Here's the context for the question: Proposition 4.6 of Freitag and Kiehl's book on etale cohomology shows that a sheaf (of sets) $\mathcal{F}$ (on the site Et(X)) is constructible if and only if it is the coequalizer of an etale equivalence relation $\mathcal{R}\rightrightarrows \mathcal{Y}$, where $\mathcal{R}$ and $\mathcal{Y}$ are representable sheaves. Here an etale equivalence relation is defined exactly as you would expect: $\mathcal{R}\rightarrow \mathcal{Y}\times \mathcal{Y}$ is injective, and for every etale $U\rightarrow X$, $\mathcal{R}(U)\subset \mathcal{Y}(U)\times \mathcal{Y}(U)$ is an equivalence relation (of sets).

Now if the quotient $\mathcal{Y}/\mathcal{R}$ "should" be represented by the quotient $Y/R$ (where $Y$ represents $\mathcal{Y}$ and $R$ represents $\mathcal{R}$), well, it sounds like constructible sheaves should be algebraic spaces, or at least there should be some relationship. On the other hand, I don't think this could be right.

So is the problem that you can't take sheafy quotients like this, or is it something more subtle?


2 Answers 2


It seems like your interpretation is correct. The bottom of page 39 reads

As we have already seen in § 1, for every sheaf of sets $\mathcal F$ on the scheme $X$ there is a family $X_\alpha$ of etale $X$-schemes and a surjective sheaf mapping $\coprod \tilde X_\alpha\to \mathcal F$ (where $\tilde X_\alpha$ is the sheaf represented by $X_\alpha$). The sheaf $\mathcal F$ is called constructible if one can get by with a finite family. In this case $\mathcal F$ actually turns out to be the quotient of a representable sheaf by a representable equivalence relation (not separated, in general). Thus one can interpret $\mathcal F$ as an etale algebraic space over $X$ (not necessarily separated) in the sense of M. Artin, Knutson [91].

Since the disjoint union of an arbitrary collection of etale $X$-schemes is an etale $X$-scheme, they must be implicitly assuming each $X_\alpha$ is finite type or something.

  • $\begingroup$ I really like this answer $\endgroup$ Nov 19, 2009 at 1:07

This is just a little side remark that in general it does make a difference whether you take quotients as representable functors or as schemes. There is a counterexample in section 4 of http://www.math.univ-toulouse.fr/~toen/cours1.pdf.

The counter example is as follows. Let R be the real number and Q the rationals. Then Q acts on R by translation. The quotiens R/Q is nothing reasonably geometric. But if we take h_R to be the functor represented by R, then the quotient h_R / Q is an algebraic space!


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