# equivalent property of open/closed immersion

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.

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.
• 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 $f$. – KReiser Apr 7 at 1:38