Induced map by $\mathbb{R}P^2\to\mathbb{C}P^2$ in homology. Let $i\colon \mathbb{R}P^2\to \mathbb{C}P^2$ the usual embedding $i( [x:y:z] )=[x:y:z]$. I want to compute the induced map by $i$ in the second homology groups with  $\mathbb{Z}/2$ coefficients. I think the easiest way should be to use the intersection pairing. Let
$$S=\{ [z_0:z_1:z_2]\in \mathbb{C}P^2\mid a_0z_0+a_1z_1+a_2z_2\}$$
be a $\mathbb{C}P^1$ in $\mathbb{C}P^2$, and 
$$N=\{ [x_0:x_1:x_2]\in \mathbb{C}P^2\mid x_i\in\mathbb{R}\}.$$
If I show that $N\cdot S=|N\cap S| \ne 0$ for some $(a_1,a_2,a_3)\in \mathbb{C}^3$, then in particular the induced map will be an isomorphism $\mathbb{Z}/2\to\mathbb{Z}/2$. 
However, I think that $N$ and $S$ must intersect transversally to have the equality  $N\cdot S=|N\cap S| $. How can I do this? I am not familiar with the intersection pairing, so any suggestions, etc will be welcome.

I like this strategy, though maybe just take $a_0 =  0$, $a_1 = 1$, $a_2 = -i$ so that we are intersecting $[x_0 : x_1 : x_2]$ with $[z_0 : z_1 : iz_1]$. The only way to scale the last two coordinates to both be real is if $z_1 = 0$; in that case we are left with the points $[z_0: 0 : 0]$. Up to scalar change, there is only one of these, $[1 : 0 : 0]$. This is of course real.
We should now check that the intersection at $[1:0:0]$ is transversal. I would write $T_{[1:0:0]} \Bbb{CP}^2$ as $\Bbb C^2$, generated by the tangent vectors changing the second or third coordinate by some complex number. $T\Bbb{RP}^2$ is $\Bbb R^2 \subset \Bbb C^2$, the real elements; the complex line $\Bbb{CP}^1$ has tangent space given by the complex subspace $\Bbb C \subset \Bbb C^2$ generated by $(1,i)$. These two vector spaces are in direct sum: their intersection is zero (you of course cannot write $(\lambda, i\lambda)$ for two reals unless $\lambda = 0$), and they sum to the dimension of the total space. So this is a transversal intersection, as desired, and the intersection product is indeed 1.
