Show that \(H_0(X, A) = 0\) iff \(A\) meets each path-component of \(X\).
Show that \(H_1(X, A) = 0\) iff \(H_1(A) \to H_1(X)\) is surjective and each path-component of \(X\) contains at most one path-component of \(A\).
Solution
Use the long exact sequence of the pair \((X,A)\).
1. One has\[ \boxed{H_0(X,A)=0 \iff A\text{ meets every path-component of }X.} \]
The tail of the long exact sequence is \[ H_0(A)\xrightarrow{i_*}H_0(X)\to H_0(X,A)\to0. \] Thus \(H_0(X,A)=0\) iff \[ i_*:H_0(A)\to H_0(X) \] is surjective.
The group \(H_0(X)\) is free on the path-components of \(X\), and a component of \(A\) maps to the generator of the component of \(X\) containing it. Hence \(i_*\) is surjective exactly when every component of \(X\) contains a point, equivalently a component, of \(A\).
2. One has\[ H_1(X,A)=0 \]if and only if\[ H_1(A)\to H_1(X) \]is surjective and\[ H_0(A)\to H_0(X) \]is injective.
Use the exact segment \[ H_1(A)\to H_1(X)\to H_1(X,A) \to H_0(A)\to H_0(X) \] and apply the five-term criterion proved in Exercise 15.
3. The map\[ H_0(A)\to H_0(X) \]is injective if and only if each path-component of \(X\) contains at most one path-component of \(A\).
Again use the component bases. Two distinct component generators of \(H_0(A)\) have the same image exactly when their components lie in the same component of \(X\). Thus injectivity is equivalent to the stated condition.
4. Hence\[ \boxed{H_1(X,A)=0} \]if and only if \(H_1(A)\to H_1(X)\) is surjective and every path-component of \(X\) contains at most one path-component of \(A\).
Combine 2 and 3.