Is being a coequalizer a target-local property in schemes? (answered: no, and no)

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 currently unanswered question about surjectivity of coequalizers, also answered by Laurent Moret-Bailly.

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 $Y_i$).

(a) local to global: 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?

(b) global to local: How about the converse?

Summary of answer by Laurent Moret-Bailly:

(a) local to global: answer is no, but yes if the maps on intersections $h_{ij}:X_{ij}\to Y_{ij}$ are epic (for example if $h$ is schematically surjective, or just universally epic).

(b) global to local: answer is simply no.

Remarks

1) The analogous statements (a) and (b) for coequalizers in the category of locally ringed spaces are true, which can be seen from the construction of coequalizers in LRS (coequalize the topological spaces, and take rings of invariants).

2) The analogous statements for coequalizers in the category of affine schemes is true: 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 $C$.

3) The analogous statements for good geometric quotients of schemes is true. 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 $i$.

4) The analogous statements for equalizers of schemes is true, 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.

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.

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.

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As with your previous question (mathoverflow.net/questions/59812/…) and related ones: This is probably not true, but it will be quite hard to give a counterexample (with proof). Let me just remark that I've worked on Anton's question for quite some time with no result and you probably know my question about colimits of schemes (mathoverflow.net/questions/9961/colimits-of-schemes). After all, we have to ask ourselves if it is worth to spend so much time with weird counterexamples (which I have done). –  Martin Brandenburg Mar 30 '11 at 22:10
A snappier way to put this question is: do regular epimorphisms descend along Zariski covers? Or more generally for other covers, but it may be unlikely. The fact that Sch is extensive may help with this (coproducts are disjoint and stable under pullback), as you can think of the Zariski cover as a regular epi $\coprod Y_i \to Y$ (Zar being subcanonical). But as Anton says, this is not abstract nonsense, and it would be interesting to see other sites in which this holds. In particular, in a topos with the canonical topology. Then you'd be asking if regular epis descend along regular epis. –  David Roberts Mar 31 '11 at 6:23
Just to be clear, we still don't know about the Zariski topology... See my comment after Anton's answer. –  Andrew Critch Mar 31 '11 at 16:45

Let me start with a remark [EDITED for clarity after Andrew's comments]. Given $h:X\to Y$, the following are equivalent:
(1) $h$ is the coequalizer of some $W\rightrightarrows X$,
(2) $h$ is the coequalizer of $X\times_Y X\rightrightarrows X$.
In other words, being a coequalizer is equivalent to being an effective epimorphism (This works in any category with fiber products).

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 this question, 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 $Y$.

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:
$\bullet$ gluing is automatic (and we get a positive answer) if we know that for each open $V\subset Y$, the restriction $h^{-1}(V)\to V$ is an epimorphism of schemes;
$\bullet$ 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 $Y_1\coprod Y_2$.

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Your first sentence is wrong: What's true is that if $W\to X\times_Y X$ is epic, then $Y$ is the coequalizer of $W\rightrightarrows X$ iff it is the coequalizer of $X\times_Y X \rightrightarrows X$ (e.g. the epic hypothesis might not be satisfied in a categorical quotient if the fibres of $Y$ do not consist of single $G$-orbits, as often happens). But epic is only needed for the converse, so your second sentence is correct, and so is everything else, so thank-you for the excellent answer! –  Andrew Critch Apr 11 '11 at 14:53
I especially like your first bullet: that for gluing the factorization maps $t_i$, one only needs that the maps on intersections $h_{ij}:X_{ij}\to Y_{ij}$ are epic, and not necessarily coequalizers themselves, as your answer to (a) shows they might not be :) –  Andrew Critch Apr 11 '11 at 14:56
Assume $h:X\to Y$ is the coequalizer of $f,g:W\rightrightarrows X$. Denoting by $p_1,p_2:X\times_Y X\rightrightarrows X$ the projections, and $(f,g):W\to X\times_Y X$ the obvious map, we have $f=p_1\circ(f,g)$ and $g=p_2\circ(f,g)$. So, every morphism $s:X\to S$ such that $sp_1=sp_2$ also satisfies $sf=sg$, whence (by the assumption on $h$) a unique $t:Y\to S$ such that $th=s$. So $h$ is the coequalizer of $(p_1,p_2)$. Am I missing something? –  Laurent Moret-Bailly Apr 11 '11 at 18:13
Yes, your answer's first sentence is the problem (your comment proves your answer's second sentence, which is correct). To check that $h$ is the coequalizer of $f,g:W\rightrightarrows X$, it is not enough to check that it is the coequalizer of $p_1,p_2:X\times_Y X\rightrightarrows X$, unless you know that $(f,g):W\to X\times_Y X$ is epic. Here's how the proof goes: Suppose $(f,g)$ is epic, and $h=coeq(p_1,p_2)$. Say $s:X\to S$ is a map such that $sf=sg$. Then by epicness of $(f,g)$ we get $sp_1=sp_2$, hence a unique factoring map from $Y$ to $S$. This proof fails without $(f,g)$ epic. –  Andrew Critch Apr 11 '11 at 19:39
I see now that this is no more than a misunderstanding of your intended meaning, which must be your second sentence, because you said "in other words" :) I am concerned with the relationship between $h$ and $(f,g)$, whereas you are concerned with the existence of an $(f,g)$ for a given $h$. –  Andrew Critch Apr 11 '11 at 19:50

[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.]

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 surprising that fiber products can be constructed locally in Sch. The reason it is surprising is that when you construct things locally, you are gluing—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.

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.

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 construct 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 saturated (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. Without special hypotheses, taking the colimit of the diagram of $Y_i$'s is just as hard as taking the colimit of the diagram $W\rightrightarrows X$.

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.

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I think you are indeed missing something... but first, thank you for thinking about this! Now, consider more precisely what "colmits commute" tells use here: Let $U$ denote the disjoint union of the cover of $Y$. I agree that $Y$ really is the coequalizer of $U\times U \rightrightarrows U$ (all products are over $Y$), that $X$ is the coequalizer of $X\times U\times U \rightrightarrows X\times U$, and that $W$ is the coequalizer of $W\times U\times U \rightrightarrows W\times U$. Now suppose $Y$ is the coequalizer of $W\rightrightarrows X$. What "colimits commutes" tells us is...(cont'd) –  Andrew Critch Mar 31 '11 at 7:23
... What "colimits commutes" tells us is that if a colimit $Y_2$ of $W\times U\times U \rightrightarrows X\times U \times U$ exists, and if a colimit $Y_1$ of $W\times U \rightrightarrows X\times U$ exists, then the colimit of $Y_2\rightrightarrows Y_1$ is $Y$. It does not tell us that $Y_1=U$, which would answer "yes" to my question (a). Likewise, if $U=Y_1$ implies that $Y$ is the coequalizer of $W\rightrightarrows X$, that would answer "yes" to question (b). –  Andrew Critch Mar 31 '11 at 7:24