Find a cell structure for, and compute a presentation for, the fundamental group of the space \(X\) described below. \(X\) is a quotient space of the disjoint union of two copies \(T_1=T_2=S^1\times S^1\) of the 2-torus. Fix a basepoint \(x_0\) in \(S^1\). \(X\) is obtained by gluing \(T_1\) and \(T_2\) together, identifying \((x_0,x)\) in \(T_1\) with \((x,x_0)\) in \(T_2\), for each \(x\in S^1\). For concreteness, use \((x_0,x_0)\) as your basepoint for \(X\).
Solution
Give \(S^1\) the CW structure with one \(0\)-cell \(x_0\) and one oriented \(1\)-cell. Then \(T_1\) has the product CW structure with vertex \((x_0,x_0)\), loops \(a=S^1\times\{x_0\}\) and \(b_1=\{x_0\}\times S^1\), and one \(2\)-cell attached along \(ab_1a^{-1}b_1^{-1}\), the boundary of the fundamental square; likewise \(T_2\) has loops \(b_2=S^1\times\{x_0\}\) and \(c=\{x_0\}\times S^1\) and a \(2\)-cell attached along \([b_2,c]\).
1. \(X\) is a CW complex with one vertex \(v\), three loops \(a,b,c\), and two \(2\)-cells attached along \([a,b]\) and \([b,c]\).
The gluing identifies the subcomplex \(b_1=\{x_0\}\times S^1\subseteq T_1\) with the subcomplex \(b_2=S^1\times\{x_0\}\subseteq T_2\) by \(x\mapsto x\), preserving orientation; call the common image \(b\). It identifies the two vertices, leaves \(a\) and \(c\) distinct, and does not touch the open \(2\)-cells, whose attaching words become \([a,b]\) and \([b,c]\).
2. \(\pi_1(X,v)\cong\langle a,b,c\mid[a,b],\ [b,c]\rangle\).
The \(1\)-skeleton is a wedge of three circles, with fundamental group free on \(a,b,c\). By the Seifert–van Kampen theorem, each \(2\)-cell of step 1 adds its attaching word as a relation.
3. Q.E.D.
Step 1 gives the cell structure and step 2 the presentation. Equivalently, \(\pi_1(X,v)\cong F(a,c)\times\ZZ\), with \(F(a,c)\) free on \(a,c\) and \(\ZZ\) generated by \(b\): the map \(F(a,c)\times\ZZ\to\pi_1(X,v)\), \((w,n)\mapsto w(a,c)b^n\), is a homomorphism because \(b\) commutes with \(a\) and \(c\), and the assignment \(a\mapsto(a,0)\), \(b\mapsto(1,1)\), \(c\mapsto(c,0)\) satisfies both relations and defines the inverse on generators.