Let $S$ be a scheme.


**Definition.** Let $X$ be an $S$-scheme and $G$ a smooth affine group $S$-scheme acting on $X.$ An $S$-scheme $Y$ is a *geometric quotient* of $X$ by $G$ if there exists a morphism $\pi\_X\colon X\rightarrow Y$ such that 


1. $\pi\_X$ is $G$-invariant,
2. the geometric fibers of $\pi\_X$ are orbits,
3. $\pi\_X$ is universally submersive, i.e., $U\subset Y$ is open iff $\pi\_X^{-1}(U)\subset X$ is open, and this property is preserved by base change,
4. $(\pi\_X)\_\*(\mathcal{O}\_X)^G=\mathcal{O}\_Y.$


Let $X$ and $X'$ be $S$-schemes with a $G$-action, where $G$ is the same introduced before. Assume that there exist geometric quotients $\pi\_X\colon X\rightarrow Y$ and $\pi\_{X'}\colon X'\rightarrow Y'.$


**Question.** Let $g\colon Y\rightarrow Y'$ be a morphism. Is there a $G$-equivariant morphism $f\colon X\rightarrow X'$ such that $\pi\_{X'}\circ f = g\circ \pi\_{X}$?



Lift of a morphism between geometric quotients
It is not true : for instance, $X$ and $X'$ might be two non-isomorphic $G$-torsors on the same $S$-scheme $T$. 


Here is a precise example. Consider $S=T=Spec(\mathbb{R})$ and $G=\mathbb{Z}/2\mathbb{Z}$, let $X=Spec(\mathbb{C})$ and $X'=Spec(\mathbb{R})\cup Spec(\mathbb{R})$ viewed as $S$-schemes, and let $G$ act on $X$ and $X'$ respectively by complex conjugation and by exchanging the connected components. Both actions admit $Spec(\mathbb{R})$ as a geometric quotient. But an isomorphism between these quotients obviously doesn't lift to an equivariant morphism between $X$ and $X'$.


Here is another example : take $S=Spec(\mathbb{C})$, $G=\mathbb{G}\_m$, $T=\mathbb{P}^1\_{\mathbb{C}}$. Let $X$ (resp. $X'$) be the total space of the line bundle $\mathcal{O}$ (resp. $\mathcal{O}(1)$) on $T$ minus the zero-section, with the natural $G$-action. Both actions admit $T$ as a geometric quotient, but there is no equivariant morphism between $X$ and $X'$ lifting identity.


