Homology Whitehead theorem for non simply connected spaces

Question

(One version of) the Whitehead theorem states that a homology equivalence between simply connected CW complexes is a homotopy equivalence. Does the following generalisation hold true?

Suppose $X,Y$ are two connected CW complexes and $f:X\to Y$ is a continuous map that induces isomorphisms of the fundamental groups and on homology. Then $f$ is a weak equivalence.

It would be enough to show that the map $\tilde{f}$ obtained by lifting $f$ to the universal covers (for some arbitrary choice of base points) is a homology equivalence, but I've never studied the homology spectral sequence, so I don't know the relationship between the homology of the universal cover $\widetilde{X}$ and that of $X$, and I don't know how to prove that $\tilde{f}$ is a homology equivalence.

Also, are there any good, reasonably self contained and short accounts of the homology of covering spaces, preferably online?

EDIT 1 I've seen some lecture notes (second paragraph on page 5) where this theorem is used (without further comment). They used this proposition to establish that a certain type of homotopy colimit is well defined up to weak equivalence.

EDIT 2 Theorem 6.71 from Kirk and Davis (page 179 of the document or 164 internally) is probably what I need, but it involves isomorphisms with local coefficient systems for homology for the base spaces.

EDIT 3 Thank you @studiosus for the reference. I don't quite understand yet, could you help out? If I get it (using the theorem of the source), there are finite connected two dimensional complexes $X,Y$ and a homotopy equivalence $X\vee S^2\simeq Y\vee S^2$, yet $X$ and $Y$ aren't homotopy equivalent. From the assumption it follows immediately that $$X\hookrightarrow X\vee S^2\to Y\vee S^2\twoheadrightarrow Y$$ gives an isomorphism of the fundamental groups. From $$\tilde{H}(X)\oplus\tilde{H}(S^2)\simeq\tilde{H}(X\vee S^2)\simeq\tilde{H}(Y\vee S^2)\simeq\tilde{H}(X)\oplus\tilde{H}(S^2)\;,$$ $\tilde{H}(S^2)\simeq \Bbb Z[2]$ and the fundamental theorem of finitely generated abelian groups, it follows at once that $\tilde{H}(X)\simeq\tilde{H}(Y)$ abstractly. The only problem I have is that it doesn't seem to me that we get a map $X\to Y$ that induces isomorphisms on $\pi_1$ and is a homology equivalence, actually, I don't see why there should be a homology equivalence $X\to Y$ at all.

Answer 1

The answer is no. An easy example is provided in Allen Hatcher's Algebraic Topology, Example 4.35.

http://www.math.cornell.edu/~hatcher/AT/ATch4.pdf

There, a CW-complex $X$ is formed by attaching an appropriate $(n+1)$-cell to the wedge $S^1 \vee S^n$ of a circle and an $n$-sphere. The inclusion of the $1$-skeleton $S^1 \to X$ induces an isomorphism on integral homology and on homotopy groups $\pi_i$ for $i < n$ but not on $\pi_n$.

Answer 2

There are two finite 2-dimensional complexes $A, B$ which are not homotopy-equivalent but are homology-equivalent, i.e., there exists a continuous map $$ f: A\to B $$ inducing isomorphism of fundamental groups and homology groups. See the last paragraph (page 522) of this paper, the actual example is due to Dunwoody.

Edit 1: The homology equivalence part follows from Dyer's paper, see the link above (read the very last paragraph of his paper). Dyer's results are by no means obvious: He proves that equality of Euler characteristics plus one extra condition ($m=1$, whatever $m$ is) imply homology equivalence of 2-d complexes with isomorphic fundamental groups. Then Dyer verifies his conditions in the case of Dunwoody's example (the only nontrivial condition is $m=1$, since equality of Euler characteristics is clear).

Edit 2: The correct version of the "homological Whitehead's theorem" is indeed requires a cohomology isomorphisms with sheaf coefficients, see the discussion and reference here.

Answer 3

...it turns out there are non-trivial high dimensional smooth knots $S^n\subset S^{n+2}$ such that $\pi_1$ of the knot complement $X$ is infinite cyclic. The inclusion of the meridian $S^1 \subset X$ is a homology isomorphism, a $\pi_1$ isomorphism but is never a weak equivalence since by a result of Levine if it were then the knot would be trivial (we should assume $n \ge 3$ here).

John Klein @ https://mathoverflow.net/a/104534/1556

Answer 4

This is meant as a question directed at @studiosus and anybody who's familiar with his example.

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M. Dyer's theorem seems to state, as a special case, that

Theorem Suppose $X,Y$ are finite, at most 2 dimensional, connected CW complexes with isomorphic fundamental groups (so called $[G,2]$-complexes, where $\pi_1(X)\simeq G\simeq\pi_1(Y)$). If $X$ and $Y$ have the same Euler characteristic $\chi$, and $\chi>\chi_{\min}(G,2)$, then $X$ and $Y$ are homology equivalent.

(From what I understood, when $\chi>\chi_{\min}(G,2)$ then automatically $m=1$, see the first sentence under his theorem)

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To the trefoil knot group presentation $G=\langle a,b\mid a^3b^{-2}=1\rangle$ corresponds a $[G,2]$-complex with one $0$ cell, two $1$ cells and one $2$ cell ($Y$ in Dyer's paper) so that $\chi(Y)=0\geq\chi_{\min}(G,2)$, and thus with $X=Y\vee S^2$, which is manifestly still a $[G,2]$-complex, $1=\chi(X)>0\geq\chi_{\min}(G,2)$. This $X$ corresponds to the presentation $G=\langle a,b\mid a^3b^{-2}=1, 1=1\rangle$.

There is another presentation for $G$ defined in Dunwoody's paper with a complicated presentation (still on two generators and two relations, it's on the last page of his paper, so with $\chi=1$), and I take it, that the realization $Z$ of this presentation has $X\not\simeq Z$ (while, I take it, $X\vee S^2\simeq Z\vee S^2$).

So since $X$ and $Z$ are $[G,2]$-spaces with identical Euler characteristic equal to $1>0\geq\chi_{\min}(G,2)$, they should automatically have $m=1$ and thus be homology equivalent by M.Dyer's therem (for him this means: there is a map that is an isomorphism on $\pi_1$ and $H_*$), yet they aren't homotopy equivalent by Dunwoody's theorem.

Is this correct?

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$\chi_{\min}(G,2)$ is the minimum over all Euler characteristics of $[G,2]$-complexes.The fact that $\chi_{\min}(G,2)$ is well defined strikes me as believable, at least not implausible, and I'm willing to accept it. Indeed, it seems that every $[G,2]$-complex $X$ should define a presentation $P$ of $G$, and a subcomplex $X'$ with the same $1$-skeleton, but possibly less $2$-cells, thus with smaller Euler characteristic, homotopy equivalent to the $[G,2]$-complex obtained from the presentation $P$. So it seems that, in order to show $\chi_{\min}(G,2)$ is well defined, one can forget about general $[G,2]$-complex and only consider those associated to presentations of $G$. Adding a new relation increases the Euler characteristic by one. Adding a new generator should introduce at least one new relation, thus at least one new $2$-cell, so it seems that ($\#$ of $2$-cells)$-$($\#$ of $1$-cells)$+1$ can only increase by adding new generators and relations to a given presentation of $G$. So it seems one can do the following:

1. From any $[G,2]$-complex $X$ extract a subcomplex $X'$ with fewer $2$ cells and identical $1$ skeleton (thus reducing $\chi$) that remains a $[G,2]$-complex and is homotopy equivalent to a $[G,2]$-complex associated to a presentation of $G$.
2. From a given presentation of $G$ extract a "minimal" presentation of $G$. Intuitively one that can't be reduced by either considering less generators or relations. From the intuitive argument above it follows that adding new generators or relations to a presentation can only increase teh Euler characteristic, so passing to "minimal" presentations should reduc the Euler characteristic of the associated $[G,2]$-complex.
3. Only study the Euler characteristics of $[G,2]$-complexes associated to "minimal" presentations of $G$.
4. Somehow show that they all have the same Euler characteristic, which then must be $\chi_{\min}(G,2)$.

EDIT 1 Proposition 1 on page 14 of these notes gives a lower bound for the Euler characteristic of any $[G,2]$-complex associated to any presentation of (finitely presentable) $G$ in terms of the rank of its first and second homology groups. So steps 3 and 4 aren't necessary, 4 might even be wrong.