In hatcher example $0.8$ we have the following :
Let $X$ be the space obtained from $S^2$ by attaching the two ends for an arc $A$ to two distinct points on the sphere, say the north and south poles. Let $B$ be an arc in $S^2$ that joins the two points. Then $X$ can be given a $CW$-complex structure with the two endpoints of $A$ as $0$-cells, the interiors of $A$ and $B$ two $1$-cells and the rest of $S^2$ a $2$-cell.
Now my doubt is how is this attachment of this $2$-cell being made? I am not quite sure what is the map from the boundary of the disk ,i.e. $S^1$, which will give us this desired attachment. Any help is appreciated, I just want to be able to visualize it I don't need explicit formulas.
Best Answer
Think of $B$ as what happens when you use a knife to make an incision in an orange from north orange town to orange-southville. What's left ($S^2\setminus B$) is homeomorphic to $D^2$ (proof: peel it off). This is meant by „Rest of $S^2$“.
Under this homeomorphism, the boundary of this disk gets mapped to $B$, so the attaching map $\partial D^2 \to X^1$ (where $X^1$ is our 1-skeleton) of this cell is precisely the map induced by our homeomorphism which sends the boundary to $B$ and the poles to the endpoint of $B$. Note that since the image is an arc, this visually „squishes“ $\partial D^2 \simeq S^1$ together, leaving the poles as-is.
$A$ is only indirectly relevant for this attaching map as it touches the poles as well, but it does not directly lie in the part of the 1-skeleton where our 2-cell attaches.