How to determine the homology groups of $\{(v,w)\in\mathbb{C}^2\mid vw \neq 0\}$ I want to determine the homology groups of $\{(v,w)\in\mathbb{C}^2\mid vw \neq 0\}$. I googled this problem and found that there's a general theory called arrangements of hyperplanes. But this theory is quite involved. So I'm wondering if there's some simpler way to do this.
I was considering Mayer-Vietoris sequence, but I was sort of confused. Since $\{(v,w)\in\mathbb{C}^2|vw \neq 0\}$ is the complement of two complex planes, I need to find an open cover with good intersections. But I don't know how to do this. Can anyone enlighten me or give me some hints? Thanks!
 A: Define the map $\Gamma:X \to X$ by $\Gamma(v,w)=(\frac{v}{|v|}, \frac{w}{|w|})$. This is a continues map as a composition of continues maps. This is a deformation retraction by the homotopy:
$$t (\frac{v}{|v|}, \frac{w}{|w|})+(1-t)(v,w)$$
By direct calculation you can show that for $t \in [0,1]$ the norm of each coordinate is different then $0$. This shows that the homotopy is well defined. Let the retract subspace be $A$ then observe that it’s homeomorphic to $S^1 \times S^1$. Now compute this by Kunneth if you know, else you can use a  Mayer-Vietoris but it’s harder.
To make this answer more complete here is how you can compute the torus($S^1 \times S^1$) using Mayer-Vietoris. Fi
I will Cover the Torus by 2 Open sets of the Torus $A=U$ and $B=V$ as seen in the figure }. Observe that $U$ and $V$ are cylinders and by the retraction trick their homology groups are the same as $S^1$. Hence will get:
$$H_n(V)=H_n(U)= \begin{cases}
\mathbb{Z} & \text{if}\ n=0,1 \\
0          & \text{otherwise}
\end{cases}$$
Observe that $U \cap V$ is a disjoint union of $S^1$ so by the following theorem :
Corresponding to the decomposition of a space $X$ into its path components
$X_\alpha $there is an isomorphism of $H_n(X)$ with the direct sum $\oplus_\alpha X_\alpha $.
\
We get: $$H_n(U \cap V)= \begin{cases}
\mathbb{Z} \oplus \mathbb{Z} & \text{if}\ n=0,1 \\
0          & \text{otherwise}
\end{cases}$$
So by Mayer-Vietoris the following sequence is exact ( $i,j$ be the inclusions from $U \cap V$  to $U ,V$ $k,l$  and  be the inclusions from $U,V$ to  the torus):
$$\dots \longrightarrow H_n(U)\oplus H_n(V) \longrightarrow H_n(S^1 \times S^1) \longrightarrow H_{n-1}(U \cap V) \longrightarrow \dots $$
So for $n > 2 $ We immediately have $H_n(S^1 \times S^1)=0$. For $n=2$ We have the sequence:
$$\dots 0 \longrightarrow H_2(S^1 \times S^1) \longrightarrow \mathbb{Z} \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \oplus \mathbb{Z} \longrightarrow \dots $$
Observe that $H(U) \cong H(v) \cong \mathbb{Z} $  $a$ and $b$ are different basis element in homology class  $U \cap V$ they will be both sent to the same homology class element in the inclusion function. So the kernel of $(i_*, j_*)$ in the exact sequence is $<a-b>$. This is an exact sequence and the image to $H_2(S^1 \times S^1)$ is $\{0\}$ so the Kernal from $H_2(S^1 \times S^1)$ is $\{0\}$. This implies that $H_2(S^1 \times S^1) \cong \mathbb{Z}<a-b> \cong \mathbb{Z} $. \
As for $n=1$ the Mayer-Vietoris sequence will be:
$$\dots \longrightarrow \mathbb{Z}\oplus\mathbb{Z} \stackrel{(i_*, j_*)}{\longrightarrow} \mathbb{Z}\oplus\mathbb{Z} \stackrel{k_* - l_*}{\longrightarrow} H_1(S^1 \times S^1) \stackrel{\partial}{\longrightarrow} \mathbb{Z}\oplus\mathbb{Z}  \stackrel{(i_* , j_*)}{\longrightarrow}\mathbb{Z}\oplus\mathbb{Z}   \longrightarrow \dots $$
So we get a short exact sequence:
$$0\longrightarrow \mathrm{ker}\ \partial \longrightarrow H_1(S^1 \times S^1) \longrightarrow
\mathrm{Im}\ \partial 
\longrightarrow 0$$
Now $\mathrm{Im}\ \partial = \mathrm{Ker}\ (i_* , j_*)$ which we have seen that is $\mathbb{Z}$. To compute $\mathrm{Ker}\ \partial$ We use:
$$\mathrm{ker}\ \partial = \mathrm{Im}\ (k_*-l_*) =(\mathbb{Z}\oplus \mathbb{Z}) / \mathrm{Ker}(k_*-l_* ) = (\mathbb{Z}\oplus \mathbb{Z})/\mathrm{Im}(i_*, j_*) = \mathbb{Z}$$
Using the fact that the long sequence is exact for the identity's in the formula.\
So we Get:
$$H_1(S^1 \times S^1)/ \mathbb{Z} \cong \mathbb{Z}$$
This implies that $H_1(S^1 \times S^1) \cong \mathbb{Z} \oplus \mathbb{Z} $
For $n=0$ we know that the torus is connected so $H_0(S^1 \times S^1) \cong \mathbb{Z}$
