(Wisconsin Aug ’10) The graph \(K\) has six vertices \(a_1,a_2,a_3,b_1,b_2,b_3\) and nine edges \(a_i b_j\) for \(i,j=1,2,3\). The space \(X\) obtained from \(K\) by attaching a \(2\)-cell along each loop formed by a cycle of four edges in \(K\). Find \(\pi_1(X)\).
Solution
Choose the spanning tree \[ T=\{a_1b_1,a_1b_2,a_1b_3,a_2b_1,a_3b_1\} \] of the graph \(K=K_{3,3}\). Since \(K\) has \(9\) edges and \(6\) vertices, its fundamental group is free of rank \[ 9-6+1=4. \] The four edges outside \(T\) are \[ a_2b_2,\quad a_2b_3,\quad a_3b_2,\quad a_3b_3. \] For each non-tree edge \(a_i b_j\) (\(i,j\in\{2,3\}\)), the associated fundamental loop is the four-edge cycle \[ a_i-b_j-a_1-b_1-a_i. \] Thus these four four-cycles form a free basis of \(\pi_1(K)\).
By construction, \(X\) is obtained by attaching a \(2\)-cell along every four-edge cycle, in particular along each of these four basis loops. Attaching a \(2\)-cell quotients the fundamental group by the normal closure of its attaching loop. Hence all four free generators are killed, and \[ \boxed{\pi_1(X)=1}. \]