- \(S^2 \union_f D^2\), where \(f\) attaches to the equator
Solution
1. Give \(S^2\) a CW structure with two \(0\)-cells (the poles), two \(1\)-cells forming the equator, and two \(2\)-cells (the hemispheres), then attach the additional disk along the equator.
This CW structure is adapted to the attaching circle.
2. Equivalently, regard \(S^2\) as two disks glued along their common boundary. Adding a third disk along that same circle produces three disks with common boundary.
This is exactly the quotient specified by the attaching map.
3. The union of any two of the disks is \(S^2\), and the third disk is attached along a null-homotopic loop in this \(S^2\). Hence\[\boxed{S^2\cup_fD^2\simeq S^2\vee S^2.}\]
The common boundary circle bounds either of the first two disks, so its inclusion into their union \(S^2\) is null-homotopic. Attaching a \(2\)-cell by a null-homotopic map wedges on an \(S^2\).
4. Consequently \(H_0\cong\mathbb Z\), \(H_2\cong\mathbb Z^2\), and all other reduced homology vanishes.
Use the homology of a wedge of two \(2\)-spheres.