Create a cell complex for \(S^2\) and a cell complex for \(T=S^1\times S^1\).
Solution
1. \(S^2\) is a CW complex with one \(0\)-cell \(v_0\) and one \(2\)-cell attached by the constant map \(\partial D^2\to\{v_0\}\).
The resulting space is \(D^2/\partial D^2\), and \(D^2/\partial D^2\cong S^2\): the map \(D^2\to S^2\) sending the point at radius \(r\) and angle \(\theta\) to the point at polar angle \(\pi r\) and longitude \(\theta\) is a continuous surjection that identifies exactly \(\partial D^2\) to the south pole, so it induces a continuous bijection from the compact space \(D^2/\partial D^2\) to the Hausdorff space \(S^2\).
2. \(T\) is a CW complex with one \(0\)-cell \(v_0\), two \(1\)-cells \(a,b\), and one \(2\)-cell attached along the loop \(aba^{-1}b^{-1}\).
Both ends of each \(1\)-cell are attached to \(v_0\), so the \(1\)-skeleton is \(S^1\vee S^1\) with loops \(a\) and \(b\). The \(2\)-cell is the square \(I\times I\), whose boundary, read counterclockwise from \((0,0)\), traverses \(a\), \(b\), \(a^{-1}\), \(b^{-1}\). The resulting space is \(I\times I/\bigl((x,0)\sim(x,1),\ (0,y)\sim(1,y)\bigr)\), which is homeomorphic to \(S^1\times S^1\) via \((x,y)\mapsto(e^{2\pi ix},e^{2\pi iy})\).