Question

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?

Answer 1

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.)

Answer 2

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

Answer 3

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.

Answer 4

Along the same lines: you can produce examples involving simply connected finite CW-complexes, by using S¹ actions: CP^m x S²ⁿ⁺¹ and S^2m+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?

Answer 5

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.)