State van Kampen’s theorem.
Calculate the fundamental group of the space obtained by taking two copies of the torus \(T=S^1\times S^1\) and gluing them together along a circle \(S^1\times\{p\}\), where \(p\) is a point in \(S^1\).
Solution
1. A based form of van Kampen’s theorem is the following.
Suppose \[ X=U\cup V, \] where \(U,V\), and \(U\cap V\) are open and path connected, and choose a basepoint \[ x_0\in U\cap V. \] Then the inclusion maps induce a surjection \[ \pi_1(U,x_0)*\pi_1(V,x_0) \longrightarrow \pi_1(X,x_0), \] whose kernel is the normal closure of \[ (i_U)_*(\gamma)(i_V)_*(\gamma)^{-1} \qquad (\gamma\in\pi_1(U\cap V,x_0)). \] Equivalently, \[ \pi_1(X,x_0) \cong \pi_1(U,x_0)*_{\pi_1(U\cap V,x_0)}\pi_1(V,x_0). \]
2. Let \(T_1\) and \(T_2\) be the two tori and let \(C\) be the circle along which they are identified.There is a van Kampen cover with\[ U\simeq T_1, \qquad V\simeq T_2, \qquad U\cap V\simeq C\simeq S^1. \]
The images of \(T_1\) and \(T_2\) in the glued space are closed subspaces, so enlarge each slightly across a collar of the common circle. The resulting sets \(U\) and \(V\) are open and path connected. They deformation retract onto \(T_1\) and \(T_2\), while their intersection deformation retracts onto the common circle \(C\). Thus 1 applies with the displayed homotopy types.
3. Choose generators so that\[ \pi_1(T_1)=\langle a,b\mid[a,b]=1\rangle, \qquad \pi_1(T_2)=\langle c,d\mid[c,d]=1\rangle, \]and the generator \(z\) of \(\pi_1(C)\cong\ZZ\) maps to \(a\) in the first torus and to \(c\) in the second.
In each torus \(S^1\times S^1\), take the first-factor loop as \(a\) or \(c\) and the second-factor loop as \(b\) or \(d\). The gluing circle is \(S^1\times\{p\}\) in each copy, so its fundamental-group generator is exactly the first-factor generator in both tori.
4. Van Kampen gives\[ \pi_1(X) \cong \langle a,b,c,d\mid[a,b]=1,\ [c,d]=1,\ a=c\rangle. \]
The free product of the two torus groups has presentation \[ \langle a,b,c,d\mid[a,b]=1,\ [c,d]=1\rangle. \] By 1 and 3, amalgamating over \(\pi_1(C)=\langle z\rangle\) imposes precisely the relation \[ a=c. \] No other cross-relation is introduced.
5. Therefore\[ \pi_1(X) \cong \langle a,b,d\mid[a,b]=1,\ [a,d]=1\rangle \cong \ZZ\times F(b,d). \]
Eliminate \(c\) from the presentation in 4 using \(c=a\). The remaining relations say exactly that \(a\) commutes with both free generators \(b\) and \(d\), while there is no relation between \(b\) and \(d\). But a presentation of the direct product of the infinite cyclic group \(\langle a\rangle\) with the free group \(F(b,d)\) is precisely \[ \langle a,b,d\mid[a,b]=1,\ [a,d]=1\rangle. \] Hence \[ \pi_1(X)\cong\ZZ\times F_2. \]