Let $X$ be a scheme and $G$ be a group acting on $X$. Suppose the action is not free. Consider the quotient sheaf $X/G.$ Can we directly prove that the sheaf quotient is not an algebraic space?
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$\begingroup$ The sheaf quotient is a sheaf of sets, and in some cases may be represented by an algebraic space. The stack quotient is a stack. $\endgroup$– S. Carnahan ♦Commented Jul 15, 2023 at 6:15
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$\begingroup$ @S. Carnahan Thank you for the clarification. $\endgroup$– S.D.Commented Jul 15, 2023 at 11:08
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1 Answer
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Say the action is free if $G \times X \to X \times X$, $(g, x) \mapsto (g\cdot x, x)$ is a monomorphism. (Convince yourself this is the right definition.)
This condition is actually equivalent to the quotient stack $[X/G]$ being an algebraic space. Indeed recall an algebraic stack is an algebraic space iff its diagonal is a monomorphism (meaning that it is a sheaf of sets). Now that condition can be checked smooth-locally on the target, and if you pull back the diagonal along $X \times X \to [X/G] \times [X/G]$ (square of the quotient map), you get precisely the map $G\times X \to X\times X$ above.
