I calculated the singular cohomology with coefficients in $\mathbb{Z}$: $$H^n(S^1 \times S^1) = \begin{cases} \mathbb{Z} & \text{if } n = 0,2 \\ \mathbb{Z} \oplus \mathbb{Z} & \text{if } n = 1 \\ 0 & \text{if } n > 2 \end{cases}$$

I want to know how to prove, that one of these cup-products

$$\cup:H^1(S^1\times S^1)\otimes H^1(S^1\times S^1)\to H^2(S^1\times S^1),\; [\alpha]\cup [\beta]=[\alpha\cup \beta]$$ or

$$\cup:H^0(S^1\times S^1)\otimes H^1(S^1\times S^1)\to H^1(S^1\times S^1),\; [\alpha]\cup [\beta]=[\alpha\cup \beta]$$ is nonzero?

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    $\begingroup$ You're going to have trouble proving this because it's not true... like you said, $H^2(S^1) = 0$. $\endgroup$ – user98602 Nov 16 '15 at 5:47
  • $\begingroup$ oh!! you are right, thanks!! I will change this! $\endgroup$ – direct-product Nov 16 '15 at 5:50
  • $\begingroup$ oh no, I meant an other topological space... I'm too tired $\endgroup$ – direct-product Nov 16 '15 at 5:53
  • $\begingroup$ Suppose $M$ is a path-connected space. Then $H^0(M) \cong \Bbb Z$, and you can canonically identify the cup product map $H^0(M) \otimes H^k(M) \to H^k(M)$ with multiplication. $\endgroup$ – user98602 Nov 16 '15 at 5:53
  • $\begingroup$ ok, iff I consider $M=S^1\times S^1$ instead of $S^1$, can I argue that this multiplication is nonzero? $\endgroup$ – direct-product Nov 16 '15 at 5:58

I would suggest that you carefully read Example 3.7. in Hatcher's Algebraic Topology. And by carefully I mean really carefully and multiple times. It explains the case for a closed surface of genus $\geq 1$, and for $=1$ you get $S^1 \times S^1$. It covers the topic really well. If you want more intuition you can also read his introduction to cohomology (ch. 3).

  • $\begingroup$ ok, thanks. I will do that!! $\endgroup$ – direct-product Nov 16 '15 at 12:02

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