TITLE: Homology group on the complex projective space
QUESTION [6 upvotes]: I need to find the homology group of 
$$ X=l_1 \cup l_2 \cup ... \cup l_n  \ \ \subset \mathbb{CP}^2 $$ where $ l_i $ are distinct projective lines.
I think this is the direct sum of the homology groups of each projective line but not sure still.
Am I right ?

REPLY [2 votes]: First let's see what is the intersection of two distinct projective lines. Say we have
$$l_i= \{[z_1:z_2:z_3] \in \mathbb C \mathbb P^2: a^i_1z_1 +a_2^iz_2+a^i_3z_3=0 \} , \ i=1(1)n$$
From here it is more or less clear that $l_i \cap l_j$ is a point in $\mathbb C \mathbb P^2 \ \forall i\neq j$ 
Now let's see what happens when you take $n=2$ i.e. you have only $2$ projective lines. 
Then using the relative Meyer-Vietoris sequence gives you the long exact sequence in homology 

$$...\rightarrow \tilde{H}_k(l_1\cap l_2)\rightarrow\tilde H_k(l_1)\oplus \tilde H_k(l_2)\rightarrow\tilde H_k(l_1\cup l_2)\rightarrow \tilde H_{k-1}(l_1\cap l_2)\rightarrow...$$

But $l_1\cap l_2\cong*$ and hence in the above sequence we get the first and the last homology groups are $0$. So the middle one is an isomorphism. 
Thus for $n=2$ we got $\tilde H_k(l_1 \cup l_2)\cong \tilde H_k(\mathbb C\mathbb P^1) \oplus\tilde H_k(\mathbb C\mathbb P^1)$
Now for general we use induction. As before we have by the relative Meyer-Vietoris sequence a long exact sequence as follows

$$...\rightarrow \tilde{H}_k(\cup_{i=1}^{n-1}l_i\bigcap l_n)\rightarrow\tilde H_k(\cup_{i=1}^{n-1}l_i)\oplus \tilde H_k(l_n)\rightarrow\tilde H_k(\cup_{i=1}^{n-1}l_i\bigcup l_n)\rightarrow \tilde H_{k-1}(\cup_{i=1}^{n-1}l_i\bigcap l_n)\rightarrow...$$.

Now $\cup_{i=1}^{n-1}l_i\bigcap l_n$ is just a finite set of points and hence have relative homology groups all $0$. So the middle one is an isomorphism and you have by induction 

$$\tilde H_k (\bigcup_{i=1}^{n}l_i)\cong\bigoplus_{i=1}^n\tilde H_k(\mathbb C\mathbb P^1)$$