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Could someone show an example of two spaces X and Y which are not of the same homotopy type, but nevertheless \pi_q(X)=\pi_q(Y) for every q? Is there an example in the CW complex or smooth category?

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5 Answers 5

up vote 23 down vote accepted

There are such spaces, for example X = S^2 \times RP^3, Y = S^3 \times RP^2. (These are both smooth and CW-complexes.)

Whitehead's Theorem says that for CW-complexes, if a map f : X \to Y induces an isomorphism on all homotopy groups then it is a homotopy equivalence. But, as the example above shows, you need the map. Such a map is called a weak homotopy equivalence.

(Whitehead's Theorem is not true for spaces wilder than CW-complexes. The Warsaw circle has all of its homotopy groups trivial but the unique map to a point is not a homotopy equivalence.)

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Also the long line has zero homotopy groups, but is not contractible. It's too long! (but it is also a manifold, sort of...) –  Chris Schommer-Pries Oct 31 '09 at 15:37
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All of these examples involve a bit of cleverness, so I thought I'd point out a more straightforward way to construct counterexamples. If X is any space, we can build a space X' = K(π0X, 0) × K(π1X, 1) × K(π2X, 2) × ... which has the same homotopy groups as X, but which is usually not weakly equivalent to X. Pretty much any invariant of spaces other than the homotopy groups will distinguish them. For instance, if X = S2, then H3(X) = 0, but X' contains K(π3S2, 3) = K(Z, 3) as a retract, so H3(X') contains H3(K(Z, 3)) = Z as a retract and cannot be 0.

Of course, this doesn't produce geometrically nice spaces like smooth manifolds.

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S3 x RP2 and RP3 x S2 are both smooth 5-manifolds with fundamental group Z/2 and universal cover S3 x S2, so their homotopy groups are all the same. On the other hand, only the latter is orientable since RP3 is orientable but RP2 isn't, so they have different values on H5 and therefore can't be homotopy equivalent. (I think this example is in Hatcher somewhere.)

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Beaten by 21 seconds -- I guess this really is a standard example! –  Steven Sivek Oct 31 '09 at 13:54
I used the Wikipedia for this example... (But, yes it is the standard example, I think.) –  Daniel Groves Oct 31 '09 at 13:59
Here's another example for the sake of originality: two 3-dimensional lens spaces L(p,q1) and L(p,q2) are homotopy equivalent iff q1*q2 = \pm n^2 (mod p) for some n, so L(5,1) and L(5,2) are not homotopy equivalent. But they both have fundamental group Z/5Z and universal cover S^3, so their homotopy groups are the same. –  Steven Sivek Oct 31 '09 at 14:00
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Along the same lines: you can produce examples involving simply connected finite CW-complexes, by using S1 actions: CPm x S2n+1 and S2m+1 x CPn have the same homotopy groups, for instance.

It's easy to produce examples with arbitrarily high connectivity: take the homotopy fiber X of any non-trivial map K(A,m)->K(B,n+1), with n>m. Then X and Y=K(A,m) x K(B,n) have the same homotopy groups, but are not homotopy equivalent.

Here's a new question: for given k, can you find a pair of k-connected finite CW-complexes which have the same homotopy groups, but aren't homotopy equivalent?

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It's easy to construct myriads of such examples using finite $T_0$ topological spaces. These are equivalent in a natural way to finite posets (check this wikipedia article). With a poset P we may associate an abstract simplicial complex K(P) whose simplices are the chains of P. It turns out that the geometric realization of K(P) and P are weakly homotopy equivalent (there is a continuous map from |K(P)| to P inducing isomorphisms on homotopy groups). However, these spaces are not homotopy equivalent (unless they are both homotopy equivalent to a discrete space).

Moreover, examples of weakly homotopy equivalent finite spaces that are not homotopy equivalent are also easy to give.

The following notes by J.P. May are a nice introduction to this topic: http://www.math.uchicago.edu/~may/MISC/FiniteSpaces.pdf , http://www.math.uchicago.edu/~may/MISC/SimpCxes.pdf

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