Prove by induction on dimension the following facts about the homology of a finite-dimensional CW complex \(X\), using the observation that \(X^n/X^{n-1}\) is a wedge sum of \(n\) spheres:
If \(X\) has dimension \(n\) then \(H_i(X) = 0\) for \(i > n\) and \(H_n(X)\) is free.
\(H_n(X)\) is free with basis in bijective correspondence with the \(n\)-cells if there are no cells of dimension \(n-1\) or \(n+1\).
If \(X\) has \(k\) \(n\)-cells, then \(H_n(X)\) is generated by at most \(k\) elements.
Solution
Let \(X^r\) denote the \(r\)-skeleton. Since \[ X^r/X^{r-1}\cong\bigvee_{\alpha\in I_r}S^r, \] where \(I_r\) indexes the \(r\)-cells, one has \[ H_i(X^r,X^{r-1})\cong \begin{cases} \mathbb Z^{(I_r)},&i=r,\\ 0,&i\ne r. \end{cases} \]
1. If \(X\) has dimension \(n\), then \(H_i(X)=0\) for \(i>n\).
Induct on \(n\). The assertion is immediate for a \(0\)-complex. Assume it for \((n-1)\)-complexes and apply the long exact sequence of \((X,X^{n-1})\). For \(i>n\) both \[ H_i(X,X^{n-1})=0 \] and \(H_i(X^{n-1})=0\) by induction. Exactness gives \(H_i(X)=0\).
2. If \(X\) has dimension \(n\), then \(H_n(X)\) is free abelian.
The relevant part of the long exact sequence is \[ 0=H_n(X^{n-1})\longrightarrow H_n(X) \longrightarrow H_n(X,X^{n-1})\cong\mathbb Z^{(I_n)}. \] Hence \(H_n(X)\) is isomorphic to a subgroup of a free abelian group. Every subgroup of a free abelian group is free abelian, so \(H_n(X)\) is free.
3. If there are no \((n-1)\)-cells, then\[ H_n(X^n)\cong\mathbb Z^{(I_n)}. \]
Now \(X^{n-1}=X^{n-2}\) has dimension at most \(n-2\), so by 1 \[ H_n(X^{n-1})=H_{n-1}(X^{n-1})=0. \] The long exact sequence of \((X^n,X^{n-1})\) therefore gives \[ H_n(X^n)\cong H_n(X^n,X^{n-1})\cong\mathbb Z^{(I_n)}. \] The standard relative generators correspond bijectively to the \(n\)-cells.
4. If in addition there are no \((n+1)\)-cells, then the inclusion \(X^n\hookrightarrow X\) induces an isomorphism on \(H_n\).
Passing from \(X^n\) to \(X\) attaches cells only in dimensions at least \(n+2\). For each \(r\ge n+2\), the relative group \[ H_i(X^r,X^{r-1}) \] vanishes for \(i=n,n+1\). The long exact sequence therefore shows inductively that attaching these cells does not change \(H_n\). Hence \[ H_n(X)\cong H_n(X^n)\cong\mathbb Z^{(I_n)}. \] This proves part (b).
5. If \(X\) has \(k\) \(n\)-cells, then \(H_n(X)\) is generated by at most \(k\) elements.
By the argument of 2, \[ H_n(X^n)\hookrightarrow H_n(X^n,X^{n-1})\cong\mathbb Z^k. \] Thus \(H_n(X^n)\) is free of rank at most \(k\). Attaching \((n+1)\)-cells can only quotient \(H_n\), while attaching cells of dimension at least \(n+2\) does not alter it. Therefore \(H_n(X)\) is a quotient of \(H_n(X^n)\) and hence is generated by at most \(k\) elements.
This proves (a), (b), and (c).