# How to compute degree of morphism given by a canonical divisor?

Let $X$ be a smooth projective curve of genus $2$. Let $K$ be a canonical divisor. It is known $K$ has degree $2$ and it is not hard to show that $|K|$ is base point free so induces a morphism into $\mathbb P^{1}$ (up to linear isomorphism of $\mathbb P^1$). How do I show the degree of this morphism is $2$?

I've thought about the formula $$deg(f^*D) = deg(f) \cdot deg(D)$$ and taking $D$ to be a point, but I don't know enough about the preimage to say anything.

You took $D=p$ for some $p\in \mathbb{P}^1$, which is a good idea. Thinking in terms of line bundles you have: $$\mathcal{O}_{\mathbb{P}^1}(p)=\mathcal{O}_{\mathbb{P}^1}(1)$$ hence: $$f^*\mathcal{O}_{\mathbb{P}^1}(p)=f^*\mathcal{O}_{\mathbb{P}^1}(1)$$ You also have: $$f^*\mathcal{O}_{\mathbb{P}^1}(1)=\mathcal{O}_C(K_C)$$ since your morphism is determined by linear system $K_C$. This means that: $$f^*p=K_C$$ Now apply your formula in order to obtain $\mathrm{deg}(f)=2$.
• It might be a little dangerous to write $f^*p = K_C$, but it's true up to linear equivalence and that's what matters. – Hoot Jul 2 '16 at 23:53
• $K_X$ is defined as any divisor such that, $\Omega^n_X=\mathcal{O}_X(K_X)$ so from the formal viewpoint there is always this linear equivalence matter, when one speaks about canonical divisor. – Slup Jul 2 '16 at 23:56
Here is a more low brow approach (I'm not comfortable with pulling back locally free sheaves of rank 1). Because $l(K)=2$, $|K|$ is nonempty so $K$ is linearly equivalent to some effective divisor $K'$. $l(K')$ is also $2$ so let $\{ 1, f\}$ be a basis of $L(K')$. $f$ either has a double pole or two simple poles (and not just one simple pole because that would give an isomorphism to $\mathbb P^1$), so $f$ induces a map of degree $2$ into $\mathbb P^1$ (just look at $f^*(\infty)$ and use the formula I mentioned in the question).