# Closed immersion definition

Hartshorne defines a closed immersion as a morphism $f:Y\longrightarrow X$ of schemes such that

a) $f$ induces a homeomorphism of $sp(Y)$ onto a closed subset of $sp(X)$, and furthermore

b) the induced map $f^\#:\mathcal{O}_X\longrightarrow f_* \mathcal{O}_Y$ of sheaves on $X$ is surjective.

My doubt is that why is the condition a) necessary?

Atleast for affine schemes, if $f:(\textrm{Spec}\ B,\mathcal{O}_{\textrm{Spec}\ B})\longrightarrow (\textrm{Spec}\ A,\mathcal{O}_{\textrm{Spec}\ A})$ is a morphism such that condition (b) holds, then we have the stalk level maps are all surjective. But the stalk level maps are all localization maps which will mean that $f^\#_{\mathcal{O}_{\textrm{Spec}\ A}}:A\longrightarrow B$ is itself surjective. But that itself will mean that $f$ is a closed immersion right?

The definition made in Hartshorne must be because this does not carry through for general schemes. But I am not able to think of any example. Any help in this respect will be appreciated.

I figured I would make an answer out the comments to the first answer addressing a point which confused me when I first learned this stuff. A morphism $$f:X\to Y$$ (I have to write it in this direction or else I'll confuse myself) is said to be a closed immersion if $$f$$ induces a homeomorphism of $$X$$ onto a closed subset of $$Y$$, and $$f^\sharp:\mathscr{O}_Y\to f_*(\mathscr{O}_X)$$ is surjective.

In some references I've seen it is casually remarked that the second condition is equivalent to surjectivity of the map $$f_x^\sharp:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$$ for all $$x\in X$$. But is this really trivial? No! This map, which might reasonably be called the stalk of the morphism $$f$$ at $$x$$, is not literally the same as the stalk of $$f^\sharp$$ at $$f(x)$$. Indeed, that is a map $$f_{f(x)}^\sharp:\mathscr{O}_{Y,f(x)}\to(f_*\mathscr{O}_X)_{f(x)}$$. In general, there is always a natural map $$\varphi_x:(f_*\mathscr{O}_X)_{f(x)}\to\mathscr{O}_{X,x}$$, but it isn't in general an isomorphism. The map $$f_x^\sharp$$ is equal to $$\varphi_x\circ f_{f(x)}^\sharp$$. So while it is standard that the map $$f^\sharp$$ of sheaves (on $$Y$$!) is surjective if and only if $$f_y^\sharp:\mathscr{O}_{Y,y}\to (f_*\mathscr{O}_X)_y$$ is surjective for all $$y\in Y$$, this does not obviously say anything about surjectivity of the maps $$f_x^\sharp:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$$ for $$x\in X$$. If however $$f$$ is a homeomorphism onto a closed subset $$f(X)\subseteq Y$$, then the stalks of $$f_*\mathscr{O}_X$$ at points of $$Y$$ are easy to compute: they are zero at points outside of $$f(X)$$, and at a point $$f(x)\in f(X)$$, we have that the natural map $$(f_*\mathscr{O}_X)_{f(x)}\to\mathscr{O}_{X,x}$$ is an isomorphism. So in that case, surjectivity of each $$f_x^\sharp$$, $$x\in X$$, actually will imply surjectivity of $$f_y^\sharp$$, $$y\in Y$$, and hence of $$f^\sharp$$.

Without the condition that $$f$$ is a closed topological immersion on the underlying topological spaces, it is not going to be true that $$f^\sharp$$ is surjective if and only if $$f_x^\sharp$$ is surjective for all $$x\in X$$. To make this clearer, let's assume $$X=\mathrm{Spec}(B)$$ and $$Y=\mathrm{Spec}(A)$$, so $$f=\mathrm{Spec}(\alpha)$$ for $$\alpha:A\to B$$ a ring homomorphism. The stalk map of $$f$$ at $$x=\mathfrak{q}\in\mathrm{Spec}(B)$$ is the ring map $$A_\mathfrak{p}\to B_\mathfrak{q}$$, where $$\mathfrak{q}=\alpha^{-1}(\mathfrak{p})$$. In general, surjectivity of this map for all $$\mathfrak{q}\in\mathrm{Spec}(B)$$ does not imply surjectivity of $$\alpha$$ itself.

I think the simplest example that will illustrate this is when $$B=A_g$$ is a principal localization of $$A$$. Then in fact the stalk map in the previous paragraph is an isomorphism for every prime ideal of $$A_g$$ (the set of which are in natural bijection with the set of primes of $$A$$ not containing $$g$$, i.e. $$D(g)$$). But the localization map $$A\to A_g$$ (i.e. the map on global sections of $$f$$) is not usually surjective. Note that in this case $$f$$ is a homeomorphism onto the open subset $$D(g)$$ of $$\mathrm{Spec}(A)$$, but $$D(g)$$ is not generally closed in $$A$$.

I think maybe this illustrates why the first condition is important, and why, if one wants to think about surjectivity of $$f^\sharp$$ in terms of the stalks of $$f$$, $$f_x^\sharp$$, for $$x\in X$$, the topological condition is needed, and logically “precedes” the condition on $$f^\sharp$$.

Lastly, I should note that the maps which I have been calling the “stalks of $$f$$,” $$f_x^\sharp$$, for $$x\in X$$, are in fact the stalks of the map of sheaves on $$X$$ (in the usual sense) $$f^\flat:f^{-1}\mathscr{O}_Y\to \mathscr{O}_X$$ corresponding to $$f^\sharp$$ under the adjunction between $$f^{-1}$$ and $$f_*$$. So surjectivity of all $$f_x^\sharp$$, $$x\in X$$, is logically equivalent to surjectivity of $$f^\flat$$. There is no reason to believe that $$f^\flat$$ is surjective if and only if $$f^\sharp$$ is, or even that there is an implication in either direction in general.

• Very complete:+1. Mar 15, 2014 at 8:29
• For the record, the general result is: Let $f:X\to Y$ be a continuous and closed map of topological spaces, and suppose there is $x\in X$ is such that $f^{-1}(f(x))=\{x\}$. Then, for any sheaf $\mathcal{F}$ over $X$, the natural map $$(f_*\mathcal{F})_{f(x)}\to\mathcal{F}_x$$ is an isomorphism. Feb 18 at 17:50
Here is where the confusion lies: Let $\varphi : \mathcal{F} \rightarrow \mathcal{G}$ be a morphism of sheaves on a scheme $X$. Then $\varphi$ is surjective does not mean that for all open $U \subset X$, the morphism $\varphi(U) : \mathcal{F}(U) \rightarrow \mathcal{G}(U)$ is surjective.
• What you say is correct, only the stalk level maps are surjective. But even then the map $A\longrightarrow B$ will be surjective, and my question will make sense, right? Mar 14, 2014 at 17:39
• @poorna: The stalks being surjective does not imply that the map on global sections is surjective. This is indeed the content of the Caution I referenced in my answer. Think about what happens for the open immersion $\mathrm{Spec}(k[x^\pm]) \hookrightarrow \mathrm{Spec}(k[x])$. The stalks at any closed point of $\mathrm{Spec}(k[x^\pm])$ is surjective, but the global sections $k[x] \rightarrow k[x^{\pm}]$ are not. Mar 14, 2014 at 18:09
• Dear @poorna, The maps on stalks are of the form $A_\mathfrak{p}\to B_\mathfrak{q}$, where $\mathfrak{q}$ is a prime of $B$ lying over the prime $\mathfrak{p}$ of $A$. These can be surjective for all $\mathfrak{q}\in\mathrm{Spec}(B)$ without $A\to B$ being surjective. You're thinking of the assertion that if $A_\mathfrak{p}\to B_\mathfrak{p}$ is surjective for all primes $\mathfrak{p}\in\mathrm{Spec}(A)$, then $A\to B$ is surjective. Mar 14, 2014 at 18:24