Let \(D\) be a closed disk embedded in the torus \[ T=S^1\times S^1, \] and let \(X\) be the result of removing the interior of \(D\) from \(T\). Let \(B\) be the boundary of \(X\), i.e. the circle boundary of the original closed disk \(D\). Compute \(H_i(T,B)\) for all \(i\).
Solution
All homology groups below have coefficients in \(\ZZ\). Since \(B\cong S^1\) and \(T\cong S^1\times S^1\), \[ H_i(B)\cong \begin{cases} \ZZ,&i=0,1,\\ 0,&\text{otherwise}, \end{cases} \qquad H_i(T)\cong \begin{cases} \ZZ,&i=0,2,\\ \ZZ^2,&i=1,\\ 0,&\text{otherwise}. \end{cases} \]
1. The inclusion\[ j:B\hookrightarrow T \]induces the zero map\[ j_*:H_1(B)\longrightarrow H_1(T). \]
The circle \(B\) is the boundary of the embedded disk \(D\subset T\). Hence the inclusion \(j:B\hookrightarrow T\) extends over the disk: \[ \begin{array}{ccc} B&\hookrightarrow&D\\ &\searrow j&\downarrow\\ &&T. \end{array} \] Since \(D\) is contractible, the inclusion \(B\to T\) is nullhomotopic. Therefore it induces the zero homomorphism on positive-dimensional homology, in particular on \(H_1\).
2. One has\[ H_2(T,B)\cong\ZZ^2. \]
The degree-\(2\) portion of the long exact sequence of the pair \((T,B)\) is \[ H_2(B) \longrightarrow H_2(T) \longrightarrow H_2(T,B) \longrightarrow H_1(B) \xrightarrow{j_*} H_1(T). \] Using \(H_2(B)=0\), \(H_2(T)\cong\ZZ\), \(H_1(B)\cong\ZZ\), and 1 gives a short exact sequence \[ 0\longrightarrow\ZZ \longrightarrow H_2(T,B) \longrightarrow\ZZ \longrightarrow0. \] Since \(\ZZ\) is free abelian, this sequence splits. Thus \[ H_2(T,B)\cong\ZZ\oplus\ZZ. \]
3. One has\[ H_1(T,B)\cong\ZZ^2 \qquad\text{and}\qquad H_0(T,B)=0. \]
The next part of the long exact sequence is \[ H_1(B) \xrightarrow{j_*} H_1(T) \longrightarrow H_1(T,B) \longrightarrow H_0(B) \longrightarrow H_0(T) \longrightarrow H_0(T,B) \longrightarrow0. \] Both \(B\) and \(T\) are path-connected, so the inclusion induces an isomorphism \[ H_0(B)\xrightarrow{\cong}H_0(T). \] By 1, the preceding map \(j_*:H_1(B)\to H_1(T)\) is zero. Exactness therefore forces \[ H_1(T)\xrightarrow{\cong}H_1(T,B), \] so \[ H_1(T,B)\cong\ZZ^2. \] Exactness at \(H_0(T)\) and the fact that \(H_0(B)\to H_0(T)\) is surjective give \[ H_0(T,B)=0. \]
4. For every \(i\ge3\),\[ H_i(T,B)=0. \]
For \(i\ge3\), both \(H_i(T)\) and \(H_{i-1}(B)\) vanish. The long exact sequence of the pair therefore forces \(H_i(T,B)=0\).
5. Hence\[ \boxed{ H_i(T,B)\cong \begin{cases} \ZZ^2,&i=1,2,\\ 0,&\text{otherwise}. \end{cases}} \]
Combine 2, 3, and 4.