4
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ 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. $\endgroup$
    – Daniel Fischer
    Jan 19, 2014 at 13:32
  • $\begingroup$ See also math.stackexchange.com/q/99302 and math.stackexchange.com/q/88943 $\endgroup$
    – Grigory M
    Jan 19, 2014 at 13:33
  • $\begingroup$ ...and math.stackexchange.com/q/401505 $\endgroup$
    – Grigory M
    Jan 19, 2014 at 13:43
  • $\begingroup$ @DanielFischer thanks alot $\endgroup$
    – Eli Elizirov
    Jan 20, 2014 at 19:03
  • $\begingroup$ 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. $\endgroup$
    – user584285
    Feb 24, 2019 at 16:58

2 Answers 2

Reset to default
8
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Just looking at the map induced on the 2nd Homology group should suffice to show that they aren't homotopy equivalent. $\endgroup$
    – Brozovic
    Jul 19, 2020 at 12:03
  • 1
    $\begingroup$ @Brozovic, yes, I'd say that homology counts as a tool ;) $\endgroup$
    – Carsten S
    Jul 19, 2020 at 13:32
4
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .