Let \(X\) be the space obtained from the sphere \(S^2\) by joining the North and South poles together with a straight line segment.
Describe a structure of a CW complex on \(X\).
Compute the homology of \(X\) using cellular homology, with decomposition from (i).
Solution
Let \(N,S\) be the north and south poles. Choose two distinct meridian arcs \(a,b\) on \(S^2\) from \(S\) to \(N\), so that \(a\cup b\) is a simple closed curve, and let \(c\) be the added segment from \(S\) to \(N\).
1. (i) \(X\) is a CW complex with \(0\)-cells \(N,S\), \(1\)-cells \(a,b,c\), and \(2\)-cells \(u,v\), the two components of \(S^2\setminus(a\cup b)\).
Each of \(a,b,c\) is an arc with endpoints in the \(0\)-skeleton \(\{N,S\}\), and their interiors are disjoint. The closures of the two components of \(S^2\setminus(a\cup b)\) are discs attached along the circle \(a\cup b\).
2. Orient \(a,b,c\) from \(S\) to \(N\) and \(u,v\) so that \(\partial_2(u)=a-b\) and \(\partial_2(v)=b-a\). In the bases \((u,v)\), \((a,b,c)\), and \((S,N)\), the cellular chain complex is\[0\to\ZZ^2\xrightarrow{\partial_2}\ZZ^3\xrightarrow{\partial_1}\ZZ^2\to0,\qquad [\partial_2]=\begin{pmatrix}1&-1\\-1&1\\0&0\end{pmatrix},\qquad [\partial_1]=\begin{pmatrix}-1&-1&-1\\1&1&1\end{pmatrix}.\]
The two \(2\)-cells have the same boundary circle \(a\cup b\), with opposite induced orientations. Each \(1\)-cell runs from \(S\) to \(N\), so its cellular boundary is \(N-S\).
3. \(H_0(X)\cong\ZZ\), \(H_1(X)\cong\ZZ\), and \(H_2(X)\cong\ZZ\).
By step 2, \(\operatorname{im}\partial_1=\ZZ(N-S)\), so \(H_0(X)=\ZZ^2/\langle N-S\rangle\cong\ZZ\).
\(\ker\partial_1=\{(x,y,z)\in\ZZ^3:x+y+z=0\}\) is free of rank \(2\), and \(\operatorname{im}\partial_2=\ZZ(1,-1,0)\) is generated by a primitive vector of it, which extends to the basis \((1,-1,0),(-1,0,1)\). Hence \(H_1(X)\cong\ZZ\), generated by the class of \(c-a\).
There is no \(3\)-cell, so \(H_2(X)=\ker\partial_2\). Since \(\partial_2(m,n)=(m-n)(1,-1,0)\), the kernel is \(\ZZ(u+v)\cong\ZZ\).
4. Q.E.D.
Step 1 answers (i). For (ii), step 3 and the absence of cells above dimension \(2\) give \[H_k(X)\cong\begin{cases}\ZZ,&k=0,1,2,\\0,&k\ge3.\end{cases}\]