Let $C$ be a smooth genus 2 curve, with canonical divisor W. From Riemann-Roch's theorem, we know that $l(K)=2$ so there exist two functions $g_1,g_2$ that span $L(W)$.
Since $g_1$and $g_2$ are linearly independent, the map $\varphi:C⟶\mathbb P^1$ defined by $[g1(P):g2(P)]$ is a non-constant morphism of smooth curves, so it is surjective.
We also know that $\deg \varphi\geq 2$, else $C$ would be birrational to $\mathbb P^1$, contradicting the assumption on its genus.
I want to show with elementary arguments that $\deg\varphi=2$. Using that $\deg\varphi$ is either the degree of the zero part of a $g_i$ or its pole part, I tried to write $W$ as $D_0+Q_1+Q_2$, with $D_0\in\mbox{Pic}^0(C)$ and the $Q_i$ possibly equal to obtain a contradiction with the numbers of poles/zeros that $g_1$ or $g_2$ should have, but it seems it does not work.
If $D_0=0$, then as $\mbox{div}(g_i)+K=\mbox{div}(g_i)+Q_1+Q_2≥0$ and that $g_i$ has poles at $Q_1$ and $Q_2$, then this means the $g_i$ has 2 zeroes and poles and I get the result. But I fail to see why $D_0$ would be 0 be zero.
So my questions are: is my reasoning correct ? Why would $D_0$ be zero ? For my curiosity also, is there a more elegant way using elementary arguments ?
I am not familiar with sheaves and linear bundles, etc, and sadly I cannot find any proof of this well-known result with the most elementary language. Also, I do not want to consider a possible equation for $C$.
Thanks in advance !