MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
In the early days of math overflow this question would have been welcomed. If this question were asked today it would almost certainly be sent to stack exchange. I propose we close it so that standards are kept consistent and so that no one else comes along to bump it to the front page. I don't think it'll garner any new answers and even if it did this is not the right place to record such examples. – David White Aug 17 '14 at 22:04
I think this is still a perfectly fine question, and should remain open. – Charles Rezk Aug 18 '14 at 2:40
up vote 38 down vote accepted

There are such spaces, for example $X = S^2 \times \mathbb{R}P^3$, $Y = S^3 \times \mathbb{R}P^2.$ (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.)

share|cite|improve this answer
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

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 $X' = K(\pi_0 X, 0) \times K(\pi_1 X, 1) \times K(\pi_2 X, 2) \times \dots$ 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 $X = S^2$, then $H_3(X) = 0$, but $X'$ contains $K(\pi_3 S^2, 3) = K(\mathbb{Z}, 3)$ as a retract, so $H_3(X')$ contains $H_3(K(\mathbb{Z}, 3)) = \mathbb{Z}$ as a retract and cannot be $0$.

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

share|cite|improve this answer
that's pretty cool – bananastack Aug 17 '14 at 17:18

$S^3 \times \mathbb{R}P^2$ and $\mathbb{R}P^3 \times S^2$ are both smooth 5-manifolds with fundamental group $\mathbb{Z}/2$ and universal cover $S^3 \times S^2$, so their homotopy groups are all the same. On the other hand, only the latter is orientable since $\mathbb{R}P^3$ is orientable but $\mathbb{R}P^2$ isn't, so they have different values on $H^5$ and therefore can't be homotopy equivalent. (I think this example is in Hatcher somewhere.)

share|cite|improve this answer
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

Along the same lines: you can produce examples involving simply connected finite CW-complexes, by using $S^1$ actions: $\mathbb{C}P^m \times S^{2n+1}$ and $S^{2m+1}\times \mathbb{C}P^n$ 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 $K(A,m)\to K(B,n+1)$, with $n>m$. Then $X$ and $Y=K(A,m) \times K(B,n)$ 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?

share|cite|improve this answer

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 $|K(P)| \to P$ 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: ,

share|cite|improve this answer

Consider $X=S^1\vee S^3$ and its double cover $X_2$ i.e, attach two copy of $S^3$ one in north pole and one in south pole of $S^1$. Then $\pi_1(X) =\mathbb{Z} = \pi_1(X_2)$. And covering map induced isomorphism in $\pi_n$ for all $n\geq 2$. But they are not homotopically equivalent since their Eular Characteristics are different.

share|cite|improve this answer

The simplest examples I know are the $3$-dimensional lens spaces $L(p,q)$. They display many oddities.

Consider the lens spaces $L(p,q_0)$ and $L(p,q_1)$, $\gcd(p,q_0)=\gcd(p,q_1)=1$. Their fundamental groups are isomorphic to the cyclic group $\newcommand{\bZ}{\mathbb{Z}}$ $\bZ/p\bZ$. Since both these lens spaces have the same universal cover $S^3$, their higher homotopy groups are also isomorphic.

A theorem of Franz-Rueff-Whitehead (see Theorem 2.60 of these notes) shows that $L(p,q_0)$ and $L(p,q_1)$ are homotopy equivalent if and only

$$q_1\equiv \pm \ell^2 q_0\bmod p, $$

for some $\ell\in\bZ$. This reduces the problem to a number theoretic one. For example, $L(5,1)$ is not homotopy equivalent to $L(5,2)$ since $\pm 2$ is not a quadratic residue mod $5$.

On the other hand, the lens spaces $L(7,1)$ and $L(7,2)$ are homotopy equivalent, but they are not homeomorphic.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.