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.