# $\mathcal{O}_Y \to f_*\mathcal{O}_X$ is an Isomorphism

Let $$f: X \to Y$$ be a morphism of ringed spaces. I'm looking for criteria for $$f, X,Y$$ such that the morphism

$$\mathcal{O}_Y \to f_*\mathcal{O}_X$$

is an isomorphism.

Obviously, a necessary condition is that $$f_*\mathcal{O}_X$$ is locally free of dimension $$1$$ (in other words an invertible sheaf). But does it suffice to get an isomorphism?

I heard often there are criteria like that $$Y$$ has to be normal or $$X$$ projective.

Could anybody explain if there exist a intuitive access to understand when one can expect that $$\mathcal{O}_Y \cong f_*\mathcal{O}_X$$ hold and what are the obstacles which could prevent this from being isomorphic.

Is there a geometric intuition behind this phenomenon when it occurs?

My considerations:

Locally $$\mathcal{O}_Y \to f_*\mathcal{O}_X$$ for affine $$U \subset Y$$ it is given by ring maps $$\mathcal{O}_Y(U)=: R \to f_*\mathcal{O}_X(U)=:A$$.

So for example if $$R$$ is normal and $$f$$ finite this gives an isomorphism.

What about the intuition if $$X$$ is projective? Can the concepts and these two examples be considered from a much more sophisticated viewpoint?