# Calculate the homology of $X$.

Let $S^3 = \{(z_1, z_2) \in \mathbb{C}^2 : \lVert z_1\rVert ^2 + \lVert z_2\rVert^2 = 1 \}$, $D^2 = \{z \in \mathbb{C} : \lVert z\rVert \leq 1\}.$ Let $X$ be the quotient of the disjoint union of $D$ and $S^3$ by the smallest equivalence relation identifying $z \in D^2$, $\lVert z\rVert = 1$, with $(z^2, 0) \in S^3$. Calculate the homology of $X$.

I thought about putting a $CW$-structure on this thing but couldn't find one.

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Well, making $D^2$ a $2$-cell would be a good start. For that to work, you'll need to find a CW structure on $S^3$ with the right $1$-skeleton. – Chris Eagle Jun 28 '13 at 18:01
$S^3$ was my problem. – kiwi Jun 28 '13 at 18:03
I wanted to do a 0-cell, a 1-cell, 2 2-cells, and 1 3 cell. Does that sound right? – kiwi Jun 28 '13 at 18:05
Friends do not let friends use ||z|| instead of \lVert z\rVert!!! – Mariano Suárez-Alvarez Jun 28 '13 at 18:35
The spacing you get with two \verts is as bad as with two |s. I am assuming norms are wanted, as opposed to absolute values, of course. – Mariano Suárez-Alvarez Jun 28 '13 at 18:55

There is no need to construct a CW structure on your space, really. $X$ is constructed by attaching a $D^2$ to $S^3$ along the map $$D^2\supset S^1\ni z\stackrel\phi\longmapsto(z^2,0)\in S^3.$$ The pair $(X,S^3)$ is a collared pair (see Greenberg+Harper, Chapter 19, for example) It follows easily that the natural map $H_\bullet(D^2,S^1)\to H_\bullet(X,S^3)$ is a an isomorphism. The long exact sequence for homology for the pair $(X,S^3)$ then allows you to compute what you want.
A remarkable consequence of this computation is that it does not depend at all on what the attacing map is (that is, on the specific way the $2$-cell was attached to the $3$-sphere) The result would have been the same if the attaching map were constant, so that the resulting space would have been $S^2\vee S^3$. Can you decide if all the spaces that we get in this way are homotopy equivalent? – Mariano Suárez-Alvarez Jun 28 '13 at 19:24
You don't need to know what it is, really. Show that the relative homology of the pair $(X,S^3)$ is the same as that of $(D^2,S^1)$ using excision and the long exact squences of the pairs. – Mariano Suárez-Alvarez Jun 28 '13 at 19:41