Consider a complex line $L$ in $\mathbb{P}^2$ which is given by $ax+by+cz=0$ with $b\neq0$. I want to show that $L$ is isomorphic to $\mathbb{P}^1$.
For that, take the map $\phi:L\to \mathbb{P}^1$ which is given by $\phi[x:y:z]=[x:z]$. We want to show that $\phi$ is bijective, and $\phi$, $\phi^{-1}$ are holomorphic maps. I will take the approach where I consider $L$ and $\mathbb{P}^1$ to be Riemann Surfaces.
First, bijectivity follows as $\phi^{-1}[x:z]=[x:\frac{ax-cz}{b}:z]$.
Next, to check that $\phi$, $\phi^{-1}$ are holomorphic maps, we need to check it on local charts. For $\phi$, it follows from the fact that $F:\mathbb{P}^2\to \mathbb{P}^1$ with $F[x:y:z]=[x:z]$ is a holomorphic map and $F|_L=\phi$. To show that $\phi^{-1}$ is holomorphic, we show that local expression of $\phi^{-1}$ is holomorphic.
Are there other ways to argue this and do my arguments sound legitimate? I skipped some details on purpose.