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

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?

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

## 2 Answers

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.

• Thank you for the correction :) I don't remember where I read that name; maybe on the Hatcher's book: Algebraic Topology. – 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.