Cohomology-Mayers-Vietoris Long exact sequence.

The exact sequence of cochain complex
$$\begin{equation}   0 \rightarrow \Omega^{*}(M) \xrightarrow{i} \Omega^{*}(U) \oplus \Omega^{*}(V) \xrightarrow{j} \Omega^{*}(U \cap V)\rightarrow 0 \end{equation}$$
for some open cover $U,V$ of manifold $M$, yields a Long exact sequence in cohomology called 'Mayer-Vietoris' sequence.
$$\begin{equation}   \cdots H^{k-1}(U \cap V) \xrightarrow{d^{*}} H^{k}(M) \xrightarrow{i^{*}} H^k(U) \oplus H^k(v) \xrightarrow{j^{*}} H^{k}(U \cap V) \xrightarrow{d^{*}} H^{k+1}(M)\cdots \end{equation}$$
In this complex, $i^{*},j^{*}$ are induced from $i,j$. Since $H^k$ is quotient, the elements of $H^k$ are in cohomoloous classes. If we take a representative element $\sigma \in \Omega^{*}(M)$, then the map $i$ sends this element to $i\sigma$. Based on this, one can define $i^{*}$ as follows:
$$\begin{equation}   i^{*}([\sigma]) = ([i\sigma]) = ([i^{*}_U\sigma],[i^{*}_V\sigma])\in H^k(U)\oplus H^k(V) \end{equation}$$
Do similar thing to $j^{*}$, that is drop in equivalence classes instead of differential forms directly.
$$\begin{equation}   j^{*}([\omega],[\tau]) = ([j^*_V\tau-j^*_U\omega]) \in H^k(U \cap V) \end{equation}$$


To make all this work, we need a connecting homomorphism map $d^*$ defined as follows:
$$\begin{equation}   d^*[\eta]=[\alpha] \in H^{k+1}(M) \end{equation}$$
Since for $k \leq -1$, $\Omega^{k}(M)=0$, the sequence can be written as,
$$\begin{equation}   0 \rightarrow H^{0}(M)  \rightarrow H^0(U) \oplus H^0(v) \rightarrow H^{0}(U \cap V) \rightarrow H^{0}(M) \rightarrow \cdots \end{equation}$$


For a connected manifold, above sequence is exact.
$$\begin{equation}   0 \rightarrow H^{0}(M)  \rightarrow H^0(U) \oplus H^0(v) \rightarrow H^{0}(U \cap V) \rightarrow H^{0}(M) \rightarrow 0 \end{equation}$$