Subadditivity of Betti number for relative homology

172 Views Asked by At

This question stems from page 28 of the "Morse Inequalities" chapter of Milnor's Morse Theory.

Given a pair of topological spaces, $Y \subset X$ and any field $F$, define $R_\lambda(X,Y)$, or the $\lambda$th Betti number, to be the rank over $F$ of $H_\lambda(X,Y;F)$. According to Milnor, the Betti number is subadditive in the following sense. Given the triple of spaces $Z \subset Y \subset X$, $R_\lambda(X,Z) \leq R_\lambda(X,Y) + R_\lambda(Y,Z)$.

Milnor says the subadditivity follows from the exact homology sequence for a triple: $\begin{align*} \cdots \rightarrow H_\lambda(Y,Z) \rightarrow H_\lambda(X,Z) \rightarrow H_\lambda(X,Y) \rightarrow \cdots \end{align*}$ But I don't see how it follows. Could someone explain?

1

There are 1 best solutions below

0
On

Let $0 \to M_n \to \ldots \to M_0 \to 0$ be an exact sequence of $A$-modules. Then $\mathrm{rank}_A(M_n) - \mathrm{rank}_A(M_{n-1}) + \mathrm{rank}_A(M_{n-2}) - \ldots + \mathrm{rank}_A(M_0) =0$. Also obviously $\mathrm{rank}_A(M_k) \geq 0$.