It is stated that for $\chi(X)$ the Euler characteristic of a space $X$, and $Z \subset Y \subset X$, it holds that:
$$ \chi (X,Z) = \chi(X,Y) + \chi(Y,Z)$$
How does this follow from the exact sequences associated to the relative homologies? I assume one uses the exact sequences $$\cdots \to H_j (Y) \to H_j (X) \to H_j (X,Y) \to \cdots$$ $$\cdots \to H_j (Z) \to H_j (Y) \to H_j (Y,Z) \to \cdots$$and the excision theorem, but I am not sure how to show this properly.
Best Answer
Given an exact sequence $\dots \to V_k \to V_{k-1} \to \dots$, assuming that all of the groups involved are finitely generated, then the alternating sum of the ranks is zero: $\sum (-1)^k \text{rank}\, V_k = 0$. Apply that formula to the long exact sequence of the triple $Z \subset Y \subset X$: $$ \dots \to H_n(Y,Z) \to H_n(X,Z) \to H_n(X,Y) \to H_{n-1}(Y,Z) \to ... $$ One way to prove the fact about ranks in an exact sequence is to tensor with $\mathbb{Q}$, turning it into an exact sequence of vector spaces and converting rank to dimension. Then apply the rank-nullity theorem. For the vector space case, see the answer here, for example.