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I'm studying morphisms of ringed topological spaces. The definition of open/closed immersion in Qing Liu's "Algebraic Geometry and Arithmetic Curves" is as follows:

Definition 2.22. We say that a morphism $(f,f^\#):(X,\mathcal O_X)\to(Y,\mathcal O_Y)$ is an open immersion (resp. closed immersion) if $f$ is a topological open immersion (resp. closed immersion) and if $f^\#_x$ is an isomorphism (resp. if $f^\#_x$ is surjective) for every $x\in X$.

But in many occasions, I think the author is using the following equivalent (not sure) property of open/closed immersion:

My conjecture. A morphism $(f,f^\#):(X,\mathcal O_X)\to(Y,\mathcal O_Y)$ is an open immersion (resp. closed immersion) if and only if $f$ is a topological open immersion (resp. closed immersion) and $f^\#_y:\mathcal O_{Y,y}\to(f_*\mathcal O_X)_y$ is an isomorphism (resp. $f^\#_y$ is surjective) for every $y\in Y$.

Is it really true? I know that $f^\#_x = g\circ f^\#_{f(x)}$, where $g$ is the canonical map $g:(f_*\mathcal O_X)_{f(x)}\to\mathcal O_{X,x}, \ s_{f(x)}\mapsto s_x$, for each $x\in X$ (see the diagram below). $$\mathcal O_{Y,f(x)}\to(f_*\mathcal O_X)_{f(x)}\to\mathcal O_{X,x}$$ But I can't go any further.

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No, this is not correct - if $y$ is not in the image of $f$, then $(f_*\mathcal{O}_X)_y$ is zero. This isn't a problem for the case of a closed immersion, but it is for an open immersion.

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  • $\begingroup$ So the conjecture in the qustion is correct for a closed immersion? $\endgroup$ – zxcv Apr 7 at 1:27
  • $\begingroup$ Sure, but I don't see a reason to try and reframe it in terms of points of $Y$ like you're doing. You already know what happens at the points of $Y$ that aren't in the image of $f$. $\endgroup$ – KReiser Apr 7 at 1:38

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