Suppose I have an Artin stack $\mathfrak M$ locally of finite-type over $\mathbb C$ with presentation $M\rightarrow \mathfrak M$. Suppose further that $\mathfrak M$ "represents" (in the stack sense) some set of objects which are bounded, i.e. roughly there is a family of such objects $\mathcal F$ over a scheme $S$ of finite type over $\mathbb C$ such that every element of our set of objects occurs as $\mathcal F\_s$ for some closed points $s\in S$. Why does it follow that $\mathfrak M$ is actually of finite-type over $\mathbb C$?


Specifically, I've seen it argued that $S$ induces a surjection $S\rightarrow M$ of schemes over $\mathbb C$, which thus makes $M$ of finite-type over $\mathbb C$. But I don't see why we get an induced map to $M$ instead of just to $\mathfrak M$. Indeed the family $\mathcal F$ should induce a surjective morphism $S\rightarrow \mathfrak M$, but I don't see why this should factor through $M$.



Finite-type Artin Stack over $\mathbb C$
Consider the fiber product $S \times\_{\frak M} M$; its projection onto $S$ is smooth and surjective, hence open. Since $S$ is quasi-compact, there exists a quasi-compact open subset $U$ of $S \times\_{\frak M} M$ surjecting onto $S$. The image of $U$ in $M$ is quasi-compact; let $V$ be a quasi-compact open subscheme of $M$ containing it. Since $S$ surjects onto $\frak M$, so does $U$. So $V \to \frak M$ is smooth and surjective, and $V$ is of finite type, so $\frak M$ is of finite type.


