposition 4.74 in my Algebraic Topology book, and the example is Example 4.35. Sorry for the self-references, but I should know the book as well as anyone!)

Answer 2

This is a modification of Anders's suggestion:

Take $X=M(\mathbb{Z}[\frac{1}{2}],2)$ to be the Moore space with $H_2(X) = \mathbb{Z}[\frac{1}{2}]$. We can have $\mathbb{Z}/2$ act on $X$ this by a sign on $H_2$. For example, we can build this by smashing $M(\mathbb{Z}[\frac{1}{2}],1)$ with $S^1$, and we can act on this $S^1$ by reflection.

If we form $E\mathbb{Z}/2 \times_{\mathbb{Z}/2} X$, this comes with a projection down to $E\mathbb{Z}/2\times_{\mathbb{Z}/2} * = B\mathbb{Z}/2$. The fiber of that map is given by $X$.

This map is clearly a $\pi_1$-iso. To see that it's also a homology iso, consider the homology Serre spectral sequence for this fibration. We have $$ E^2_{p,q} = H_p(B\mathbb{Z}/2; H_q(X)) $$ where we take cohomology with local coefficients given by the $\pi_1$-action on $X$. Now the only stuff in positive $q$-degree exists for $q=2$, and it is easy to see that $H_p(B\mathbb{Z}/2; \mathbb{Z}[\frac{1}{2}]) = 0$ for all $p$: For $p=0$, it is given by the quotient of $\mathbb{Z}[\frac{1}{2}]$ by the $\pi_1$-action, that's $0$, and for positive $p$ it is definitely zero, for example by universal coefficients (over $\mathbb{Z}$ with the sign action, everything is $2$-torsion).

So the spectral sequence is concentrated in the $q=0$-line, but that just means that the map induces an iso in homology.

Answer 3

$$ (E\mathbb{Z}^\times \times K(\mathbb{Q},3))/\mathbb{Z}^\times \to B\mathbb{Z}^\times. $$