For an exact sequence \(A \to B \to C \to D \to E\) show that \(C = 0\) iff the map \(A \to B\) is surjective and \(D \to E\) is injective. Hence for a pair of spaces \((X, A)\), the inclusion \(A \hookrightarrow X\) induces isomorphisms on all homology groups iff \(H_n(X, A) = 0\) for all \(n\).
Solution
Consider an exact sequence \[ A\xrightarrow{\alpha}B\xrightarrow{\beta}C\xrightarrow{\gamma}D\xrightarrow{\delta}E. \]
1. If \(C=0\), then \(\alpha\) is surjective and \(\delta\) is injective.
Exactness at \(B\) gives \[ \operatorname{im}\alpha=\ker\beta. \] Since \(\beta:B\to0\) is zero, \(\ker\beta=B\), so \(\alpha\) is surjective.
Exactness at \(D\) gives \[ \ker\delta=\operatorname{im}\gamma. \] Since \(\gamma:0\to D\) has zero image, \(\ker\delta=0\), so \(\delta\) is injective.
2. Conversely, if \(\alpha\) is surjective and \(\delta\) is injective, then \(C=0\).
Surjectivity of \(\alpha\) and exactness at \(B\) imply \[ \ker\beta=B, \] so \(\beta=0\). Exactness at \(C\) then gives \[ \ker\gamma=\operatorname{im}\beta=0, \] so \(\gamma\) is injective.
Injectivity of \(\delta\) gives \(\ker\delta=0\). Exactness at \(D\) gives \[ \operatorname{im}\gamma=0. \] Thus \(\gamma\) is both injective and zero, forcing \(C=0\).
3. For a pair \((X,A)\), the segment\[ H_n(A)\xrightarrow{i_*}H_n(X)\to H_n(X,A) \to H_{n-1}(A)\xrightarrow{i_*}H_{n-1}(X) \]is exact. Hence\[ H_n(X,A)=0 \]if and only if the first \(i_*\) is surjective and the second \(i_*\) is injective.
Apply <1>1–<1>2 to this five-term exact segment.
4. Therefore\[ \boxed{H_n(X,A)=0\text{ for all }n \iff i_*:H_n(A)\to H_n(X)\text{ is an isomorphism for all }n.} \]
If all relative groups vanish, 3 gives surjectivity of \(i_*\) in degree \(n\) and injectivity in degree \(n-1\) for every \(n\), hence both properties in every degree. Conversely, if every \(i_*\) is an isomorphism, the two conditions in 3 hold in every degree, so every relative group vanishes.