Hatcher Exercise 0.19 Attaching $2$-cells to $S^2$

Hatcher has the following exercise in chapter $0$:

Show that the space obtained from $S^2$ by attaching $n$ 2 cells along any collection of $n$ circles in $S^2$ is homotopy equivalent to the wedge sum of $n + 1$ 2-spheres.

So, looking at the case when $n=1$, we can "pinch" the attached disk to a point and we get $S^2 \vee S^2$. Following this pinching idea, the claim seems clear when we are attaching our disks across disjoint circles.

However, when the attaching circles intersect, for example with $n=2$, taking both circles as geodesic circles, applying the idea for the disjoint circles seems to produce the wedge sum of four spheres. Can someone point out what is wrong with my thought process?

• I'm guessing you have a picture where the two added cells divide the inside of the sphere into four pieces. This isn't right since the added cells only meet at their boundary. They don't have an arc of intersection as implied by a schematic drawing. – Cheerful Parsnip Jan 22 '16 at 23:22

If $\gamma : S^1 \to X$ is a loop on the topological space $X$, attaching a cell along $\gamma$ means constructing the quotient space $X \cup_\gamma D^2:= D^2 \coprod X/\sim$ where the equivalence relation is defined by $x\sim \gamma(x)$, $x \in \partial D^2$.
So if $\gamma_1, \gamma_2$ are two intersecting loops in $S^2$, $D^2_1 \cup_{\gamma_1} S^2 \cup_{\gamma_2} D_2^2$ is an abstract topological space where the interiors of $D_\alpha^2$'s don't intersect, since the only identifications are imposed by $\sim$, which leaves the interiors alone.
This may seem impossible if you're trying to glue disks to $S^2$ while sitting in $\Bbb R^3$. This just means you cannot do this construction while embedded in $\Bbb R^3$ without moving the loops, but certainly that doesn't mean the space does not exist.
That said, you can prove the statement of the problem as follows. If $\gamma$ is any loop on $S^2$, $\gamma$ can be homotoped to the zero loop as $S^2$ is simply connected. As homotopic attaching map gives homotopy equivalent spaces, $D^2 \cup_\gamma S^2$ is homotopic to $D^2$ attached to $S^2$ by the constant map, which is $S^2 \vee S^2$. This can be done with each of the disks. So you get $n$ spheres plus the sphere you started with, i.e., $n+1$ spheres wedged together.