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.
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 $H^1(T^2) = \mathbb{R}\oplus \mathbb{R}$