Compute the homology groups \(H_n(X, A)\) when \(X\) is \(S^2\) or \(S^1 \times S^1\) and \(A\) is a finite set of points in \(X\).
Compute the groups \(H_n(X, A)\) and \(H_n(X, B)\) for \(X\) a closed orientable surface of genus two with \(A\) and \(B\) the circles shown. [What are \(X/A\) and \(X/B\)?]
Solution
Let \(A\) first denote a finite set of \(m\ge1\) points in a connected space \(X\) equal to \(S^2\) or \(T^2=S^1\times S^1\).
1. For \(X=S^2\),\[ H_k(S^2,A)\cong \begin{cases} \mathbb Z,&k=2,\\ \mathbb Z^{m-1},&k=1,\\ 0,&\text{otherwise}. \end{cases} \]
Since \(A\) is finite, \[ H_0(A)\cong\mathbb Z^m, \qquad H_k(A)=0\quad(k>0). \] For \(k\ge2\), the long exact sequence gives \[ H_k(S^2,A)\cong H_k(S^2), \] so only degree \(2\) contributes \(\mathbb Z\). In low degrees the exact segment is \[ 0\to H_1(S^2,A)\to\mathbb Z^m \xrightarrow{\epsilon}\mathbb Z\to H_0(S^2,A)\to0, \] where \(\epsilon\) sums coordinates. Hence \[ H_1(S^2,A)\cong\ker\epsilon\cong\mathbb Z^{m-1}, \qquad H_0(S^2,A)=0. \]
2. For \(X=T^2\),\[ H_k(T^2,A)\cong \begin{cases} \mathbb Z,&k=2,\\ \mathbb Z^{m+1},&k=1,\\ 0,&\text{otherwise}. \end{cases} \]
Again \(H_k(T^2,A)\cong H_k(T^2)\) for \(k\ge2\), so degree \(2\) is \(\mathbb Z\). The low-degree exact sequence is \[ 0\to\mathbb Z^2\to H_1(T^2,A) \to\mathbb Z^m\xrightarrow{\epsilon}\mathbb Z\to0. \] Thus there is a short exact sequence \[ 0\to\mathbb Z^2\to H_1(T^2,A)\to\mathbb Z^{m-1}\to0. \] Since the quotient is free abelian, the sequence splits, giving \[ H_1(T^2,A)\cong\mathbb Z^{m+1}. \] Also \(H_0(T^2,A)=0\).
Now let \(X\) be the genus-two closed orientable surface and let \(A,B\) be the circles in Hatcher’s figure.
3. Collapsing the separating circle \(A\) gives\[ X/A\cong T^2\vee T^2. \]Therefore\[ H_k(X,A)\cong \begin{cases} \mathbb Z^4,&k=1,\\ \mathbb Z^2,&k=2,\\ 0,&\text{otherwise}. \end{cases} \]
The circle \(A\) in the figure separates the two handles. Collapsing it to a point turns each side into a torus, with the two resulting tori meeting only at the collapsed point. Since \((X,A)\) is a good pair, \[ H_k(X,A)\cong\widetilde H_k(X/A). \] The reduced homology of \(T^2\vee T^2\) is the direct sum of the reduced homologies of the two tori, giving the displayed groups.
4. Collapsing the nonseparating circle \(B\) gives a torus with two distinct points identified. This space is homotopy equivalent to\[ T^2\vee S^1. \]Hence\[ H_k(X,B)\cong \begin{cases} \mathbb Z^3,&k=1,\\ \mathbb Z,&k=2,\\ 0,&\text{otherwise}. \end{cases} \]
Cutting the genus-two surface along the nonseparating circle \(B\) produces a genus-one surface with two boundary circles. Collapsing \(B\) in the original surface amounts to capping each of these boundary circles to a point and then identifying the two resulting points. Capping gives a torus, so \(X/B\) is a torus with two points identified.
For any path-connected space, identifying two distinct points is homotopy equivalent to adjoining a circle between them and then choosing a path already in the space, hence to wedging on an \(S^1\). Thus \[ X/B\simeq T^2\vee S^1. \] Again \[ H_k(X,B)\cong\widetilde H_k(X/B), \] which gives the displayed result.