In chapter II Hartshorne defines a closed immersion of schemes as a morphism $f:X \rightarrow Y$ of schemes such that is a homeomorphism between X and a closed subset of Y, and the map $f^{\#}: \mathcal{O}_Y \rightarrow f_{*}\mathcal{O}_X$ is surjective.

Is there an analogous classical definition for a morphism $f: X \rightarrow Y$ of quasiprojective varieties? It seems a pretty basic concept but I can't quite find it in chapter I. If $f$ is such a morphism is it true that the induced map $f^{\#}$ is surjective?

  • $\begingroup$ What result do you want to hold? In your first paragraph you describe a definition, not a result. $\endgroup$ Commented Aug 26, 2017 at 21:36
  • $\begingroup$ @MarianoSuárez-Álvarez I edited the question, I hope it makes more sense. $\endgroup$
    – Jesse
    Commented Aug 26, 2017 at 22:04

1 Answer 1


The morphism of sheaves $f^{\#}: \mathcal{O}_Y \rightarrow f_{*}\mathcal{O}_X$ is surjective if and only if the associated morphisms on stalks $f^{\#}_y: \mathcal{O}_Y,y \rightarrow (f_{*}\mathcal{O}_X)_y$ are surjective for all $y\in Y$.
However for $x\in X$ with image $f(x)=y\in Y$ we have an identification $ (f_{*}\mathcal{O}_X)_y\stackrel {\to }{=}\mathcal O_{X,x}$.
So the surjectivity of $f^{\#}$ means that the germs of regular functions at $x\in X$ are exactly the germs obtained by restricting (or to be more formal, pulling back) germs of regular functions at $y$ in $Y$. This is extremely natural, and exactly the same explanation works in other geometric categories like algebraic varieties, smooth manifolds, complex analytic spaces,...

NB Needless to say, for $y\in Y\setminus f(X)$ we have $(f_{*}\mathcal{O}_X)_y=0$ and surjectivity of $f^{\#}_y: \mathcal{O}_Y,y \rightarrow (f_{*}\mathcal{O}_X)_y=0$ is then automatic.


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