# A basic query on canonical map associated to a line bundle

Let $$L$$ be a line bundle on a smooth complex projective curve $$C$$ of genus atleast two such that $$h^0(L)=2$$. Then consider the associated canonical map $$\varphi_L: C \to \mathbb P^1$$. Since we know that the image of the canonical map is always a nondegenerate curve in the target projective space, does this imply that in our situation the image of $$\varphi_L$$ is precisely $$\mathbb P^1$$ and hence the canonical map is surjective in this case?

Please correct me if this is not necessarily the case.

Yes, exactly and it holds in a more general setting under some added hypothesis:

If $$X$$ is an algebraic variety of dimension $$n$$ and $$L$$ is a line bundle over $$X$$ such that $$h^0(L)=n+1$$, $$H^n\neq 0$$, where $$H$$ is the mobile part of $$L$$, then the map $$\phi_L: X \to \mathbb{P}(H^0(L)^*)\cong \mathbb{P}^{n}$$ associated to $$L$$ is surjective.

A proof of the statement maybe can be the following one:

Let us suppose by contradiction that there is a point $$p$$ that is not in the image of the canonical map. Then you can choose $$n$$ hyperplanes of $$\mathbb{P}^n$$ whose common intersection is the point $$p$$. Then

$$H_1.\cdots .H_n=0$$ and so

$$H^n=\phi_L^*(H_1).\cdots . \phi_L^*(H_n)=0$$

Thus the image of the map associated to $$L$$ has to be surjective.
In the case $$n=1$$ the hypothesis are always satisfied. In fact taking the mobile part $$H$$ of $$L$$, then $$H\neq 0$$ otherwise $$h^0(L)<2$$, and it is effective, by definition. This means $$deg(H)>0$$ and we have done.
Please note that if the map is a morphism then there isn't a base point in which any global section of $$L$$ vanishes on it, so that $$L=H$$. Pay attention that the viceversa does not hold. Thus you get also a direct corollary:
If $$h^0(L)=n+1$$ and $$L$$ is base point free with $$L^n\neq 0$$, then the induced map $$\phi_L$$ is a surjective morphism.