The question concerns quotient stacks. I am not very comfortable with stacks, so feel free to edit the question if I am saying nonsense. It is also the reason why I may be spelling in too much detail some obvious facts...
Fix a base scheme $S$ for everything that follows. Say we have a quotient stack $[X/G]$, where $X$ is a scheme and $G$ acts on $X$ (e.g. $G=GL_n$) and consider $F$ a substack of $[X/G]$, such that $F\hookrightarrow [X/G]$ is representable and a closed immersion.
(If it helps, think of $[X/G]$ as some moduli space of some objects and $F$ a moduli space classifying those objects with an extra property $P$).
If I understand correctly, we have a map (of stacks) $X\to[X/G]$ corresponding to the trivial torsor $G\times X$: \begin{array}{c c c} G\times X & \to & X \\ \downarrow & & \\ X & & \end{array} where the vertical map is projection and the horizontal is the action.
Q1: is this map universal in any sense? (It seems so to me, but I can't make it a precise statement). In the moduli space setting, it seems to correspond to ''the most general object'' you can write, assuming the objects can be cut out by equations). Does this make sense?
Edit/addendum to Q1: In the moduli space setting, the "distinguished" element $X\to[X/G]$ corresponds to some object $A$ over $X$. What can be said about $A$?
Continuing, take the fiber product \begin{array}{c c c} Z & \to & F \\ \downarrow & & \downarrow\\ X & \to & [X/G] \end{array} gives a closed subscheme $Z$ of $X$ (by the assumption on $F$).
Q2: Under what conditions on $F$ (or is it always true), do we have $F=[Z/G]$? A priori, it's not even clear that $Z$ has a $G$-action, but I believe I checked that and I think I can write a map $[Z/G]\to F$.
And the last question: Q3: Again, is there any "universality" to this? It seems to me that it says something along the lines of: take a scheme $U$ over $S$. Then a map $U\to[X/G]$ corresponds to an object $A$ classified by our moduli problem and the assumption on $F$ shows that the fiber product $V$ is closed in $U$: \begin{array}{c c c} V & \to & F \\ \downarrow & & \downarrow\\ U & \to & [X/G] \end{array} Then $A$ (over $U$) has property $P$ if and only if it factors through $V$? Or maybe (if $P$ can be checked on points), $V$ is the ''locus'' where the object $A$ has property $P$. Does this make sense?
Thank you.