Need help with computing homology group.

Question

Let $D=$$S^2\cup$ x-axis$\cup$ y-axis be surface in $R^3$ I want to compute the homology group $H_n(D,\mathbb{Z})$ forcannot all $n\geq 0$ using Mayer-Vietoris Exact sequence. There exists many open cover of $D$ for example $U=D-\{(0,0,0)\},V=D-\{(\pm a,0,0)\cup(0,\pm a,0)\mid a\geq1\}$ etc. however I had a difficulty visualize what surface that the given $U,V,U\cap V$ is homotopy equivalent to. Can anyone give me some appropiate U,V and how to find it? Thank for you help.

Answer 1

I'd probably go with $U=S^2$ and $V=\{(x,y,0)\mid x=0\mbox{ or }y=0\}$. Then the intersection is just $U\cap V=\{(1,0,0),(0,1,0),(-1,0,0),(0,-1,0) \}$. (really you need to take small open thickenings of the $U$ and $V$ I've given here because they're not open as they stand, but up to homotopy nothing changes as they are deformation retracts of open subset of $D$.)

The only tricky group to find will be $H_1(D)$. In (the reduced version of) the long exact sequence we get $$H_1(S^2)\oplus H_1(V) \to H_1(D)\to \tilde{H}_0(U\cap V)\to \tilde{H}_0(S^2)\oplus \tilde{H}_0(V)\to \tilde{H}_0(D)$$ but we know $V$ is contractible, $D$ is path-connected and we know the homology groups of $S^2$ so this reduces to $$0 \to H_1(D)\to \mathbb{Z}^3\to 0 \to 0$$ so $H_1(D)\cong\mathbb{Z}^3$.

Answer 2

Well, $S^2$ is a deformation retract of $D - \{(0, 0\}$, so the homology groups of $U$ are easy. I'd modify $U$ to be $D - E$, where $E$ is a small closed ball around the origin, though.

If I had to pick $V$, I'd make it the union of the $x$ and $y$ axes together with four open disks in $S^2$ around the points where the axes meet the sphere. Clearly $V$ is contractible. And $U \cap V$ looks like four long needles sticking through disks. Each such "sword" is clearly contractible as well. So the homology of $U$, $V$, and $U \cap V$ are pretty easy, And the only interesting maps come in the $H_0$ terms, so you should be able to work from there.