Here $X$ is a complex manifold. The proof below is from Wells'book p.106. I have a question about the claim that to see $\delta(H^1(X,\mathcal{O}^*))=H^2_{1,1}(X,\mathbb{Z})$, it suffices to show that the image of $H^2_{1,1}(X,\mathbb{Z})$ in $H^1(X,\mathcal{O})$ is zero.
By the exactness of (4.7) (attached at last), we know $\delta(H^1(X,\mathcal{O}^*))=ker(H^2(X,\mathbb{Z})\to H^1(X,\mathcal{O}))$, but the image of $H^2_{1,1}(X,\mathbb{Z})$ in $H^1(X,\mathcal{O})$ is zero only shows $H^2_{1,1}(X,\mathbb{Z})\subset ker(H^2(X,\mathbb{Z})\to H^1(X,\mathcal{O}))$, I wonder why it proves that $\delta(H^1(X,\mathcal{O}^*))=H^2_{1,1}(X,\mathbb{Z})$?
