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A common caution about Whitehead's theorem is that you need the map between the spaces; it's easy to give examples of spaces with isomorphic homotopy groups that are not homotopy equivalent. (See Are there two non-homotopy equivalent spaces with equal homotopy groups?). It's surely also true that the pair (homotopy groups, homology groups) is not a complete invariant, but can anyone give examples? That is, I'm looking for spaces $X$ and $Y$ so that $\pi_n(X) \simeq \pi_n(Y)$ and $H_n(X;\mathbb{Z}) \simeq H_n(Y; \mathbb{Z})$ but $X$ and $Y$ are still not (weakly) homotopy equivalent.

(Easier examples are preferred, of course.)

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It seems to me that Allen Hatcher's examples in mathoverflow.net/questions/4665/… gives what you want. These are total spaces of spherical fibrations over spheres. The simplest examples I think are $S^3 \times S^3$ and the total space of the fibration $S^3 \to E\to S^3$ given by the unit sphere bundle of the Hopf bundle plus the trivial bundle. –  John Klein Jan 26 '11 at 21:25
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Since the homology groups of a closed 3-manifold are determined by the fundamental group, any pair of non-homotopy-equivalent 3-manifolds with the same homotopy groups should work, like the lens spaces L(5,1) and L(5,2) I mentioned in a comment at the above link. –  Steven Sivek Jan 26 '11 at 21:39
    
$L(5,1)$ and $L(5,2)$ are an excellent example, although I didn't see them at any link. –  Dylan Thurston Jan 27 '11 at 7:05
    
It's the last comment on my answer here: mathoverflow.net/questions/3540/… –  Steven Sivek Jan 27 '11 at 13:18
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3 Answers

up vote 22 down vote accepted

Following up on John's comment, one can consider $S^2$-fibrations over $S^2$. There are two of them since such fibrations are classified by $\pi_1(\textrm{Diff}^{+}(S^2))=\mathbb{Z}_2$. One of them is $S^2\times S^2$ while the other can be shown to be the connected sum of $\mathbb{CP}^2$ and $\overline{\mathbb{CP}}^2$. These two spaces have the same homology. They have the same homotopy groups since they both form the base of a $S^1$-fibration with total space $S^2 \times S^3$. However, the intersection forms are not equivalent and hence they are not homotopy equivalent.

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Another reason they have the same homotopy groups is that both fibrations have a section, so their long exact sequences of homotopy groups split. –  Allen Hatcher Jan 26 '11 at 22:45
    
Very nice, this is what I was looking for. –  Dylan Thurston Jan 27 '11 at 7:04
    
I realized I goofed on the dimension of my base space above. Thanks for the fix. –  John Klein Jan 27 '11 at 14:56
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In this context, I guess it's better to talk about the cup product on cohomology than the intersection form, as the cup product is manifestly homotopy invariant. –  Dylan Thurston Jan 27 '11 at 17:22
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Take a finite group $G$, a finite-dimensional $\mathbb{Q}$-vector space $V$ and two representations actions of $G$ that are inequivalent but the spaces of coinvariants both have the same dimension, say zero.

The first concrete example that comes to my mind is $G=Z/4$, $V=Q[i]$, with the two actions where the generator acts by $-1$ or $i$. These two actions are not even equivalent under outer automorphisms of $G$.

Let $n \geq 3$ be an odd integer and consider the Eilenberg-Mac-Lane space $K(V,n)$, which inherits two $G$-actions. The two Borel-constructions $EG \times_G K(V,n)$ have the same homotopy groups. But the $n$th homotopy groups are not isomorphic when considered as a $\pi_1$-module; so these two spaces are not homotopy equivalent.

The homology can be computed from the Leray-Serre spectral sequence of the fibration $EG \times_G K(V,n) \to BG$. Recall $E^{2}_{pq}= H_p (G; H_p (K(V;n)))$.

To begin with, $\tilde{H}_* (K(V,n), \mathbb{Z}) =V$ if $*=n$ and $0$ otherwise. Thus $E^{2}_{pq}=0$ unless $q=0$ (then it is the group homology $H_p(G;Z)$ or $(p,q)=(0,n)$, in which case it is $V_G=0$. Thus the projections $EG \times K(V,n) \to BG$ are homology equivalences.

Finally note that the Eilenberg Mac Lane spaces can be realized as abelian topological groups and $G$ acts fixing the basepoint. Thus the maps $EG \times_G K(V;n)) \to BG$ have sections, which are homology equivalences as well. So the construction even produces a homology equivalence between two spaces with abstractly isomorphic homotopy groups.

EDIT: since there is a homology equivalence between these two spaces, it follows that the homology with coefficients, the cohomology rings and even the actions of the Steenrod algebra for all primes are isomorphic.

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The homology with coefficients will not be the same: one can see this already in your example, taking the coefficient system given by one of the representations. –  Oscar Randal-Williams Jan 27 '11 at 15:11
    
There is a theorem as follows: $f:X \to Y$ a $\pi_1$-isomorphism and an isomorphism for homology with all local coefficient systems, then $f$ is a weak homotopy equivalence. This follows from the natural isomorphism H_p(X,Z\pi_1 (X))\cong H_p (\tilde{X}). –  Johannes Ebert Jan 27 '11 at 15:33
    
Ah, my mistake. I have a habit of understanding "with coefficients" as "with local coefficients". –  Oscar Randal-Williams Jan 27 '11 at 16:38
    
Thats a good habit, but the universal coefficient theorem does not work for all sorts of coefficients. In my example, if you take $ZG$ as coefficients, then the map won't induce an isomorphism, because $H_* (EG \times_G K(V;n); ZG) \cong H_* (K(V;n)) = V$; and my map induces $0$ on $V$. –  Johannes Ebert Jan 27 '11 at 17:11
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A fairly easy obstruction is the action of the fundamental group on the homotopy groups. Here is an example where the homotopy groups are isomorphic as abstract groups, but not preserving the action of the fundamental group.

Form a fibration over a circle with fiber a wedge of spheres $S^2\vee S^2$. That is, take the wedge, cross with an interval, and identify the ends by a homotopy equivalence of the wedge. Homotopy equivalences are parameterized by $GL_2(\mathbb Z)$, the action on homology. The homotopy type of the space remembers the monodromy as the action of the fundamental group on the homotopy groups. The homotopy groups are that of the universal cover, which does not depend on the choice of monodromy. Compute the homology by the Serre spectral sequence. This involves the homology of $\mathbb Z$ acting on $\mathbb Z^2$ by the monodromy. If the monodromy is hyperbolic, the homology vanishes and the space has the homology of the circle. Thus two different hyperbolic matrices give spaces with isomorphic homotopy and homology groups.

In a completely different direction, there are examples of pairs of simply connected spaces such that the Postnikov truncations are equivalent, but the inverse limits of their Postnikov towers are not, but I think such examples have to be pretty large.

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This is basically the same example as Johannes. The difference is that he used rational vector spaces. That makes the group homology easy at the cost of compactness. To get both advantages, one could let the group act on a vector space over a finite field of characteristic coprime to its cardinality; but then one also has to approximate $K(\mathbb Z/p,n)$ by something compact –  Ben Wieland Jan 26 '11 at 22:14
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