Is being a coequalizer a target-local property in schemes? (answered: no, and no) - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T10:39:13Z http://mathoverflow.net/feeds/question/60105 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60105/is-being-a-coequalizer-a-target-local-property-in-schemes-answered-no-and-no Is being a coequalizer a target-local property in schemes? (answered: no, and no) Andrew Critch 2011-03-30T18:58:11Z 2011-04-12T06:52:49Z <p>This question is aimed at a better understanding of GIT's "categorical quotients", which are defined as coequalizers of group actions $G\times X\rightrightarrows X$ in the category of schemes. See also Anton's <del>currently unanswered</del> question about <a href="http://mathoverflow.net/questions/63/can-a-coequalizer-of-schemes-fail-to-be-surjective" rel="nofollow">surjectivity of coequalizers</a>, also answered by Laurent Moret-Bailly.</p> <p>Suppose $f,g:W\rightrightarrows X$ and $h:X\to Y$ are scheme maps such that $hf=hg$. Let $Y_i$ be a Zariksi cover of $Y$, and let $X_i$ and $W_i$ be their pullbacks to $Y_i$ (i.e. the preimages of the open sets $Y_i$).</p> <blockquote> <p><b> (a) local to global: </b> Is it true that if $W_i\rightrightarrows X_i\to Y_i$ is a coequalizer in the category of schemes for every $i$, then $Y$ is a coequalizer in schemes?</p> <p><b> (b) global to local: </b> How about the converse?</p> </blockquote> <p><b>Summary</b> of <a href="http://mathoverflow.net/questions/60105/is-being-a-coequalizer-a-target-local-property-in-schemes/61192#61192" rel="nofollow">answer by Laurent Moret-Bailly</a>:</p> <p>(a) local to global: answer is <b>no, but yes</b> if the maps on intersections $h_{ij}:X_{ij}\to Y_{ij}$ are <b>epic</b> (for example if $h$ is schematically surjective, or just universally epic).</p> <p>(b) global to local: answer is simply <b>no</b>.</p> <p><b>Remarks</b></p> <p>1) The analogous statements (a) and (b) for coequalizers in the category of locally ringed spaces are <b>true</b>, which can be seen from the construction of coequalizers in LRS (coequalize the topological spaces, and take rings of invariants).</p> <p>2) The analogous statements for coequalizers in the category of affine schemes is <b>true</b>: That $C\to B\rightrightarrows A$ is an equalizer is equivalent to the exactness of the $C$-module sequence $0\to C \to B \stackrel{f-g}{\to} A \to 0$, which can be checked in the localizations at prime (or maximal) ideals of $C$.</p> <p>3) The analogous statements for good geometric quotients of schemes is <b>true</b>. That is, working in Schemes/$S$, if we take $W=G\times_S X$, then $X\to Y$ is a good geometric quotient iff $Y_i$ is a good geometric quotient of $W_i\rightrightarrows X_i$ for all $i$.</p> <p>4) The analogous statements for <i>equalizers</i> of schemes is <b>true</b>, because fibred products can be checked/constructed on open covers, as is essentially proved in Hartshorne chapter II.3. In fact in any category, pulling back along a morphism preserves all limits, but not colimits, and in particular not coequalizers.</p> <p>5) If $W=Spec(A),X=Spec(B)$ and $Y$ is their scheme coequalizer, then $Y$ is usually not affine (e.g. when gluing along opens), but $Spec(\cal{O}_Y(Y))$ is the coequalizer in the category of affine schemes. That is, $\cal{O}_Y(Y)$ is canonically isomorphic to the equalizer $C$ of $f^\sharp, g^\sharp:B\rightrightarrows A$ in rings, whose underlying set is the equalizer in sets.</p> <p>6) If in (5) $B$ is a local ring, then $Y$ is affine, $Y=Spec(C)$, $C$ is local, and $C\to A$ is a local map.</p> http://mathoverflow.net/questions/60105/is-being-a-coequalizer-a-target-local-property-in-schemes-answered-no-and-no/60151#60151 Answer by Anton Geraschenko for Is being a coequalizer a target-local property in schemes? (answered: no, and no) Anton Geraschenko 2011-03-31T04:54:51Z 2011-04-04T16:03:06Z <p>[Edit: This answer is wrong, as Critch explains in the comments, but I'd like to leave it undeleted. Please don't vote it up.]</p> <p>The answer to both questions is yes, but it seems so obvious to me that I suspect I'm making a mistake. I think your Remark 4 is a bit confusing. It's true in any category that pulling back preserves limits, but it is <strong>surprising</strong> that fiber products can be constructed locally in <strong>Sch</strong>. The reason it is surprising is that when you construct things locally, you are gluing&mdash;that is, you are making a colimit. There is no abstract reason that fiber products (which are limits) should commute with gluing (which is a colimit). This is why Harshorne II.3 is not simply abstract nonsense.</p> <p>On the other hand, forming colimits automatically commutes with forming other colimits in any category, just as pulling back automatically respects limits. Whether you form the colimits $Y_i$ and then glue, or glue the $X_i$ and then form the colimit, it's all the colimit of one big diagram.</p> <p>Note that this is only easy because you started with an open cover of $Y$. In other words, it's easy to check locally that something is a colimit. But it is not easy to <em>construct</em> colimits locally. Indeed, it's hard to even formulate what it would mean to construct colimits locally. First of all, you need the open cover $X_i$ to be <em>saturated</em> (i.e. you need the two pullbacks to $W$ to agree). Even if you construct colimits of $W_i\rightrightarrows X_i\to Y_i$ for a saturated cover $X_i$, there is no guarantee that the maps between the $Y_i$ will be open immersions.<sup>&dagger;</sup> Without special hypotheses, taking the colimit of the diagram of $Y_i$'s is just as hard as taking the colimit of the diagram $W\rightrightarrows X$.</p> <p><sup>&dagger;</sup>For example, $\mathbb A^2\smallsetminus{0}$ is an open subscheme of $\mathbb A^2$, and is saturated with respect to the dialation action of $\mathbb G_m$. However, the map on colimits is $\mathbb P^1\to \ast$, which is not an open immersion.</p> http://mathoverflow.net/questions/60105/is-being-a-coequalizer-a-target-local-property-in-schemes-answered-no-and-no/61192#61192 Answer by Laurent Moret-Bailly for Is being a coequalizer a target-local property in schemes? (answered: no, and no) Laurent Moret-Bailly 2011-04-10T07:56:49Z 2011-04-12T06:52:49Z <p>Let me start with a remark [EDITED for clarity after Andrew's comments]. Given $h:X\to Y$, the following are equivalent:<br> (1) $h$ is the coequalizer of some $W\rightrightarrows X$,<br> (2) $h$ is the coequalizer of $X\times_Y X\rightrightarrows X$.<br> In other words, being a coequalizer is equivalent to being an effective epimorphism (This works in any category with fiber products).</p> <p>Back to the questions. Question (b) asks whether if $h$ is a coequalizer, then its restriction $h^{-1}(V)\to V$ also is, for each open $V\subset Y$. Let me recall the example I gave to answer <a href="http://mathoverflow.net/questions/63/can-a-coequalizer-of-schemes-fail-to-be-surjective" rel="nofollow">this question</a>, which provides a counterexample where $h^{-1}(V)$ is empty (and $V$ isn't): take $Y=\mathrm{Spec}\,k[[t]]$ ($k$ a field), $X=$ the disjoint sum of all subschemes $\mathrm{Spec}\,(k[[t]]/(t^n))$ ($n\geq1$), $V=$ generic point of $Y$.</p> <p>For question (a), assume each $h_i:X_i\to Y_i$ is a coequalizer and let $s:X\to S$ be a morphism such that $sf=sg$. Then for each $i$, the restriction of $s$ to $X_i$ descends uniquely to $t_i:Y_i\to S$. The question is whether $t_i$ and $t_j$ coincide on $Y_i\cap Y_j$. Composing them with (the restriction of) $f$ (or $g$) gives the same result, hence:<br> <code>$\bullet$</code> gluing is automatic (and we get a positive answer) <em>if</em> we know that for each open $V\subset Y$, the restriction $h^{-1}(V)\to V$ is an epimorphism of schemes;<br> <code>$\bullet$</code> but the above example shows that this is not true in general, and in fact we get a (nonseparated) counterexample to the question by taking two copies $X_i\to Y_i$ ($i=1,2$) of that example and putting $X=X_1\coprod X_2$, $Y=$ gluing of $Y_1$ and $Y_2$ along the generic points: here the coequalizer of $X\times_Y X\rightrightarrows X$ is $Y_1\coprod Y_2$.</p>