TITLE: Prove that the Torus is not homotopy equivalent to $S^1\vee S^1\vee S^2$
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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.