Prove that the Torus is not homotopy equivalent to $S^1\vee S^1\vee S^2$.

I need to show that a homotopy equivalence between them doesn't exist, but it seems like the homology groups of the spaces are equal. How can I show it? maybe using fixed point theorem?

  • 3
    $\begingroup$ Have you considered using homotopy groups (in particular the fundamental group)? $\endgroup$ – Amitai Yuval Jul 23 '15 at 15:21
  • $\begingroup$ How can I use them? $\endgroup$ – Dante Jul 23 '15 at 15:23
  • 4
    $\begingroup$ Homotopy equivalent spaces have isomorphic homotopy groups. Hence: If two connected spaces have non-isomorphic fundamental groups, they are not homotopy equivalent. $\endgroup$ – Daniel Fischer Jul 23 '15 at 15:46

The space $S^1\vee S^1\vee S^2$ is the wedge sum of two circles and one sphere. In particular this space is also known as mouse space.

It is quite easy to see that the fundamental group of $S^1\vee S^1\vee S^2$ is $\mathbb{Z}*\mathbb{Z}\cong\langle a,b| \emptyset \rangle$ ( you can use the Seifert-VanKampen theorem to see it ); whereas the fundamental group of the torus is $\mathbb{Z}\times\mathbb{Z}=\langle a,b | [a,b]=1\rangle$, which is not isomorphic to the previous.

If there was a homotopic equivalence between them; their fundamental groups needs to be isomorphic; but this is not true. So there is not any homotopic equivalence.

  • $\begingroup$ Thank you for the correction :) I don't remember where I read that name; maybe on the Hatcher's book: Algebraic Topology. $\endgroup$ – InsideOut Jul 23 '15 at 16:55

Working over $\mathbb{Z}$ throughout, the cohomology ring $H^*(T^2) = H^1(S^1)\otimes H^1(S^1)$ contains a nontrivial cup product (above dimension $0$), while $H^*(S^1\vee S^1\vee S^2)$ does not.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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