Hyperelliptic iff there exist points $P,Q$ such that $\dim L(P+Q)=2$? I want to prove that a curve $C$ of genus $\geq3$ is hyperelliptic iff there exist points $P,Q\in C$ such that $\dim L(P+Q)=2$. My definition of hyperelliptic curve is that it has genus $\geq2$ and it has a degree $2$ morphism into $\mathbb{P}^1$.
One direction is obvious, if there exist points $P,Q\in C$ such that $\dim L(P+Q)=2$, then any non-constant $f\in L(P+Q)$ gives a degree $2$ morphism into $\mathbb{P}^1$.
In the other direction, if $C$ is hyperelliptic, then it has a degree $2$ morphism into $\mathbb{P}^1$, I want to choose the preimage of $\infty$ as $P,Q$, but I can't see why $\dim L(P+Q)=2$. By using the general bound of $L(D)\leq1+\deg D$ for effective divisors, I know $\dim L(P+Q)\leq3$, but why $\dim L(P+Q)\neq3$?
Could anyone help me?
 A: $\newcommand\P{\mathbb P}\newcommand\msL{\mathscr L}\newcommand\pull[1]{#1^*}\newcommand\msO{\mathscr O}\newcommand\msF{\mathscr F}$
Say $C$ is hyperelliptic with morphism $f:C\xrightarrow2\P^1$, and let $\msL=\pull f\msO(1)$. In other words $H^0(\msL)=L(P+Q)$ where $P,Q$ are the preimages of $\infty\in\P^1$, as you suggest. Our strategy will be to use $f$ to related cohomology groups on $C$ to those on $\P^1$, and then stare at a diagram until the answer pops out.
We have a short exact sequence $0\to\msO_{\P^1}\to\msO_{\P^1}(1)\to\msO_\infty\to0$. Further, it is a general fact that the morphism $f$ induces a (unique) map $H^\bullet(\P^1,-)\to H^\bullet(C,\pull f(-))$ of $\delta$-functors extending the natural map in degree $0$ (i.e. the map $H^0(\P^1,\msF)=\msF(\P^1)\to (\pull f\msF)(C)=H^0(C,\pull f\msF)$). The upshot is that our short exact sequence induces a homomorphism of long exact sequences
$$\require{AMScd}\begin{CD}
0 @>>> H^0(\msO_{\P^1}) @>>> H^0(\msO_{\P^1}(1)) @>>> H^0(\msO_\infty) @>>> H^1(\msO_{\P^1}) @>>> H^1(\msO_{\P^1}(1)) @>>> 0\\
& @VVV @VVV @VVV @VVV @VVV\\
0 @>>> H^0(\pull f\msO_{\P^1}) @>>> H^0(\pull f\msO_{\P^1}(1)) @>>> H^0(\pull f\msO_\infty) @>>> H^1(\pull f\msO_{\P^1}) @>>> H^1(\pull f\msO_{\P^1}(1)) @>>> 0
\end{CD}$$
This is a bit of a mouthful, so let's rewrite this a little. We know that $\pull f\msO_{\P^1}=\msO_C$, that $\pull f\msO_{\P^1}(1)=\msL$, and that $\pull f\msO_\infty=\msO_{P+Q}$. We also know $h^1(\msO_{\P^1})=g(\P^1)=0$ and $h^1(\msO_{\P^1}(1))=h^0(\msO_{\P^1}(-3))=0$ by Serre duality. Letting $k$ be our ground field, we have
$$\begin{CD}
0 @>>> k @>>> k^{\oplus2} @>>> k @>>> 0 @>>> 0 @>>> 0\\
& @VVV @VVV @VVV @VVV @VVV\\
0 @>>> k @>>> H^0(\msL) @>>> k^{\oplus2} @>>> k^{\oplus g} @>>> H^1(\msL) @>>> 0
\end{CD}$$
which now has a shorter scroll bar. Now, we're in luck. The map
$$k=H^0(\msO_\infty)\to H^0(\msO_{P+Q})=k^{\oplus2}$$
above is injective; it sends a function on the point $\infty\in\P^1$ to the induced function on the subscheme $P+Q\subset C$. Thus, commutativity of the square (with top row $k\to0$) shows that
$$\dim\mathrm{im}(H^0(\msL)\to k^{\oplus2})=\dim\ker(k^{\oplus2}\to k^{\oplus g})\ge1$$
and
$$\dim\ker(k^{\oplus g}\to H^1(\msL))=\dim\mathrm{im}(k^{\oplus2}\to k^{\oplus g})\le1.$$
Since $k^{\oplus g}\twoheadrightarrow H^1(\msL)$ is surjective, we see that $h^1(\msL)\le g-1$. Hence,
$$h^0(\msL)=\chi(\msL)+h^1(\msL)=(3-g)+h^1(\msL)\le2.$$
At the same time, exactness of the bottom row of our diagram + the bound $\dim\mathrm{im}(H^0(\msL)\to k^{\oplus2})\ge1$ shows that $h^0(\msL)\ge2$
Edit: To anyone looking at this answer, user10354138 has a comment on the question giving a much simpler proof that $h^0(\msL)\le2$.
