Let $X$ be a scheme over $S$, and $G$ be an affine group scheme over $S$ acting on $X$. This [Wikipedia article](https://en.wikipedia.org/wiki/Quotient_stack) (or also this related [MO question](https://mathoverflow.net/questions/159279/understanding-the-definition-of-the-quotient-stack-x-g)) defines a quotient stack $[X/G]$ as a category of principal $G$-bundles fibred over Sch/S.


When they discuss the map $[X/G] \to X/G$ (when $X/G$ exists as an algebraic space, or let's say scheme for convenience if one prefers), they say "complete the diagram" i.e. if $D\to T$ is a principal $G$-bundle corresponding to a $T$-point $T\to [X/G]$, then we can complete the diagram
$T \leftarrow D\to X \to X/G$ (which I understand as induing map $T \to X/G$)


It is not clear to me how one can induce a map $T\to X/G$ using the fact that $D\to X$ is $G$-equivariant.


**Q1. How do we get the induced map $T\to X/G$?**


**Q2. If we have $T$-point of $X/G$, pull back of the diagram $T\to X/G \leftarrow X$ gives a principal $G$-bundle $D\to T$ and $D\to X$ which is $G$-equivariant. Then what is the obstruction of this map $X/G \to [X/G]$ being the quasi-inverse of $[X/G]\to X/G$?**



basic question on quotient stacks
The map $D\to X$ induces a map $D/G\to X/G$. Since $D\to T$ is a principal $G$-bundle the induced map $D/G\to T$ is an isomorphism (e.g. by checking locally over $T$ and reduce to the case of the trivial bundle) and so we get a map $T \to X/G$. This is the map they refer to. 


For the second, in general the map $X\to X/G$ is not a principal bundle. In general some fibers of the map $X\to X/G$ might not have a free transitive action of $G$. A prototypical example is the action of $\mathbb{G}\_m$ on $\mathbb{A}^1$ homotheties, where the quotient shceme is a single point and carry no corresponding principal bundle of course. 


