Suppose $g,h:Z\to X$ are two morphisms of schemes. Then we say that $f:X\to Y$ is the coequalizer of $g$ and $h$ if the following condition holds: any morphism $t:X\to T$ such that $t\circ g=t\circ h$ factors uniquely through $f$. The question is whether it is possible for a coequalizer $f:X\to Y$ to fail to be surjective.

Remark: $f$ must hit all the closed points of $Y$. To see this, suppose $y\in Y$ is a closed point that $f$ misses. Then $f$ factors through the open subscheme $Y\smallsetminus\{y\}$. It is easy to check (using the fact that $Y$ is the coequalizer) that $Y\smallsetminus\{y\}$ satisfies the universal property of the coequalizer. But coequalizers are unique, so we get $Y=Y\smallsetminus\{y\}$.

Background: A categorical quotient of a scheme $X$ by a group $G$ is the same thing as a coequalizer of the two maps $G\times X\rightrightarrows X$ (given by $(g,x)\mapsto x$ and the action $(g,x)\mapsto g\cdot x$) in the category of schemes. In Geometric Invariant Theory, Mumford defines the notion of a geometric quotient of a scheme by a group (Definition 0.6), which is stronger than the notion of a categorical quotient (Proposition 0.1). Part of the definition is that a map $f:X\to Y$ must be surjective in order to be a geometric quotient. In subsequent pages, he suggests strongly (but doesn't explicitly state) that a categorical quotient need not be surjective.


Let $k$ be a field. Take $Y=\mathrm{Spec}\,k[[t]]$, and take for $X$ the disjoint sum of the closed subschemes $X_n:=\mathrm{Spec}\,k[[t]]/(t^n)$ ($n>0$). Put $Z=X\times_Y X$ with the two obvious maps to $X$. A coequalizer is just a direct limit of the system $X_1\hookrightarrow\dots X_n\hookrightarrow X_{n+1}\hookrightarrow\dots$ in the category of schemes (look at the definition!). Clearly, $Y$ is a direct limit in the category of affine schemes, hence also in the category of schemes since the $X_n$'s are one-point schemes and every compatible system of morphisms $(X_n\to T)_{n>0}$ must factor through an affine open subscheme of $T$.

So, $X\to Y$ is a coequalizer of $pr_1, pr_2 :Z\to X$, but its set-theoretic image is the closed point.

  • $\begingroup$ This is a beautiful example. Thank you! $\endgroup$ – Anton Geraschenko Apr 3 '11 at 22:57
  • $\begingroup$ Awesome! I was waiting for an answer to this :) $\endgroup$ – Andrew Critch Apr 4 '11 at 5:18
  • $\begingroup$ Ah, finally an easy example! :) $\endgroup$ – Martin Brandenburg Apr 12 '11 at 7:46
  • $\begingroup$ @Martin: unfortunately it involves an infinite disjoint sum. I have another example which is separated of finite type, but not so "easy". $\endgroup$ – Laurent Moret-Bailly Apr 14 '11 at 6:26

This might lead to a pedestrian example: Let $R$ be a local ring with maximal ideal $m$, $X$ a scheme, and $f:Spec(R) \to X$ a map. If $U$ is an open subscheme of $X$ containing $f(m)$ then $f$ factors through a map $Spec(R) \to U$. Thus if we have two arrows $g,h:Y \to Spec(R)$, the coequalizer $c: Spec(R)\to C$ must be an affine scheme (else $c$ would factor strictly through an affine neighborhood $U$ of $c(m)$, and $U$ would be a "better coequalizer" than $C$). So If $Y=Spec(B)$ is also affine, then the coequalizer of $g,h$ is just $Spec(A)$ where $A$ is the equalizer of $g^\sharp,h^\sharp$.

let R be $k[x\_i]\_{(x\_i)}$ and $S'=k[y\_i,z\_i]\_{(y\_i,z\_i)}.$ there are two maps $g',h':R \to S$ given by

$$g'(x\_i) = y\_i$$

$$h'(x\_i) = z\_i$$

suppose that $I$ is an ideal of $R$. let $I\_y, I\_z$ be the ideals generated by $g(I)$ and $h(I)$ respectively and let $S = S'/(I\_y + I\_z)$. write $g$ and $h$ for the induced maps $R \to S$. the equalizer of $g,h$ is just $A = k + I \subset R$. it seems unlikely to me that $spec R \to spec A$ is surjective for all choices of $I$.

  • $\begingroup$ Can you explain how you see that A/J is isomorphic to k[z]? $\endgroup$ – Anton Geraschenko Oct 13 '09 at 23:45
  • $\begingroup$ it isn't. now i don't know whether this approach can work. $\endgroup$ – Ian Shipman Oct 14 '09 at 1:32
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    $\begingroup$ This feels very promising. I really like the trick to show that the coequalizer of the spectrum of a local ring in the category of affine schemes is the coequalizer in the category of schemes. That alone gives a much better handle on the problem than I've ever had before. $\endgroup$ – Anton Geraschenko Oct 14 '09 at 22:57
  • $\begingroup$ Something perhaps useful to note... if $E\to R$ is the equalizer of $f,g:R\rightrightarrows B$ and $R$ is local, then $E$ is local and $E\to R$ is a local map: if an element of $E$ is invertible in $R$, its inverse is in $E$. $\endgroup$ – Andrew Critch Mar 30 '11 at 18:14
  • $\begingroup$ The approach using local schemes has led me to this mathoverflow.net/questions/24066/… unanswered question. $\endgroup$ – Martin Brandenburg Mar 30 '11 at 22:48

I think that if G is a simple complex Lie group, and N is its unipotent radical, then the natural map G -> G//N is not surjective. The categorical quotient G//N is the Spec of a ring which is the direct sum of each irreducible representation of G once (this follows from algebraic Peter-Weyl). This ring is graded by the positive Weyl chamber intersected with the weight lattice (its multi-proj is the flag variety). Consider the point defined by the ideal generated by all non-trivial representations. This isn't in the image of G (the group element above it would have to send all highest weight vectors to 0, which is impossible since G acts invertibly), but is a perfectly good element of G//N.

Though now that I think about it, this seems to suggest that the categorical quotient isn't the coequalizer in the category of schemes, but just in the category of affine schemes. Which may be what's confusing you: sometimes you have to add more points to affinize the coequalizer of two maps between affine schemes.

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    $\begingroup$ As Anton said, a categorical quotient is by definition the coequalizer in the category of schemes. This is a problem with your second paragraph, no? $\endgroup$ – Andrew Critch Mar 30 '11 at 17:42

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