I'm always having trouble understanding the orientation we consider for a given homology or cohomology group when it is isomorphic to $\mathbb{Z}$.
Take specifically the case the tautological line bundle $\mathcal{O}(-1)$ of $\mathbb{CP}^n$. Here the $-1$ denotes its first Chern class is negative of the generator of $H^2(\mathbb{CP}^n,\mathbb{Z})$. But how do we first choose the orientation of $H^2(\mathbb{CP}^n,\mathbb{Z})$?
One possible approach that I tried to understand this is the following: Consider the case especially for $\mathbb{CP}^1$. We first choose orientation of $H_2(\mathbb{CP}^1,\mathbb{Z})$, which is a choice of fundamental class $[\mathbb{CP}^1]$, this is determined naturally by its complex structure. Then by the isomorphism of $H^2(\mathbb{CP}^1,\mathbb{Z})$ and $H_2(\mathbb{CP}^1,\mathbb{Z})$, we therefore have a canonical orientation for $H^2(\mathbb{CP}^1,\mathbb{Z})$, i.e. we define the canonical generator $[\omega]$ to be the one that $<[\mathbb{CP}^1],[\omega]>=1$.
The above argument works fine for $\mathbb{CP}^1$. For $\mathbb{CP}^n$, I'm thinking we should choose the canonical generator of $H_2(\mathbb{CP}^n,\mathbb{Z})$ to be the pushforward of the fundamental class of $\mathbb{CP}^1$ to $\mathbb{CP}^n$ under the natural embedding $$[z_0,z_1] \rightarrow [z_0,z_1,0,...,0]$$
Is this the right way to think about in this setting?