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

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

2011-03-30T18:58:11Z

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.

Answer by Anton Geraschenko for Is being a coequalizer a target-local property in schemes? (answered: no, and no)

2011-03-31T04:54:51Z

[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.

Answer by Laurent Moret-Bailly for Is being a coequalizer a target-local property in schemes? (answered: no, and no)

2011-04-10T07:56:49Z

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 $Y_1\coprod Y_2$.