# Show $S^2$ with 2 cells attached is equivalent to a wedge of spheres

Show that a 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 of n+1 spheres.

I'm a little confused here. I'm imagining a sphere, and putting 2 dimensional discs inside of it. Attaching one disc along one circle of $$S^2$$ and then collapsing that disc to a point would appear to me to create a wedge of two spheres.

However, if I attached 2 discs along 2 different circles of $$S^2$$ and collapsed each to a point, to me it seems I would have created a wedge of four spheres, one sphere for each division created by the two discs inside the sphere.

What is wrong with my thinking here? Thanks!!

• Consider what happens if the circles are parallel. What, if anything, changes when we move the circles to not be parallel? Jan 18, 2019 at 15:25
• You can use an inductive argument: attaching $n$ 2-cells to $S^2$ is equivalent to attaching one 2-cell to the space obtained by attaching $(n-1)$ 2-cells to $S^2$. Then if you agree that attaching one 2-cell is equivalent to wedging with a new sphere, the claim follows. Jan 18, 2019 at 16:07
• Specifically, I don't understand why you are imagining four spheres instead of three. This is what I have in mind. Jan 18, 2019 at 16:29

If you form $$Z= S^2 \cup_f e^2$$ then the attaching map- $$f: S^1 \to S^2$$ is null homotopic, as $$S^2$$ is simply connected. So $$Z \simeq S^2 \vee S^2$$. The general result you need is that if $$Z= B \cup_f X$$ where $$f: A \to B$$, and the inclusion $$i : A \to X$$ is a closed cofibration, then the homotopy type of $$Z$$ depends only on the homotopy class of $$f$$.