Let $G$ be an algebraic group acting on an $S$-scheme $X$. Let $T\in \operatorname{Sch}/S $.
Given a principle $G$-bundle $P\to T$, the contracted product is defined as $(P\times_SX)/G$ where $G$ acts by $g.(p,x) = (pg^{-1}, gx)$.
Show that giving an equivariant map $P\to X$ is equivalent to giving a section of the projection $P\times^G X\to T$.
- First direction
Suppose we have an equivariant map $\pi:P\to X$.
If we construct a map $T\to P$, then composing with $\pi$ we obtain a map $T\to X$, then we clearly have a map $T\to P\times_S X$ by the universal property of the pullback. Finally we can compose with the quotient map $P\times_S X \to P\times^G X$ to obtain $PT \to \times^G X$
Given an fppf covering $\{U_i\}_i$ of T, we have $P|_{U_i}\cong U_i\times G|_{U_i}$ but I don't see how to construct canonically the map $T\to P$.
- Second direction
I don't know where to start.
Any help is much appreciated.