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.
 A: Recall that pulling back Cartier divisors is compatible with pulling back locally free sheaves of rank one. 
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 $\mathrm{deg}(f)=2$.
A: 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).
