Let $\{U,V\}$ be an open cover of torus $M$, the Mayer–Vietoris sequence on $M$ and $\{U,V\}$ is:
$0\rightarrow H^0(M)\rightarrow H^0(U\sqcup V)\rightarrow H^0(U\cap V)\rightarrow H^1(M)\rightarrow H^1(U\sqcup V)\rightarrow ...$
Denote the map between $H^1(M)\rightarrow H^1(U\sqcup V)$ as $i^*$. In "An Introduction to Manifold" it is said $H^1(M)\simeq ker(i^*)\oplus im(i^*)$
Can anyone shed me some light on why the $H^1(M)$ can be decomposed as direct sum of $ker(i^*)$ and $im(i^*)$?
By the first isomorphism theorem we know that $$\mathrm{im}\ i^* \simeq H^1(M)/\mathrm{ker}\ i^*.$$
Choose a subspace $E \subset H^1(M)$ complementing $\mathrm{ker}\ i^*$; i.e. such that $H^1(M) = E \oplus \ker i^*.$ Then the above says $$ \mathrm{im}\ i^* \simeq \frac{E\oplus \ker i^*}{\mathrm{ker}\ i^*} \simeq E,$$
so we have the desired decomposition. Note that the isomorphism here is not canonical - as far as I know there is no natural choice of $E$.