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I am trying to understand why $H^1(T^2) \cong \mathbb{R} \oplus \mathbb{R}$. $U$ and $V$ are open sets such that when $T^2$ is immersed in a cup of coffee vertically, the submerged portion, which is a little more than half of $T^2$ correspond to $U,V$. Thus $U\cap V = S^1\times(0,1) \sqcup S^1\times(2,3)$.

Consider the sequence $0\rightarrow H^0(U)\oplus H^0(V)\xrightarrow{f} H^0(U\cap V)\xrightarrow{d_1^*} H^1(T^2)\xrightarrow{g} H^1(U)\oplus H^1(V) \xrightarrow{f^{'}} H^1(U\cap V)\xrightarrow{d^*_{2}}H^2(T^2)\xrightarrow{g^{'}}H^2(U)\oplus H^2(V)\rightarrow H^2(U\cap V)\rightarrow 0$

I know that $H^0(T^2)=\mathbb{R}, H^0(U)\oplus H^0(V) = \mathbb{R}\oplus \mathbb{R}, H^0(U\cap V) = \mathbb{R}\oplus \mathbb{R}, H^1(U)\oplus H^1(V) = \mathbb{R}\oplus \mathbb{R}, H^2(U)\oplus H^2(V) = 0$

$d_2^{*}$ is surjective and by thus by isomorphism theorem, $H^1(U\cap V)/\mathrm{Ker}(d_2^*)\cong \mathbb{R}$. But $\mathrm{Ker}(d_2^*)=\mathrm{Im}(f^{'})=\mathbb{R}$ and thus $H^1(U\cap V) = \mathbb{R}\oplus \mathbb{R}$. Also I know that $\mathrm{Ker}(d_1^*)=\mathrm{Im}(f)=\mathbb{R}$ and $\mathrm{Im}(d_1^*)=\mathrm{Ker}(g)$ by exactness but am not sure how that leads to the answer? Thanks.

trickymaverick
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    Could you clarify what are your opens U and V here? –  May 28 '19 at 08:30
  • To understand it, you have to go a bit further and compute what the map $H^1(U)\oplus H^1(V)\to H^1(U\cap V)$ really is – Maxime Ramzi May 28 '19 at 08:33
  • @RobinCarlier Added details, thanks. – trickymaverick May 28 '19 at 08:53
  • @Max The image of that map is $\mathbb{R}$ and not sure what the kernel is. Thanks. – trickymaverick May 28 '19 at 08:55
  • What can the kernel be ? What does the rank nullity theorem tell you ? – Maxime Ramzi May 28 '19 at 10:05
  • @Max True, the kernel is $\mathbb{R}$ by rank-nullity. $\therefore \mathrm{Im}(g) = \mathbb{R}$ but not sure how to arrive at the answer? – trickymaverick May 28 '19 at 10:23
  • Now you have to do the same thing in the other direction : what is the image of $H^0(U)\oplus H^0(V)$, and what is the quotient of $H^0(U\cap V)$ by this image ? Then you'll have a short exact sequence of vector spaces, with $H^1(T^2)$ standing in the middle – Maxime Ramzi May 28 '19 at 10:31
  • I see, I get that the image of $H^0(U)\oplus H^0(V) = \mathbb{R}$ and the quotient of $H^0(U\cap V)$ by this image is $\mathbb{R}\oplus \mathbb{R}/\mathbb{R} = \mathbb{R}$. I have seen cutting a long sequence short a while ago, so am still not getting it, let me think for a bit. – trickymaverick May 28 '19 at 10:38

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As pointed out in the hint, if I cut the long exact sequence short at $H^0(U\cap V)$, the left side of the cut would be $0\rightarrow \mathbb{R}\rightarrow \mathbb{R}^2\rightarrow \mathrm{Ker}(d_1^*)\rightarrow 0$ and the right side would be $0\rightarrow \mathrm{Coker}(d_1^*)\rightarrow H^1(T^2)\rightarrow \mathbb{R}^2\rightarrow \mathbb{R}^2\rightarrow \mathbb{R}\rightarrow 0$ and if we cut at $\mathbb{R}^2$ again, the left of this cut would be $0\rightarrow \mathbb{R}\rightarrow H^1(T^2)\rightarrow \mathrm{Ker}(f^{'})\rightarrow 0$ and the right side is $0\rightarrow \mathrm{Cokernel}(f^{'})\rightarrow \mathbb{R}^2\rightarrow \mathbb{R}\rightarrow 0$ and thus $\mathrm{Cokernel}(f^{'})= \mathbb{R}$ and $\mathrm{Ker}(f^{'})=\mathbb{R}$ and thus finally $H^1(T^2) = \mathbb{R}\oplus \mathbb{R}$

trickymaverick
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