Betti Numbers' inequality and Mayer-Vietoris sequence

Question

Let $U, V\subset M$ be open. How the Mayer-Vietoris sequence, $$\to H^{i-1}(U \cap V) \to H^i(U \cup V)\to H^i(U)\oplus H^i(V)\to$$

leads immediately to the $$b^i(U \cup V) \leq b^i(U) + b^i(V) + b^{i-1}(U \cap V)?$$

Answer 1

The Meyer-Vietoris sequence for homology gives you

$$\cdots\rightarrow H_k(U\cap V)\rightarrow H_k(U)\oplus H_k(V)\rightarrow H_k(U\cup V)\rightarrow H_{k-1}(U\cap V)\rightarrow\cdots.$$

This gives you a short exact sequence

$$0\rightarrow K\rightarrow H_k(U)\oplus H_k(V)\rightarrow I\rightarrow 0,$$

where $K=\ker(H_k(U\cap V)\rightarrow H_k(U)\oplus H_k(V))$ and $I=\mathrm{Im}(H_k(U\cup V)\rightarrow H_{k-1}(U\cap V))$

This gives $$b_k(U\cup V)=\operatorname{rk}H_k(U\cup V)=\operatorname{rk}I + \operatorname{rk}K\leq \operatorname{rk}H_k(U)\oplus H_k(V)+\operatorname{rk}H_{k-1}(U\cap V)$$ $$\implies b_k(U \cup V) \leq b_k(U)+b_k(V)+b_{k-1}(U\cap V).$$