Suppose $X$ is a complex manifold, we have the map $H^0(X,K\_X^\*/O\_X^\*)\to H^1(X,O\_X^\*)$, is the canonical line bundle $\wedge^n{\Omega}$ always in the image of the map?



Must a canonical line bundle be associated to a cartier divisor?
The answer is no. The group $H^0(X, \mathcal{K}\_X^\*/\mathcal{O}\_X^\*)$ is isomorphic to the divisor group $\mathrm{Div}(X)$ (see e.g. Huybrechts' *Complex Geometry*, Prop. 2.3.9), so its image in $\mathrm{Pic}(X)$ is the group of line bundles associated to some divisor. Now there are complex compact surfaces which do not contain any curve but with a nontrivial canonical bundle, for instance the Inoue surfaces (see Barth et al., ch. V, §19).


