I am just beginning learning algebraic topology after finishing point-set topology. I did not have background in group theory(only know some basic fact about it). For this question, can I just claim if torus and klein bottle do not have same fundemental group,then they are not homotopy equivalent? This is the first time I have encountered such questions and if anyone can give explanations in details, then it may help me to learn greatly. Thank you.

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    $\begingroup$ $\pi_1$ of torus is abelian, but $\pi_1$ of a Klein bottle is not. Alternatively, you can calculate the homology groups. Or show that Klein bottle is not orientable. .... $\endgroup$ – user10354138 Jul 30 '20 at 14:57
  • $\begingroup$ In group-theory land, your two-sheeted cover just says that $\pi_1$ of your Klein bottle contains $\pi_1$ of your torus with index two. But this doesn't help you that much, as in fact both groups ($\pi_1$ of your Klein bottle and of your torus) each contain a copy of $\pi_1$ of your torus with index two. $\endgroup$ – user1729 Jul 30 '20 at 15:21
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    $\begingroup$ @user10354138 showing non-orientability is not enough on its own. For example, the annulus is homotopy equivalent to the Möbius band, but the former is orientable while the latter is not. $\endgroup$ – Tom Sharpe Jul 30 '20 at 15:39
  • $\begingroup$ The fundamental group is homotopy invariant, meaning that homotopy equivalent spaces have the same (isomorphic) fundamental group. That is not the same as saying non-homotopy equivalent spaces have non-isomorphic fundamental groups, which is not true. $\endgroup$ – Javi Jul 30 '20 at 21:53

It seems like the other answers are trying to prove that the Klein bottle and the torus don't have the same fundamental group which, as far as I understand, is not your question (EDIT : per Jason's comment : it's no longer your question, after an edit to the original one, which explains the other answers)

If I understand correctly, you know that and want to conclude from this that they are not homotopy equivalent. You are correct that one can conclude that way, indeed :

Let $f:X\to Y$ be a homotopy equivalence between spaces. Then for any $x\in X$, the induced morphism $f_* : \pi_1(X,x)\to \pi_1(Y,f(x))$ is an isomorphism

In particular, the contrapositive states that if the fundamental groups are different, the spaces are not homotopy equivalent.

The proof of the claim can be found in any introductory algebraic topology text. The idea is fairly easy, the only possible problem being a problem of basepoints. Indeed, the homotopy equivalences need not be basepoint preserving, and the homotopies $gf\simeq id_X$ need not be a pointed homotopy either.

There are various ways of dealing with that issue, a particularly nice one is to work with the fundamental groupoid instead of the fundamental group, and with this tool, the basepoint problem kind of goes away.

  • $\begingroup$ To be fair to the other answers, it looks like the precise question has been edited. $\endgroup$ – Jason DeVito Jul 30 '20 at 20:29
  • $\begingroup$ @JasonDeVito : you are correct, I hadn't noticed that ! $\endgroup$ – Maxime Ramzi Jul 30 '20 at 20:31
  • $\begingroup$ @user1729 : yes, I noted Jason's comment and then went to check the edit history of the question, and my first paragraph was no longer adequate. I didn't alter it but added an edit to clarify the situation, I hope that's ok $\endgroup$ – Maxime Ramzi Jul 30 '20 at 20:32
  • $\begingroup$ Whoops, didn't see the other comment. Yes, that's better. $\endgroup$ – user1729 Jul 30 '20 at 21:07

One way to think about this problem is by using cut and paste constructions of the torus and the Klein bottle to compute their fundamental groups and distinguish them as not homotopy equivalent. The torus (left) and Klein bottle (right) are built like so: enter image description here

From the picture, we see that the non-trivial loops we can build on these spaces are given by "linear combinations" of $a$ and $b$, and hence these are the generators of the fundamental groups. To figure out their relations, we again turn to the picture and pay attention to the directions of the arrows to tell us whether we are using a generator or its inverse.

On the left, walking around the boundary over the top gives us that $bab^{-1} = a$. On the right however, that same walk over the boundary on the top gives us the relationship $bab^{-1} = a^{-1}$. Writing out these fundamental groups' presentations of the torus $T^2$ and Klein bottle $K$:

$$\pi_1(T^2) = \langle a,b | ba = ab\rangle \text{ and } \pi_1(K) = \langle a,b | bab^{-1}= a^{-1}\rangle$$

From the presentations, the torus has an abelian fundamental group because the generators commute, but this is not the case for the Klein bottle, hence they are not the same group and thus they are not homotopy equivalent. I hope this helped you visualize the problem better!


As far as I know, the only way to do this is to calculate some algebraic invariant of $T$ and $K$. I don't know if you've covered any algebraic topology, but there are very simple algorithms to calculate both the fundamental group and the homology groups of $T$ and $K$, by realising them as squares with side identifications. See the Wikipedia page on cellular homology, and this page on fundamental groups of CW complexes. It is a standard result of algebraic topology that if two spaces are homotopy equivalent, then their fundamental groups are isomorphic, and their homology groups are isomorphic. But we can check that

$$H_1T\cong\mathbb{Z}^2\ncong\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\cong H_1K.$$

You can check $\pi_1$ and $H_2$ as well, but this is enough to show they cannot be homotopy equivalent.


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