I'm using Hatcher's Algebraic Topology, and he gives:
Let's give $S^2$ the standard CW complex structure consisting of
- two $0$-cells: $\{e^0_1, e^0_2\}$,
- two $1$-cells: $\{e^1_E, e^1_W\}$,
- two $2$-cells: $\{e^2_N, e^2_S\}$.
I'm trying to compute $d_2$.
We have $d_2(e^2_N)=d_{NE}e^1_E + d_{NW}e^1_W$ where $d_{NE}$ is the degree of the map $S^1_N \to X^1 \to S^1_E$ and similarly for $d_{NW}$.
However, I am not sure how to compute the degree of these maps. How is this done?
Fix an orientation of the two intervals so that the orientations are in opposite directions (think an arrow on the 1-cell). Now when we look at the maps used to compute the boundary, one will map $S^1 \rightarrow S^1$ via a map that has winding number 1. The other map will map $S^1 \rightarrow S^1$ via map that has winding number -1 (draw the quotient maps including the arrows). From here you can calculate the homology.