$ \newcommand{\OXD}{\mathcal{O}_X(D)} \newcommand{\OXDD}{\mathcal{O}_X(D')} $ Let $X$ be a smooth projective curve over $k=\bar{k}$ an effective $D$ a divisor on $X$. Associated to $D$ we have a line bundle $\OXD$, and it turns out that we have $$ D\sim D' \iff \OXD \cong \OXDD, $$ where $\sim$ denotes linear equivalence of divisors. For every $D$ we have the cup-product map on the cohomology groups $$ \mu: H^0(X,\OXD)\otimes H^0(X,K(-D))\to H^0(K) $$ which, if $\OXD = L$, can also be rewritten equivalently as $$ \mu: H^0(X,L)\otimes H^0(X,K\otimes L^{-1})\to H^0(K). $$ Now we come to my question: I'm reading this book about curves and Brill-Noether theory and, at page $190$, they say that
Given a non zero section $s \in H^0(L) $ such that $(s)=D$, the image of $$ s\otimes H^0(X,K\otimes L^{-1})\to H^0(K) $$ under $\mu $ is equal to $H^0(X,K(-D))$.
Could you help me understanding this claim? I'm confused, especially because I was convinced that $H^0(X,K(-D))$ and $H^0(X,K\otimes L^{-1})$ were the same spaces... but from the above statement it looks like this is not the case.
Dear Abramo: I think the issue here is just a slight confusion between equality and isomorphism of sheaves. When you say that $H^0(X,K(−D))$ and $H^0(X,K \otimes L^{−1})$ are "the same" space, you are implicitly using an isomorphism between the sheaves $K \otimes L^{−1}$ and $K(−D)$. (Note that it doesn't really make sense to assert that these sheaves are "the same", since $K \otimes L^{−1}$ is not a priori a subsheaf of $K$.) The statement you quote is just making explicit where that isomorphism comes from: namely, from multiplying by a nonzero section of $L$ which vanishes along $D$.
For completeness, let me say that one thing that might have made the quoted passage clearer is to point out that multiplication by s gives an injective map from $H^0(X,K \otimes L^{−1})$ to $H^0(X,K)$, hence an isomorphism onto its image.