Let $\pi: C^2\to S$ be the canonical quotient morphism. It is étale outside of the diagonal $\Delta$ of $C^2$. Let $K$ be the base field. The canonical map of differentials
$$ \pi^*\Omega_{S/K}^1\to \Omega^1_{C^2/K}$$
is then an isomorphism outside of $\Delta$, and induces an injective homomorphism $\pi^*\omega_{S/K}\to \omega_{C^2/K}$ of canonical sheaves. So $\pi^*\omega_{S/K}=\omega_{C^2/K}(-D)$ for some effective divisor $D$ on $C^2$, with support in $\Delta$, hence $D=r\Delta$ for some integer $r\ge 0$. The multiplicity $r$ can be computed locally for the Zariski topology, and even for étale topology on $C$. So we can work with Spec ($K[x]$) and find $r=1$:
$$ \pi^*\omega_{S/K}=\omega_{C^2/K}(-\Delta).$$
This should be enough to describe a canonical divisor on $S$ interms of the pushforward of a canonical divisor of $C^2$ and of $\pi(\Delta)$. The self-intersection should be easily computed with the projection formula. Otherwise ask for more details.

**Edit** Computation of the multiplicity $r$. Let $\xi$ be the generic point of $\Delta$ and $\eta=\pi(\xi)$. Then
$$ \omega_{S/K, \xi}\otimes O_{C^2,\xi}=(\pi^*\omega_{S/K})_{\xi}=\omega_{C^2/K}(-r\Delta)_{\xi}=\omega_{C^2/K,\xi}(-r\Delta).$$
This explains why $r$ can be computed Zariski locally.
Let $U$ be a dense open subset of $C$. Then one can compute $r$ on $U^2\to U^{(2)}$. If we can write $U$ as an étale cover $U\to V\subseteq \mathbb A^1_K$, then the map
$$ \pi^*\Omega_{U^{(2)}/K}^1\to \Omega^1_{U^2/K}$$
is just the pull-back of the map
$$ \pi^*\Omega_{V^{(2)}/K}^1\to \Omega^1_{V^2/K}.$$
If you take a local basis $dx, dy$ for $\Omega^1_{V^2/K}$, then $d(x+y), d(xy)$ is a local basis for $\Omega^1_{V^{(2)}/K}$, and their pull-backs to $U^2$ (resp. $U^{(2)}$) are local bases, and $r$ can be computed with these local bases. Now $\omega_{V^2/K}$ is generated by $dx\wedge dy$, and $\omega_{V^{(2)}/K}$ is generated by $d(x+y)\wedge d(xy)$ who image in $\omega_{V^2/K}$ is $(x-y)(dx\wedge dy)$. As $x-y$ generates locally the ideal of $\Delta$, we see that $r=1$.

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

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

that is a contradiction.

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.

## Best Answer

The image of canonical map is not contained in any hyperplane follows by the definition: If it were, then the hyperplane corresponds to a 1-form $\omega\in H^0(\Omega)$ which vanish on $X$ everywhere.