Cohomology-Mayers-Vietoris Long exact sequence.

The exact sequence of cochain complex
$$0\to {\mathrm{\Omega }}^{\ast }(M)\stackrel{i}{\to }{\mathrm{\Omega }}^{\ast }(U)\oplus {\mathrm{\Omega }}^{\ast }(V)\stackrel{j}{\to }{\mathrm{\Omega }}^{\ast }(U\cap V)\to 0$$
for some open cover $U,V$ of manifold $M$, yields a Long exact sequence in cohomology called 'Mayer-Vietoris' sequence.
$$\cdots H^{k-1}(U\cap V)\stackrel{d^{\ast }}{\to }{H}^k(M)\stackrel{i^{\ast }}{\to }{H}^k(U)\oplus H^k(v)\stackrel{j^{\ast }}{\to }{H}^k(U\cap V)\stackrel{d^{\ast }}{\to }{H}^{k+1}(M)\cdots$$
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:
$$i^{\ast }([\sigma ])=([i\sigma ])=([i_U^{\ast }\sigma ],[i_V^{\ast }\sigma ])\in H^k(U)\oplus H^k(V)$$
Do similar thing to $j^{*}$, that is drop in equivalence classes instead of differential forms directly.
$$j^{\ast }([\omega ],[\tau ])=([j_V^{\ast }\tau -j_U^{\ast }\omega ])\in H^k(U\cap V)$$


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


For a connected manifold, above sequence is exact.
$$0\to H^0(M)\to H^0(U)\oplus H^0(v)\to H^0(U\cap V)\to H^0(M)\to 0$$