The following is theorem III.8.7. in Farkas & Kra Riemann surfaces
Theorem. Let $M$ be a compact Riemann surface of genus $4$. Then one and only one of the following holds:
a. $M$ is hyperelliptic.
b. $M$ has a function $f$ of degree $3$ such that $(f) = \mathcal{U}/D$ with $D,\mathcal{U}$ integral and $D^2\sim Z$ where $Z$ is the canonical divisor. Any other function of degree $3$ on $M$ is a fractional linear transformation of $f$.
c. $M$ has exactly two functions of degree $3$ that are not Mobius transformations of each other.
I won't write the full proof but the part I can't understand.
Proof. ... We know that every compact surface of genus $4$ admits a non-constant function of degree $\leq 3$. Note that for any divisor $D$ of degree $3$ on $M$ of genus $4$, we have $$r(D^{-1}) = i(D).$$ Suppose we are in case (b). Let us assume there is an integral divisor $D_1\neq D$ with $\deg D_1 =3$ and $r(D_1^{-1}) = 2$. Let $f_1\in L(D_1^{-1})\setminus\Bbb C$. If $D_1$ is equivalent to $D$, then $f_1$ is a Mobius transformation of $f$. Thus we assume that $f_1$ is not a Mobius transformation composed with $f$, which implies that $D_1$ is not equivalent to $D$. We therefore have $1,f,f_1,ff_1$ are $4$ linearly independent functions in $L(D^{-1}D_1^{-1})$. Then $r(D^{-1}D_1^{-1})\geq 4$ and hence $$i(DD_1) = r(D^{-1}D_1^{-1})-6+4-1\geq 1.$$ $\color{red}{\text{Because }\deg DD_1 = 6, DD_1\text{ must be canonical, and hence }i(DD_1)=1.}$
I can't understand the red part. It looks like if $D$ is a divisor on genus $g$ compact Riemann surface with $\deg D =2g-2$ then $D$ is a canonical class. I know the converse is true in general. Also I can't understand why $DD_1$ canonical implies $i(DD_1)=1$. Any explanation on these?
I'll add some notational detail if one asks in the comment.
$1$. If $D$ is a special divisor over a compact Riemann surface with degree $2g-2$, then $D\sim Z$ (equivalent). So $D$ is contained in a canonical class. In particular, $D$ is a canonical divisor.
Note that any integral divisor of degree $\leq g-1$ is special by Riemann-Roch theorem.
$2$. If $Z = (\omega)$ for some abelian differential $\omega$, then $\Omega(Z) = \{\eta:\eta\text{ is an abelian differential with }(\eta)\geq(\omega)\} = \langle\omega\rangle_{\Bbb C}$ as $(\eta/\omega)\geq 0$ so $\eta/\omega$ is a holomorphic function so $\eta/\omega\in\Bbb C$. So, $\dim\Omega(Z) = i(Z) =1$.