# basic question on quotient stacks

Let $$X$$ be a scheme over $$S$$, and $$G$$ be an affine group scheme over $$S$$ acting on $$X$$. This Wikipedia article (or also this related MO question) 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 $$[X/G]\to X/G$$?

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.