Cohomology-Computations 1

Let $U,V$ be open cover of circle - $S^1$. Let $X,Y$ be arcs on the circle that correspond to the open cover with disjoint overlaps at top of the circle and at the bottom of the circle.

Each arc $X,Y$ is diffeomorphic to an interval and thus to $\mathcal{R}$.

Here instead of writing one forms and seeking existence of integral solutions, we can use Mayer-Vietoris sequence to make computations. The Mayer-Vietoris sequence for $S^1$ is as follows:
$$\begin{equation}   \label{eq:MV1}   0 \rightarrow H^{0}(M) \xrightarrow{i^*} H^{0}(U) \oplus H^{0}(V) \xrightarrow{j^*} H^{0}(U \cap V) \xrightarrow{d^*} H^1(M) \rightarrow 0. \end{equation}$$
Using dimensional formula for sequence of vector spaces $\sum_{k=0}^n(-1)^k d^k$, we can figure out dimension of $H^1(M)=d^1=1$ as follows:
$$\begin{equation}   1 - 2 + 2 - d^1 = 0 \end{equation}$$
Since $S^1$ is connected, $H^0(S^1)=\mathcal{R}$. As shown before $H^0(U)=H^0(V)=\mathcal{R}$. Since overlaps are disjoint we have $\mathcal{R} \oplus \mathcal{R}$. All this results in the following sequence.
$$\begin{equation}   \label{eq:MV2}   0 \rightarrow \mathcal{R} \xrightarrow{i^*} \mathcal{R} \oplus \mathcal{R} \xrightarrow{j^*} \mathcal{R} \oplus \mathcal{R} \rightarrow 0 \end{equation}$$
Notice $j^*:H^{0}(U) \oplus H^{0}(V) \xrightarrow{j^*} H^{0}(U \cap V)$ is given as follows. Since, we are dealing with $0$ dimensional space, corresponding vectors are $0$ dimensionals - that is scalars or real numbers.
$$\begin{equation}   j^*(m,n) = (n-m,n-m) \end{equation}$$
That is these are elements of diagonal in $\mathcal{R}\times\mathcal{R}$. $d^*: H^{0}(U \cap V) \xrightarrow{d^*} H^{1}(M)$ sends this to one dimensional space which is isomorphic to $\mathcal{R}$, hence points $m \neq n$ will result in this element. Among many, we can choose one.