# Non-homotopy equivalent spaces with isomorphic fundamental groups

I know that if two spaces $X,Y$ have the same fundamental group : $\pi_1(X,a) \cong \pi_1(Y,b)$ does not imply that they are homotopy equivalent.

But I can't find an example. I was thinking about the Moebius strip and the circle, as both of them have the fundamental group $\mathbb Z$. I can't show that they are not homotopy equivalent, the circle is the Moebius strip boundary so it is not a deformation retract of it, but it also does not necessary means that the are not homotopy equivalent.

Can some one give me an example.

Is there a way to proof that two spaces are not homotopy equivalent or not deformation retract?

• You can retract the Möbius strip to its central circle. Choose spaces with different higher homotpoy groups, for example the two-dimensional sphere $S^2$ and a disk. – Daniel Fischer Jan 19 '14 at 13:32
• – Grigory M Jan 19 '14 at 13:33
• – Grigory M Jan 19 '14 at 13:43
• @DanielFischer thanks alot – Eli Elizirov Jan 20 '14 at 19:03
• FWIW, I want to mention that the fact that two spaces can have the same fundamental group yet not be homotopy equivalent has the same flavor as the negative answer to Mark Kac’s question ‘Can you hear the shape of a drum?’, that is, that a resonator is not characterized by its eigenfrequency set. – EulerSpoiler Feb 24 at 16:58

## 2 Answers

The $2$-sphere and the point have the same fundamental group but are not homotopy equivalent. However, to show that the $2$-sphere is not contractible you need some tools.

The standard way to prove two spaces are not homotopy equivalent is to find some homotopy invariant that distinguishes them. Since they are going to have the same fundamental group, the obvious candidates are homology groups and higher homotopy groups (either of which will tell you a sphere is not homotopy equivalent to a point, as in the answer above).

If you don't have those tools available, you're going to have to do some hard work.