TITLE: Do quotients of representable sheaves represent quotients?
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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!