TITLE: Does a scheme have a "separification"?
QUESTION [38 upvotes]: Background: 
(1) If C and D are categories and there is a forgetful functor U:C→D, then a C-ification functor F:D→C is an adjoint to U. For example, the (left) adjoint to the forgetful functor from groups to monoids is "groupification" of a monoid, given by formally adjoining inverses. The (left) adjoint to the forgetful functor from presheaves to sheaves is the usual "sheafification" functor.
Note that whenever you have a (left adjoint) C-ification functor F (whenever you have an adjunction, for that matter), you get a universal property. For any object X∈D, there is a canonical morphism (called the unit of adjunction) εX:X→U(F(X)) with the property that any morphism f:X→U(Y) factors as f=U(g)\circ εX for a unique morphism g:F(X)→Y in C.
(2) A scheme X is separated if the diagonal morphism X→XxX is a closed immersion. It is enough to check that the image of the diagonal is closed. Being separated is the algebro-geometric analogue of being hausdorff, which nothing in algebraic geometry ever is. 

My question is whether there exists a "separification" functor adjoint to the forgetful functor U from the category of separated schemes to the category of schemes. Note that the forgetful functor U does not respect colimits (you can glue together separated schemes to get a non-separated scheme), so it has no hope of having a right adjoint. But U does respect limits (it's enough to show that an arbitrary product of separated schemes is separated and that fiber products of separated schemes are separated), so it might have a left adjoint.
To put it another way, given a scheme X, is there a canonically defined separated scheme Xs and a morphism X→Xs so that any morphism from X to a separated scheme factors uniquely through X→Xs?

Related questions I'd like to know the answer to:

Is there a "relative separification" functor. That is, does an arbitrary morphism of schemes f:X→Y admit a canonical factorization through a separated morphism fs:X'→Y. This would be analogous to Stein factorization, which I regard as "relative affinification". An arbitrary (quasi-compact and quasi-separated) morphism f:X→Y canonically factors through the affine morphism SpecY(f*OX)→Y
Is there a separification functor for algebraic spaces? Is it possible that the separification of a scheme is naturally an algebraic space?
Is there a separification functor for algebraic stacks? (An algebraic stack is separated if the diagonal is proper.)

REPLY [3 votes]: Doesn't a separification exist for trivial reasons?
Given a scheme $X$, consider the category of separated schemes $Y$ together with a morphism $X \to Y$. View this as a diagram, and take the inverse limit. This means we take the product of all schemes $Y$, and set $X^s$ equal to the intersection of the closed graphs of each arrow in this category.
Then every morphism from $X$ to a separated scheme factors through $X^s$. However this is unique because, were there two different morphisms, there equalizer would be a nonempty closed subscheme of $X^s$ which itself is separated and admits a map from $Y$, which is impossible because $X^s$ would have to lie in a map of this graph.
Then we get functoriality the obvious way.
The only issue is set-theoretic. However somehow I think we only need consider maps to separated schemes of bounded cardinalities. 
There's likely some obvious flaw in this thinking. If not, I think the examples David and Gro-Tsen have come up with conclusively show that there is no geometrically natural construction of the separification.