If $C$ is a curve on a surface $S$, i.e. $i:C\subset S$, and $G$ is a line bundle on $C$, then $G|_U\cong \mathcal{O}_C$ where $U$ is an open subset of $S$, that is, $G$ is trivial on the complement of finitely many points. Then is $\textrm{det}(i_*G)\cong\textrm{det}(i_*\mathcal{O}_C)$? Why is it so?
If $\textrm{det}(i_*G)\cong\textrm{det}(i_*\mathcal{O}_C)$, then from the short exact sequence $0\longrightarrow\mathcal{O}_S(-C)\longrightarrow\mathcal{O}_S\longrightarrow\mathcal{O}_C\longrightarrow 0$, we get that $\textrm{det}(i_*G)=\textrm{det}(i_* \mathcal{O}_C)=\mathcal{O}_S(-C)$.
So is $\textrm{det}(i_*G)\cong\textrm{det}(i_*\mathcal{O}_C)$? Thank you!
The determinant is a line bundle hence locally free. But it is easy to see that two locally free sheaves are isomorphic, if they are isomorphic in codimension one. (Because they are reflexive.) And i think it should be $det(i_{*}G)=\mathcal{O}_S(C)$.