The second proof in Hatcher of Lemma 2.34(c) for infinite-dimensional \(X\) runs as follows. From the long exact sequence of the pair \((X, X^n)\) it suffices to show \(H_k(X, X^n) = 0\) for \(k \leq n\); since \(H_k(X, X^n) \approx \widetilde{H}_k(X/X^n)\), this reduces to showing that \(\widetilde{H}_k(X) = 0\) for \(k \leq n\) if the \(n\)-skeleton of \(X\) is a point. The finite-dimensional case follows from the long exact sequences of the pairs \((X^m, X^{m-1})\). In general, let \(T = \bigcup_i X^i \times [i, \infty) \subset X \times [0, \infty)\); then \(X \times [0,\infty)\) deformation retracts onto \(T\), so \(T \simeq X\). With \(R = X^0 \times [0, \infty)\) and \(Z = R \cup \bigcup_i X^i \times \{i\}\), the quotient \(Z/R\) is a wedge sum of finite-dimensional complexes with \(n\)-skeleton a point and \(T/Z\) is a wedge sum of finite-dimensional complexes with \((n+1)\)-skeleton a point, and the long exact sequences of the pairs \((Z, R)\) and \((T, Z)\) give \(\widetilde{H}_k(T) = 0\) for \(k \leq n\).
Show \(H^k(X, X^n; G) = 0\) if \(X\) is a CW complex and \(k \leq n\), by using the cohomology version of the second proof of the corresponding result for homology in Lemma 2.34.
Solution
We follow the second proof of Lemma 2.34(c), replacing homology by cohomology. Cohomology with coefficients in \(G\) is suppressed from the notation.
1. It suffices to show that \(\widetilde H^k(Y)=0\) for \(k\le n\) whenever \(Y\) is a CW complex whose \(n\)-skeleton is a point.
\((X,X^n)\) is a good pair, so \(H^k(X,X^n)\cong\widetilde H^k(X/X^n)\) [@Hat02, §3.1]. The quotient \(X/X^n\) is a CW complex whose \(n\)-skeleton is a point.
2. The claim holds when \(Y\) is finite-dimensional.
\(H^k(Y^m,Y^{m-1})\cong\widetilde H^k(Y^m/Y^{m-1})\) vanishes for \(k\ne m\), since \(Y^m/Y^{m-1}\) is a wedge of \(m\)-spheres and the reduced cohomology of a wedge is the product of the reduced cohomologies of the summands [@Hat02, §3.1]. For \(m>n\) and \(k\le n\), the exact sequence \[ H^k(Y^m,Y^{m-1})\to\widetilde H^k(Y^m)\to\widetilde H^k(Y^{m-1}) \] has first term \(0\), so restriction \(\widetilde H^k(Y^m)\to\widetilde H^k(Y^{m-1})\) is injective. Since \(\widetilde H^k(Y^n)=0\) for all \(k\), induction on \(m\) gives \(\widetilde H^k(Y^m)=0\) for all \(m\) and \(k\le n\), and \(Y=Y^m\) for \(m\) large.
3. \(\widetilde H^k(Z)=0\) for \(k\le n\).
\(R=Y^0\times[0,\infty)\) is contractible because \(Y^0\) is a point, so \(\widetilde H^k(Z)\cong H^k(Z,R)\cong\widetilde H^k(Z/R)\) by the long exact sequence of the good pair \((Z,R)\). The quotient \(Z/R\) is a wedge of finite-dimensional complexes with \(n\)-skeleton a point, so its reduced cohomology in degrees \(k\le n\) is a product of groups that vanish by 2.
4. \(\widetilde H^k(T)=0\) for \(k\le n\).
The quotient \(T/Z\) is a wedge of finite-dimensional complexes with \((n+1)\)-skeleton a point, so \(H^k(T,Z)\cong\widetilde H^k(T/Z)=0\) for \(k\le n+1\) by 2 applied with \(n+1\). In the exact sequence \[ H^k(T,Z)\to\widetilde H^k(T)\to\widetilde H^k(Z) \] both outer terms vanish for \(k\le n\), by this and 3.
5. \(H^k(X,X^n;G)=0\) for \(k\le n\).
Apply the construction to \(Y=X/X^n\). \(T\simeq Y\), so \(\widetilde H^k(Y)\cong\widetilde H^k(T)=0\) for \(k\le n\) by 4 and homotopy invariance, and 1 concludes.