# Deducing characteristics of a map induced by a divisor

Given a divisor $D$ on an algebraic curve $X$, there is a corresponding map $\phi_D$ from $X$ to the projective space (of dimension $\dim L(D)-1$). In particular, we know that if $D$ is a very ample divisor, then this is a holomorphic embedding into projective space. When $D$ is not very ample, how does one read off information about the map?

Here is a specific problem from Miranda's textbook to provide context:

Let $X$ be an algebraic curve of genus $2$. Let $K$ be a canonical divisor on $X$, and let $\phi_{2K}$ be the associated map to projective space. Show that the image of $\phi_{2K}$ is a smooth projective plane conic, and that the map has degree $2$.

Here are my thoughts so far. We know that $\deg(2K) = 4g-4 = 4$. By Riemann-Roch, we also have $\deg L(2K) = \deg(2K) + 1 - g = 3$. This would suggest that the image is in $\mathbb(P)^2$, and by the degree it could be a double covering of a conic, an embedding as a degree $4$ curve, or perhaps a quadruple covering of a line?

Continuing your reasoning, from the formula $deg(\phi^*(H))=deg(\phi)deg(\phi(X))$, where H is an hyperplane and we denote $\phi(X)=Y$, we see that by definition of hyperplane divisor $deg(\phi^*(H))=deg(2K)=4$. So we have to exlude two possibilities, the cases when $deg(\phi)$ is 1 or 4.
In the case $deg(\phi)=1$ we have $deg(Y)=4$, so the genus of Y would be $(4-1)(4-2)/2=3>2=g(X)$, which is not possible by Hurwitz's formula. Hence we may assume $deg(\phi)=4$ and that $Y\cong \mathbb{P}^1\cong \mathbb{C}_{\infty}$. Therefore $\phi$ comes from a meromorphic function $f:X\rightarrow \mathbb{C}$ which belongs to $L(2K)$. Since $dim(L(2K))=3$ we may assume $2K>0$, so we may write $2K=P+Q+R+S$ with possibly two or more points equal. Since $div(f)+R+Q+S+T\geq0$ and $deg(\phi)=4$ we must have $deg(div_{\infty}(f))=4$. From the fact that $deg(div_{\infty}(f))=deg(div_{0}(f))$ we get that $deg(R_{\phi})=8$. In this way we can achive a contraddiction using Hurwitz's formula:
since $deg(Y)=1$ we have $g(Y)=0$ and so we get $deg(R_{\phi})=10$, a contaddiction.
Hence we must have $deg(\phi)=2$ and that $Y$ is a conic.