# are closed subfunctors complimentary to open subfunctors?

I apologize if the following question has already been asked and settled. I couldn't find any thread.

Say, $\mathcal{C} = (Sch/k)$, the category of schemes over $k$ (a field). Let $\mathcal{F} \in \mathcal{C}^{\wedge}$, be an object of $\mathcal{C}^{\wedge}$ - the category of contravariant functors from $\mathcal{C}$ to $(Sets)$. One has the set of points:

$$|\mathcal{F}| := \lim_{\to} \mathcal{F} (K),$$

the limit taken over fields $K/k$. Given a subfunctor $\mathcal{G} \hookrightarrow \mathcal{F}$ one gets a subset $|\mathcal{G}| \subset |\mathcal{F}|$ (ie. a canonical map from $|\mathcal{G}| \to |\mathcal{F}|$ that is injective). In particular, $|\mathcal{U}|$ for the open subfunctors $\mathcal{U} \hookrightarrow \mathcal{F}$ form a topology on $|\mathcal{F}|$.

Question: Given a closed subset $Z \subset |\mathcal{F}|$ does there exist a closed subfunctor (possibly non-unique) $\mathcal{Z} \hookrightarrow \mathcal{F}$ so that $Z = |\mathcal{Z}|$ (as subsets of $|\mathcal{F}|$)?

In some sense, are open subfunctors and closed subfunctors really "complimentary"?

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*complementary. Interesting question. – Qfwfq Apr 17 '10 at 13:29
If you don't restrict the form of $\mathcal{F}$ (e.g., to left Kan extensions of functors on schemes of bounded size), it seems that $| \mathcal{F} |$ can be too large to be a set. – S. Carnahan Jul 22 '11 at 8:17
@S. Carnahan: Yes. Usually these size issues in functorial algebraic geometry are ignored by using universes. – Martin Brandenburg Jul 22 '11 at 8:19
Yes, they certainly do make the open subfunctors look good, don't they.. =D – Harry Gindi Jul 22 '11 at 8:22
Regarding the size issue, one can completely bypass it by replacing sheaves on Sch/X with sheaves on $\mathrm{Aff_{fin}}/X$ the small category of finite-presentation affine schemes over $X$. Any adjustment of the size conditions should match this case. – Harry Gindi Jul 22 '11 at 8:43

You are asking if for every open subfunctor $U \to F$ there is a closed subfunctor $Z \to G$ such that $|F|$ is the disjoint union of $|U|$ and $|Z|$. The answer is yes.

For every $A$-valued point $a \in F(A)$, the pullback $U \times_F \text{Spec}(A)$ is an open subfunctor of $\text{Spec}(A)$. Thus there exists a unique reduced ideal $I \subseteq A$ such that $U \times_F \text{Spec}(A) = \text{Spec}(A)_I$. The uniqueness implies that these ideals are compatible when we vary $a$, i.e. we get a quasi-coherent ideal $I \subseteq \mathcal{O}_F$. Now $\text{Spec}(\mathcal{O}_F / I) \to F$ is the desired closed subfunctor.

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I am inclined to think that by "compatible", you mean a functorial association (of course, only then do they stitch together to give a closed subfunctor)-- but that's exactly where the problem is! Given an extension $A\to B$, the pullback of $Spec (A/I)$ to $Spec (B)$ is given by the extension $J = I^e \subseteq B$, but this only has $J \subseteq I_B$, or in fact $\sqrt$J = $I_B$, and need not coincide with $I_B$. For example, even if $\mathcal{F}$ is representable, by, say $Spec (R)$, and $Z$ actually corresponds to $|spec R/{0}| = Spec (R)$, for a non-reduced R-algebra A (corresponding .. – amaanush Jul 25 '11 at 12:54
...to an $A$-valued point of $\mathcal{F}$), the unique reduced ideal in A would be the nil-radical of A, and not the extension of $\{0\}$, which is $\{0\}$! In other words, the association above, as you describe do not associate $I_A \subseteq A$ functorially, thus $\mathcal{F}(Spec (A/I_A)) \subseteq \mathcal{F}$ functorially, ie. does not give a subfunctor. While it may still be possible to augment your method to correctly get an ideal in "$\mathcal{O}_\mathcal{F}$". Please let me know if I have understood you correctly. thanks. – amaanush Jul 25 '11 at 13:14

There is a precise formulation of this fact in Toen's master course on stacks, although a proof is not given:

First, we define the complementary subfunctor of a closed immersion $Y\hookrightarrow X$ to be a particular subfunctor $X - Y\hookrightarrow X$, which is then proven to itself be an open immersion (see example 4 section 4 of Cours 4). It is then noted (without proof) in proposition 1 of section 1 of Cours 4.5 that for every open immersion $U\hookrightarrow X$, there exists a closed immersion $V\hookrightarrow X$ such that $U\cong X - V$, which is, if I remember correctly, not unique (although I believe there is a universal (either initial or terminal) such closed immersion (this universal object is analogous to the "reduced induced" closed subscheme structure on a closed subset of a scheme).

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Yes. The way you state it, if proven will precisely answer my question. I had a copy of "Champs algébriques" -- and while I went through it, trying to give a comprehensible (English) lecture introducing stacks to working mathematicians, I detected many such statements. Independently, these questions are naturally important in studying schemes, or in studying the problem of constructing fine or coarse moduli spaces for a given functor (in a non-GIT specific way). – amaanush Jul 25 '11 at 13:20
Also to note that he may be largely using algebraicity (the diagnal is representable etc) to prove this claim. Even his definition of $X-Y$ seems to depend upon an atlas! which is absolutely unnecessary, and in fact confusing! one can simply observe that the set of points of an open subfunctor rediscovers the subfunctor, as: $\mathcal{U}(T) := \{ \xi \in \mathcal{F}(T) : |\xi| (|T|) \subseteq |\mathcal{U}| \}$. In some sense, the open subfunctors assume the largest possible structure of a subfunctor, while closed ones rarely do so. – amaanush Jul 25 '11 at 13:53