24
$\begingroup$

(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.

$\endgroup$
12
  • $\begingroup$ This generalisation seems unlikely (mainly because the homotopy group version of Whitehead's theorem seems close to a 'sharp' result), but I can't think of a counterexample at the moment. $\endgroup$
    – Dan Rust
    Apr 13, 2014 at 12:26
  • $\begingroup$ Could you outline why it's enough to show the lift $\tilde{f}$ is a homology equivalence? I understand that this would show $\tilde{f}$ is a homotopy equivalence, but why would there be a homotopy inverse $\tilde{g}$ to $\tilde{f}$ which respects the projection mappings? Is there an equivariant Whitehead's theorem? $\endgroup$ Apr 14, 2014 at 23:45
  • $\begingroup$ @JasonDeVito If $\tilde{f}$ is a homology equivalence, it follows from simple connectedness and Whitehead's homology theorem that it is a homotopy equivalence $\widetilde{X}\to\widetilde{Y}$. Since a covering projection induces an isomorphisms $p_*:\pi_n(\widetilde{X})\simeq\pi_n( X)$ for all $n\geq 2$, the commutative diagram $$\begin{array}{ccc}\pi_n(\widetilde{X})&\xrightarrow{\tilde{f}_*}&\pi_n( \widetilde{Y} )\\\downarrow&&\downarrow\\\pi_n(X)&\xrightarrow{f_*}&\pi_n(Y)\end{array}$$ that $f_*$ is an isomorphism for all $\pi_n$, $n\geq 2$. $\endgroup$ Apr 15, 2014 at 0:23
  • $\begingroup$ @JasonDeVito Since by hypothesis $f_*$ is an isomorphism on $\pi_1$ and $\pi_0$, $f$ is a weak equivalence between CW complexes and Whitehead's theorem tells us that it is a homotopy equivalence. $\endgroup$ Apr 15, 2014 at 0:26
  • 1
    $\begingroup$ I think that a counterexample is given at the top of p. 281 in McDuff and Segal's article on group completion (available here: maths.ed.ac.uk/~aar/papers/mcdsegal.pdf) but maybe there is an issue of basepoints. They consider the shift map on the infinite symmetric group (finitely supported permutations of $\mathbb{N}$). $\endgroup$
    – Dan Ramras
    Apr 15, 2014 at 3:04

5 Answers 5

16
$\begingroup$

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 $\pi_n$.

$\endgroup$
1
  • $\begingroup$ Thanks for your answer, I somehow hadn't noticed it until today ... $\endgroup$ Oct 22, 2016 at 15:17
9
+50
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Thanks for the reference, I have a question though, too long for a comment. I added it in the main body of the question, I'd appreciate it if you could take a look at it! $\endgroup$ Apr 15, 2014 at 16:54
  • $\begingroup$ Just in case you already read my question, there was a mistake in my EDIT to the original question, the source I linked to wasn't the article by Dunwoody. $\endgroup$ Apr 15, 2014 at 19:16
  • $\begingroup$ @OlivierBégassat: See the edit. $\endgroup$ Apr 15, 2014 at 20:11
  • $\begingroup$ Ok, thank you for the explanation. I'll try to understand Dyer's result tomorow. $\endgroup$ Apr 15, 2014 at 20:44
  • $\begingroup$ I left an answer to tell you what I understood from your examples, could you take a look at it and correct my understanding of it? $\endgroup$ Apr 20, 2014 at 16:20
7
$\begingroup$

...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

$\endgroup$
4
  • $\begingroup$ That's very interesting. I asked John Klein for a reference in the comment section of his answer; do you know of a reference for this? $\endgroup$ Apr 18, 2014 at 12:38
  • $\begingroup$ @Olivier The first part (the inclusion of the meridian is a homology iso) follows from Alexander duality. The reference for the second part (codimension 2 sphere is unknotted iff its complement is homotopy equivalent to $S^1$) is, I believe, J. Levine, Unknotting spheres in codimension 2, Topology 4, 9–16 (1965) $\endgroup$
    – Grigory M
    Apr 18, 2014 at 12:52
  • 2
    $\begingroup$ Actually, the second part is due to John Stallings "On topologically unknotted spheres", Annals of Math., 1963. $\endgroup$ Apr 18, 2014 at 16:39
  • $\begingroup$ @studiosus Oh, looks like you're right $\endgroup$
    – Grigory M
    Apr 18, 2014 at 16:45
4
$\begingroup$

You know already that the answer for your question is no, nevertheless, there is a generalisation which you might like:

Suppose that $X$, $Y$ are path-connected, simple CW-complexes. If $f:X\rightarrow Y$ induces isomorphism in $H_n$ for all $n\geq 0$, then $f$ is a homotopy equivalence.

$\endgroup$
9
  • $\begingroup$ Hi Mihail, I only noticed your answer today, sorry about that. What is the definition of a simple CW complex ? $\endgroup$ Jul 31, 2019 at 12:39
  • $\begingroup$ @OlivierBégassat A topological space is called simple if the action of a fundamental group on higher homotopy groups is trivial. For example, this is the case when the fundamental group is trivial. $\endgroup$
    – Mihail
    Aug 1, 2019 at 22:54
  • $\begingroup$ Thanks! ${}{}{}$ $\endgroup$ Aug 1, 2019 at 22:55
  • 1
    $\begingroup$ I think that one also requires that the fundamental group is abelian usally. At least, that is the generalization that I know $\endgroup$ Nov 17, 2020 at 23:47
  • 1
    $\begingroup$ It acts on itself my inner automorphisms $\endgroup$ Mar 15, 2021 at 3:43
3
$\begingroup$

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


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)


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?



$\chi_{\min}(G,2)$ 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.

$\endgroup$
5
  • $\begingroup$ I did not look at the staff before Edit 1, but what you say after Edit 1 is correct. The trick is to show that $0=\chi_{min}(G,2)$ in Dunwoody's example. You can see this from the fact that the presentation complex for $G$ is $K(G,1)$. $\endgroup$ Apr 20, 2014 at 18:13
  • $\begingroup$ @studiosus But as I understand it, it is only required that the Euler characteristics be equal and $m=1$, and it seems to me he is saying that $m=1$ whenever $\chi>\chi_{\min}(G,2)$. Since we know by example (the complex associated to the presentation $\langle a,b\mid a^3=b^2\rangle$) that there is a $[G,2]$-complex with $\chi=0$ we have $0\geq\chi_{\min}(G,2)$, and so, for any $[G,2]$-complex with $\chi\geq 1$, $m=1$. My point is that it apparantly doesn't matter that $\chi_{\min}(G,2)=0$ or not, it only matters that it be less than our $\chi$. Am I correctly interpreting the article? $\endgroup$ Apr 20, 2014 at 18:59
  • $\begingroup$ Also, are the spaces $X$ and $Z$ the actual spaces he uses to find homology equivalent (in $\pi_1$ and $H_*$), non homotopy equivalent spaces? $\endgroup$ Apr 20, 2014 at 19:02
  • $\begingroup$ True on both counts. $\endgroup$ Apr 20, 2014 at 19:40
  • $\begingroup$ Ok, if you say so ^^ I awarded you the bounty, I guess I need to read the articles more in depth to believe their claims and my own explanation ^^ $\endgroup$ Apr 20, 2014 at 19:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .