De Rham cohomology of connected components

Question

I am studying a bit about De Rham cohomology and, in the reference I am using, it says that is clear that:

If M is a compact, orientable and differenciable manifold of dimension $n \geq 1$ with connected components $M_1, M_2, ..., M_r$, then, $$ H^k (M) \simeq H^k (M_1) \oplus ... \oplus H^k(M_r)$$

but I can not see why this is 'clear'. Could someone give me a proof of that?

Answer 1

I think that I may have found the answer.

Since $M_1,...,M_r$ are the connected components of $M$, we may write $M=\bigcup_{i=1}^r M_i$, where this union is disjoint. Then, consider the following Mayer-Vietoris sequence: $$0 \rightarrow H^k(M) \rightarrow^\alpha H^k(M_1) \oplus ... \oplus H^k(M_r) \rightarrow H^k\left(\bigcap_{i=1}^r M_i\right) \rightarrow 0$$ In one hand, we have that $H^k\left(\bigcap_{i=1}^r M_i\right)=0$, since $\bigcap_{i=1}^r M_i = \emptyset$, so, the sequence can be reduced to $$0 \rightarrow^{0_1} H^k(M) \rightarrow^\alpha H^k(M_1) \oplus ... \oplus H^k(M_r) \rightarrow^{0_2} 0$$ In the other hand, we have that this sequence is exact (by construction of the Mayer-Vietoris sequence), thus, $$Im(0_1) = ker (\alpha)$$ and, since $Im(0_1) = \{0\}$, comes that $ker(\alpha) = \{0\}$ and, so, $\alpha$ is injective. By the other hand, $$Im(\alpha) = ker(0_2)$$ and, since $ker(0_2) = H^k(M_1) \oplus ... \oplus H^k(M_2)$, comes that $\alpha$ is surjective, then, is an isomorphism, which concludes the proof that $$H^k(M) \simeq H^k(M_1) \oplus ... \oplus H^k(M_r)$$