are closed subfunctors complimentary to open subfunctors? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:27:09Z http://mathoverflow.net/feeds/question/13136 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13136/are-closed-subfunctors-complimentary-to-open-subfunctors are closed subfunctors complimentary to open subfunctors? amaanush 2010-01-27T15:11:00Z 2011-07-22T08:59:04Z <p>I apologize if the following question has already been asked and settled. I couldn't find any thread.</p> <p>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:</p> <p>$$|\mathcal{F}| := \lim_{\to} \mathcal{F} (K),$$</p> <p>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 <em>canonical</em> 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 $|\mathcal{F}|$.</p> <p>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 $|\mathcal{F}|$)?</p> <p>In some sense, are <em>open subfunctors</em> and <em>closed subfunctors</em> really "complimentary"?</p> http://mathoverflow.net/questions/13136/are-closed-subfunctors-complimentary-to-open-subfunctors/70973#70973 Answer by Martin Brandenburg for are closed subfunctors complimentary to open subfunctors? Martin Brandenburg 2011-07-22T08:05:00Z 2011-07-22T08:05:00Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/13136/are-closed-subfunctors-complimentary-to-open-subfunctors/70975#70975 Answer by Harry Gindi for are closed subfunctors complimentary to open subfunctors? Harry Gindi 2011-07-22T08:33:37Z 2011-07-22T08:59:04Z <p>There is a precise formulation of this fact in Toen's <a href="http://ens.math.univ-montp2.fr/~toen/m2.html" rel="nofollow">master course on stacks</a>, although a proof is not given:</p> <p>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). </p>