Question

Suppose I have a diagram of schemes, and I know that the colimit exists in the category of schemes. How does this colimit compare with the colimit of the corresponding sheaves (I'm being nonspecific about the topology on purpose)? We always have a map from the colimit of sheaves to the colimit of schemes. Are then any conditions I can impose on my diagram so that this map is an isomorphism? Is there any reference where this issue is discussed?

Answer 1

My understanding is that it typically requires hard projective geometry to show that the sheaf-theoretic quotient is a scheme. (See Grothendieck's Seminaire Bourbaki talks in FGA.) I also think that showing it is an algebraic space is much easier, or at least can be done under wide and reasonable generality, but still requires some work. (See the work of M Artin.) But I think it is usually not hard to check whether a given scheme or algebraic space is the sheaf-theoretic quotient. In fact, unless I'm mistaken, it's easy enough that no one ever bothered to write it down.

First consider the (main) case where the colimit diagram is of the form $E\rightrightarrows X$, where the two maps are the projections of an equivalence relation on $X$. Then the quotient $X/E$ in the category of sheaves has the properties that (i) the map $X\to X/E$ is an epimorphism and (ii) the induced map $E\to X\times_{X/E} X$ is an isomorphism. Further, these properties characterize the quotient up to unique isomorphism. This is just general sheaf theory.

So if $f\colon X\to Y$ is a map such that the two compositions $E\rightrightarrows X \to Y$ agree, the induced map $X/E\to Y$ is an isomorphism if and only if $Y$ satisfies the properties above. So the candidate quotient map $f\colon X\to Y$ mush be a sheaf epimorphism, and we must have $E=X\times_{Y} X$. These are easy to check in many cases. For (i), a map $f\colon X \to Y$ is an epimorphism if and only if it has a section locally on a cover of $Y$ with respect to the Grothendieck topology in question. This is again just general sheaf theory. If $X$ and $Y$ are schemes, this is often easy to check -- for instance any smooth surjection has a section etale locally. And for (ii), you just need to check that two subobjects of $X\times X$ agree. Often they'll be closed subschemes of $X\times X$, so you just need to show they have the same ideal sheaf.

For example, the map $\mathbf{A}^1\to\mathbf{A}^1$ given by $z\mapsto z^2$ is the sheaf-theoretic quotient of $\mathbf{A}^1$ by the involution $z\mapsto -z$ in the fppf or fpqc topology, but not in the etale or Zariski topology.

Now consider the case where you're forming the sheaf-theoretic coproduct of a family of schemes. This is easy. It's representable by the disjoint union of the schemes in the family. (Maybe I using some weak properties of the topology here...)

For a general diagram of schemes, I think we can just combine the two previous cases. Represent such a diagram by a functor $I\to \mathrm{Schemes}$, where $I$ is an "index" category (small). Let $X'$ be the sheaf colimit $\mathrm{colim}_{i\in I} X_i$, and let $D$ be the disjoint union $\coprod_{i\in I} X_i$. Then the induced map $D\to X'$ is an epimorphism, and so $X'$ is the quotient of $D$ by the equivalence relation $D\times_{X'} D$. So to check whether a candidate quotient $Y$ is the actual quotient, you need to show that (i) the map $D\to Y$ is a sheaf epimorphism and (ii) the corresponding equivalence relation $D\times_Y D$ agrees with $D\times_{X'} D$. As above, I'd say it's typically not hard to show (i). To show (ii), you'd need to have some hands-on understanding of $D\times_{X'}D$. But I'm going to guess that this is not hard. I bet you just take the usual definition of colimits in the category of sets (disjoint unions modulo and explicit equivalence relation given by the maps in the diagram) and then repeat the definition in the category of sheaves, making sure to replace existential quantifiers by their local versions with respect to the topology in question. (This is getting a bit more abstract than I enjoy, so I'm calling it quits. But if you have a specific diagram, I might be willing to give it a shot.)