# Isomorphic fundamental groups result in homeomorphism?

I know that the fundamental group of homeomorphic spaces are isomorphic. Is the converse true? I mean, can we say the two spaces with isomorphic fundamental groups are homeomorphic?

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Nope, consider a point and $\mathbb{R}$. :) – Alex Youcis Mar 10 '14 at 9:26
+1 : I really like these naïve questions from beginners. And "naïve" is here meant as a compliment: I think we older mathematicians tend to forget how difficult concepts were, which we now dismiss as "trivial" . – Georges Elencwajg Mar 10 '14 at 9:39
...and easy examples like that by Alex are the best of answers: +1 for his comment too . – Georges Elencwajg Mar 10 '14 at 9:40
@GeorgesElencwajg I agree. I'm wondering why the OP didn't ask about homotopy equivalence type though instead of homeomorphism type, as one would think homotopies have been covered before the fundamental group. – Dan Rust Mar 10 '14 at 9:43

Other answers give good counterexamples ($\mathbb{R}$ and a point, $\mathbb{S}^2$ and a point), so I'm just going to write a little expository answer about searching for converses.

The fundamental group is far too weak to detect homeomorphism type. In fact, knowing the entire sequence of homotopy groups is too weak: you can always "inflate" the space, e.g., as in Daniel Rust's answer, in such a manner that the result has the same homotopy groups but is not homeomorphic to the original space.

This method of building a counterexample violates compactness, and compact spaces are nice, so you might try adding a compactness assumption. Is knowing that two compact spaces have isomorphic homotopy groups enough to conclude that they are homeomorphic? Or even homotopy-equivalent?

Nope, still not good enough. In fact, lens spaces provide examples of non-homeomorphic, non-homotopy-equivalent compact manifolds with the same dimension and homotopy groups.

To my limited knowledge, here's the best general converse available: Whitehead's theorem.

If $X$ and $Y$ have the homotopy type of CW complexes and if a map $f:X\to Y$ induces an isomorphism of all homotopy groups, then $f$ is a homotopy equivalence.

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The closest one can get to your statement is probably Whitehead's theorem:

If a map $f$ between two connected CW complexes $X$, $Y$ induces isomorphisms on all homotopy groups, then $f$ is a homotopy equivalence between $X$ and $Y$.

Homotopy groups are the higher-dimensional analogue of the fundamental group. Of course, homotopy equivalence is a weaker condition than the existence of a homeomorphism. Also note that it's note enough for homotopy groups to be isomorphic; one map must induces all isomorphisms. The only "exception" is when all homotpy groups of a CW complex $X$ are trivial. In this case, the constant map from $X$ to a point induces isomorphisms on all homotopy groups, and therefore $X$ is contractible (homotopy equivalent to a point).

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Going another direction from the answers, there are important cases where this is true.

For example, if you restrict "space" to mean "hyperbolic manifolds of dimension at least three and of finite volume", then the answer is, surprisingly, yes and goes by the name of Mostow Rigidity

If $M$ and $N$ are finite volume hyperbolic manifolds each of dimension at least 3, and if $f:\pi_1(M)\rightarrow \pi_1(N)$ is an isomorphism, then $f$ is induced by a unique isometry between $M$ and $N$.

In particular, in such a case, $M$ and $N$ are homeomorphic (and more!).

Further, a generalization of this is known as the Borel Conjecture:

Suppose $M$ and $N$ are closed manifolds with $\pi_k(M) = \pi_k(N) = 0$ for $k\geq 2$. Then $M$ and $N$ are homeomorphic.

It follows from Whitehead's theorem (see the other answers) and slightly more work that $M$ and $N$ are necessarily homotopy equivalent. The Borel conjecture asserts that for closed manifolds, they are necessarily homeomorphic. As the name implies, it is still a conjecture.

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... with some important recent advances in dimensions different from 4. (Farrel and Jones, Perelman, Lueck et al.) – studiosus Mar 10 '14 at 19:05
@studiosus: I'm actually not familiar with what you're referring to. Could you elaborate please? – Jason DeVito Mar 11 '14 at 1:56
I posted this as another answer since it is a bit long. – studiosus Mar 11 '14 at 2:36

This is an addendum to Jason DeVito's answer; it concerns the current status of Borel conjecture:

If two closed aspherical (i.e., with contractible universal covers) manifolds $M, N$ have isomorphic fundamental groups then they are homeomorphic.

The strongest to date result towards Borel conjecture in dimensions $\ge 5$ is in the paper

"The Borel Conjecture for hyperbolic and CAT(0)-groups" by Bartels and Lueck which can be found here.

Their theorem states that Borel conjecture (in dimensions $\ge 5$) holds for CAT(0) and hyperbolic groups; this result includes the earlier results by Farrel and Jones on Borel conjecture for nonpositively curved manifolds.

Furthermore, Borel conjecture holds in dimension 3 thanks to Perelman's solution of Thurston's Geometrization Conjecture. People usually are aware of the fact that Perelman proved Poincare conjecture in dimension 3, but his theorem proves much more than this. Here is how Borel conjecture in 3d follows from Perelman's work:

Perelman's result immediately implies that every closed 3-manifold $M$ with contractible universal cover is one of the following:

1. Hyperbolic.

2. Seifert.

3. Haken.

Borel conjecture for hyperbolic manifolds follows from Mostow rigidity theorem. For Haken manifolds Borel conjecture was proven by Waldhausen (see here). For Seifert manifolds with contractible universal cover Seifert invariants (which completely determine topology of the manifolds) can be read off the fundamental group (my guess is that this was known to Seifert).

Borel conjecture in dimension 4 is wide-open as far as I know.

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The Bartels-Lueck paper on page 2 claims that the Borel conjecture holds in dimension 4 for "good" CAT(0) and hyperbolic groups. "Good" is in the sense of Freedman, in the "Disk theorem" paper (I haven't read it). – Neal Mar 11 '14 at 2:48
Thanks for adding this! – Jason DeVito Mar 11 '14 at 4:03
@Neal: I do not know much about the 4-d topology, but from what I remember, the only groups known to be "good" in Freedman's sense are the ones of subexponential growth. Among hyperbolic and CAT(0) groups this leaves only virtually abelian groups. – studiosus Mar 13 '14 at 16:49

Let $X'=X\sqcup Y$ where $Y$ is a space with cardinality greater than $X$ and make the basepoint of $X'$ still lie inside the copy of $X$. Then $\pi_1(X')=\pi_1(X)$ but they are not homeomorphic because of different cardinalities.

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There are certainly much simpler examples of this, and ones where the spaces are connected. Your example makes it seem that if we impose connectivity we have a chance of the claim being true. – Alex Youcis Mar 10 '14 at 9:28
I wanted an examples which would work for any space. – Dan Rust Mar 10 '14 at 9:29
Ok, it's correct regardless, I just wanted to point out :) +1 – Alex Youcis Mar 10 '14 at 9:29
@AlexYoucis Maybe I should have taken the product instead of the coproduct. At least then for connected $Y$ that leaves $X'$ connected when $X$ is connected. – Dan Rust Mar 10 '14 at 9:31
You can wedge with a wedge of more-than-points-in-the-space intervals. – Mariano Suárez-Alvarez Mar 10 '14 at 11:38

No. In fact, there are spaces that are not homotopy equivalent (isomorphic in the homotopy category of topological spaces) but have the same fundamental group. Take $X=S^2$ and $Y=\{\operatorname{pt}\}$.

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Nitpick: Maps are homotopic; spaces are homotopy-equivalent. – Neal Mar 10 '14 at 10:33