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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!!

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    $\begingroup$ Consider what happens if the circles are parallel. What, if anything, changes when we move the circles to not be parallel? $\endgroup$ Jan 18, 2019 at 15:25
  • $\begingroup$ 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. $\endgroup$ Jan 18, 2019 at 16:07
  • $\begingroup$ Specifically, I don't understand why you are imagining four spheres instead of three. This is what I have in mind. $\endgroup$ Jan 18, 2019 at 16:29

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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 $f$.

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