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As a consequence of the Whitehead theorem, Spanier's Algebraic Topology book has on 7.6.25 the following theorem:

A weak homotopy equivalence induces isomorphisms of the corresponding integral singular homology. Conversely, a map between simply connected spaces which induces isomorphisms of the corresponding integral singular homology groups is a weak homotopy equivalence.

Is it possible that a map between (non-simply connected) topological spaces induces an isomorphism on all homology groups, and yet is not a weak homotopy equivalence? If so, I would be glad to have an example.

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4 Answers 4

up vote 7 down vote accepted

If the isomorphism in homology is meant with integral coefficients (or all constant coefficients), you can take the classifying space $BG$ of any non-trivial discrete acyclic group, for example Higman's four-generator four-relator group (see http://www.encyclopediaofmath.org/index.php/Acyclic_group).

By definition of acyclicity, $H_i(BG;\mathbb{Z})=0$ for $i > 0$ whence (similar as in B. Steinberg's answer) $f: BG \to \ast$ induces an isomorphism on integral homology in all degrees, while $\Pi_1(f): G = \Pi_1(BG) \to \Pi_1(\ast)=0$ is zero.

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Let $M$ be the Poincare sphere. It is a 3-manifold which has the homology of $S^3$, but non-trivial fundamental group (the binary icosahedral group). In particular, if we remove a point from $M$ we obtain a space $X$ which has the homology of a point (one can verify this by Mayer--Vietoris) but non-trivial fundamental group.

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Surely $M$ itself, together with the map to $S^3$ which collapses the exterior of a coordinate neighbourhood, already gives an example. –  Mark Grant Aug 11 '12 at 21:51

How about a knot complement? It's known that the inclusion of a meridian $S^1$ (a small circle which links the knot exactly once) into the knot complement $S^3 \setminus S^1$ is a homology isomorphism. It is only an equivalence when the knot is trivial. Of course, this case is detected by the fundamental group.

If you want an example which is not detected by $\pi_1$, 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).

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I like this answer because it shows that inducing an isomorphism on $\pi_1$ and $H_\ast(-;\mathbb Z)$ does not imply the map is a homotopy equivalence. –  John Pardon Aug 12 '12 at 3:54
    
Could you recommend a reference for the high dimensional example? –  Olivier Bégassat Apr 18 at 12:36
    
(related question @ MSE: math.stackexchange.com/questions/750955) –  Grigory M Apr 18 at 12:55
    
Olivier, there are many references. The papers which classify simple knots in high dimensions give a wealth of examples. See e.g., the work of Levine: An algebraic classification of some knots of codimension two. Comment. Math. Helv. 45 1970 185–198. –  John Klein Apr 28 at 14:19

Several decades ago this question led to the concept of nilpotent space, nilpotent CW complex. You can google that, along with "Emmanuel Dror", who later changed his name to Emmanuel Dror Farjoun.

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