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Let $X:=S^1 \times S^1$ be the torus. Is the zero and second singular cohomology group $\mathbb{Z}$, the first $\mathbb{Z}^2$ and otherwise vanishing?

To show this: One first computes singular homology which gives that the zero and second homology group equals $\mathbb{Z}$ and first homology group equals $\mathbb{Z}^2$. Using universal coefficient theorem and that homology groups are free over $\mathbb{Z}$, we get the desired.

I was just wondering, because I wanted to compute cohomology groups of the sheaf of locally constant function with values in $\mathbb{Z}$, which should coincide with the singular cohomology group, but in this case I get that the first cohomology group equals $\mathbb{Z}^4$.

bjn
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  • If you google "homology of torus" this MSE post http://math.stackexchange.com/questions/41284/homology-groups-of-torus is the first hit. – Pedro Aug 29 '16 at 12:29
  • @PedroTamaroff Does in this case the homology group and Cohomology group coincide? – bjn Sep 12 '16 at 12:33

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