On the homotopy type of unions of 2-spheres Here is the problem I am stuck on: If $X$ is a connected Hausdorff space that is a union of a finite number of $2$-spheres, any two of which intersect in at most one point, then show that $X$ is homotopy equivalent to a wedge sum of $S^1$ and $S^2$.
Intuitively this is pretty easy to see, but the formal details are eluding me. I think the right way to go with this problem is to realize the unions of 2-spheres as a cell complex and then shrink away contractible subcomplexes until we arrive at the desired decomposition. Is there an easy way to make this more formal?
 A: I think the result you want is stated on pg. 11 of Hatcher's Algebraic Topology, restated here for convenience.

If $(X,A)$ is a CW pair consisting of a CW complex $X$ and a contractible subcomplex $A$, then the quotient map $X\to X/A$ is a homotopy equivalence.

The exact problem that you are referring to is done on pg. 12 (with helpful pictures) in the case of a necklace of $n$ two spheres, which gives a wedge of $S^1$ with $n$ two spheres.
In the general case, I believe we can use induction.  Suppose $X$ consists of $n$ two spheres.  Distinguish any one of the spheres by 'drawing it away' from the others by lengthening its $m$ points of intersection with other spheres into line segments.  On the distinguished sphere, contract all of the $m$ points which intersect the line segments to one point.  Now the remaining $n-1$ spheres are homotopy equivalent to a wedge sum of circles and two spheres, and by contracting along subcomplexes, we can assume the $m$ line segments from the distinguished sphere all intersect at the wedge point.  Then, just contract one of the $m$ line segments to obtain the desired result (this last contraction adds one two sphere and $m-1$ circles).
