Let \(X_1\) and \(X_2\) be two copies of \(S^2\) and let \(N_1,S_1\) and \(N_2,S_2\) be the north and south poles of \(X_1\) and \(X_2\), respectively. Define \(X\) to be the quotient space obtained by identifying \(N_1\) with \(N_2\) and \(S_1\) with \(S_2\). Compute the fundamental group of \(X\) by using the Seifert–van Kampen theorem.
Solution
Let \(N\) and \(S\) denote the two points of \(X\) obtained from the identified north and south poles.
1. Give \(X\) a CW decomposition with two vertices, four edges, and four \(2\)-cells.
For each sphere \(X_i\), choose a great circle through \(N_i\) and \(S_i\). Its two semicircles give two \(1\)-cells \[ a_i,b_i:N\longrightarrow S. \] The two complementary open hemispheres are \(2\)-cells. Thus the quotient \(X\) has \[ X^{(0)}=\{N,S\} \] and a \(1\)-skeleton \(G=X^{(1)}\) consisting of four parallel edges \[ a_1,b_1,a_2,b_2 \] from \(N\) to \(S\). For the sphere \(X_i\), the boundary of either hemisphere traverses one semicircle from \(N\) to \(S\) and the other back from \(S\) to \(N\). Hence its two \(2\)-cells are attached along the mutually inverse loops \[ a_i b_i^{-1} \quad\text{and}\quad b_i a_i^{-1}. \]
2. The fundamental group of the \(1\)-skeleton is free of rank three.
Choose the single edge \(a_1\) as a maximal tree in \(G\) and use \(N\) as basepoint. By the standard graph computation from Seifert–van Kampen, \[ \pi_1(G,N)\cong F(x,y,z), \] where we may take \[ x=[a_1b_1^{-1}], \qquad y=[a_1a_2^{-1}], \qquad z=[a_1b_2^{-1}]. \] Indeed, each of the three edges outside the maximal tree gives one free generator.
3. Attaching the two hemispheres of \(X_1\) imposes only the relation \(x=1\).
The attaching loops for the two hemispheres of \(X_1\) are \[ a_1b_1^{-1} \quad\text{and}\quad b_1a_1^{-1}. \] These represent \(x\) and \(x^{-1}\). The Seifert–van Kampen theorem for attaching a \(2\)-cell says that the resulting fundamental group is the previous group modulo the normal closure of the attaching loop. Thus the two hemispheres together impose precisely \[ x=1. \]
4. Attaching the two hemispheres of \(X_2\) imposes only the relation \(y^{-1}z=1\).
The relevant attaching loop is \[ a_2b_2^{-1}. \] Using the generators from 2, \[ y^{-1}z =(a_2a_1^{-1})(a_1b_2^{-1}) =a_2b_2^{-1}. \] Hence one hemisphere of \(X_2\) imposes \[ y^{-1}z=1. \] The other hemisphere is attached along the inverse loop and contributes no additional relation.
5. Therefore\[ \pi_1(X,N)\cong\ZZ. \]
By <1>2–<1>4 and Seifert–van Kampen, \[ \pi_1(X,N) \cong \langle x,y,z\mid x,\ y^{-1}z\rangle. \] The first relation eliminates \(x\), and the second identifies \(z\) with \(y\). Thus \[ \pi_1(X,N) \cong \langle y\mid\ \rangle \cong\ZZ. \] Geometrically, a generator is represented by going from \(N\) to \(S\) along the arc \(a_1\) in the first sphere and returning from \(S\) to \(N\) along \(a_2\) in the second sphere. Hence \[ \boxed{\pi_1(X)\cong\ZZ}. \]