Let $Q$ be a smooth conic (the zero set of a homogeneous degree 2 polynomial) 
in $\mathbb{P}^2(\mathbb{C})$ and let $j:Q\rightarrow\mathbb{P}^2(\mathbb{C})$ be the
immersion in the projective plane. How can I compute $j\_\*H\_2(Q,\mathbb{Z})\subseteq H\_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$? (in $H\_i(X,\mathbb{Z})$, i means the i-th singular homology group). 


I know that $Q$ is homeomorphic to $\mathbb{P}^1(\mathbb{C})$ so their homology groups are isomorphic, I can restrict to the Veronese embedding $\phi:\mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$,
$\phi([x\_0:x\_1])=[{x\_0}^2: x\_0x\_1 :{x\_1}^2]$.
I also note that the map $\phi$ is homotopic to $\psi: \mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$, $\psi([x\_0:x\_1])=[{x\_0}^2: 0 :{x\_1}^2]$ that is a line "counted two times" so my guess is that $j\_\*H\_2(Q,\mathbb{Z})$ can be the whole $H\_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ or it can be the subgroup of $H\_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ generated by two times its generator. How can I proceed? Is there a simpler way to do it? 


Thank you in advance.



Homology of a complex projective conic
The subgroup $j\_\* H\_2(Q)$ must be generated by twice the generator of
$H\_2(P^2(\mathbb{C}))$
(I'm dropping the coefficient group from my notation).
To see this, your map $\psi$ decomposes as the embedding from $P^1$ into
$P^2$ (which induces isomorphism on $H\_2$) composed with the map $[x,y]\to[x^2,y^2]$
on $P^1$. This map has degree $2$ and so acts on $H\_2$ by multiplication by $2$.
To see this, think of it as the squaring map on the Riemann sphere. This
is homotopic to the suspension of the double covering of $S^1$ onto itself.
(That is multiplication by $2$ on $H\_1(S^1)$ so suspends to multiplication by $2$
on $H^2(S^2)$).


All this generalizes to degree $d$ rational curves in $P^2$.


