Image of $c_1: H^1(\mathbb CP^n, \mathcal O^*)\to H^2(\mathbb CP^n,\mathbb Z)$

Question

By the exact sequence $0\to \mathbb Z\to \mathcal O\to \mathcal O^*\to0$ and $H^1(\mathbb CP^n, \mathcal O)=0$ we can get the injection $$\text {Pic}(\mathbb CP^n)\cong H^1(\mathbb CP^n, \mathcal O^*)\xrightarrow{c_1} H^2(\mathbb CP^n,\mathbb Z)\cong \mathbb Z$$

Since $H^2(\mathbb CP^n, \mathcal O)=\mathbb Z$, it is not clear if $c_1$ is surjective. But I heard the $c_1(\mathcal O_{\mathbb CP^n} (1))$ is the generator of $H^2(\mathbb CP^n,\mathbb Z)$. I wonder how to see this?

Answer 1

Well the image of $c_1$ is killed by the map $H^2(\mathbb CP^n, \mathbb Z) \to H^2(\mathbb CP^n, \mathcal O)$, which is a map $\mathbb {Z\to Z}$.

However, such a map is either $0$ or injective.

If the map is $0$, then $c_1$ is surjective, and if the map is injective, then the image of $c_1$ (which is injective) is $0$ and so $H^1(\mathbb CP^n, \mathcal O^*) = 0$, which is absurd.

It follows that the map is $0$, and so $c_1$ is surjective.

EDIT : Ted Shifrin below suggests that actually $H^2(\mathbb CP^n, \mathcal O)=0$ (which seems more reasonable to me but as I explained I don't know nearly enough about complex geonetry to argue over that).

In that case, the fact that $c_1$ is surjective follows immediately from the exactnes of the sequence (note that in any case the amswer is only related to the long exact sequence)