# Why are equivariant morphisms of $G$-torsors necessarily isomorphisms?

This was something I read on the Stacks project, but whose proof was omitted.

Simply stated, if $f\colon E\to F$ is a $G$-equivariant morphism of $G$-torsors over a scheme $X$, why is $f$ necessarily an isomorphism?

• Can you do this when $G$ is simply a discrete group and $E$ and $F$ are regular $G$-sets (that is, sets on which $G$ acts simply transitively)? – Mariano Suárez-Álvarez May 10 '15 at 20:24

By definition, after base changing to a suitable cover of $Y$ of $X$ depending on the topology, the $G$-torsors become trivial, so that $E_Y$ and $F_Y$ both become isomorphic to $Y \times G$. Now any equivariant map $f': Y\times G \rightarrow Y\times G$ over $Y$ is going to be multiplication by some element of $G$ on the second factor, hence an isomorphism. This isomorphism then descends back down to $f$.

I don’t know what is their definition of $$G$$- torsor.

The definition that I know for a $$G$$-torsor is

a manifold $$X$$ with a (smooth) free and transitive action of $$G$$ on $$X$$.

Let $$X$$ and $$Y$$ be $$G$$-torsors. A morphism of $$G$$-torsors from $$X$$ to $$Y$$ is given by a smooth map $$f:X\rightarrow Y$$ such that $$f(x.g)=f(x).g$$ for all $$x\in X$$ and $$g\in G$$.

Suppose $$x_1,x_2\in X$$ such that $$f(x_1)=f(x_2)$$. As $$G$$ on $$X$$ acts transitively (and freely) there exists (unique) $$g\in G$$ such that $$x_2=x_1.g$$, which then imply $$f(x_2)=f(x_1.g)=f(x_1).g$$. As $$G$$ acts freely on $$Y$$ the conditions $$f(x_2)=f(x_1).g$$ and $$f(x_1)=f(x_2)$$ imply $$g=1$$ which further imply $$x_1=x_2$$. So, $$f$$ is one-one.

Let $$y\in Y$$. For any $$x\in X$$, we have $$f(x)\in Y$$. As $$G$$ acts transitively on $$Y$$, there exists $$g\in G$$ such that $$f(x).g=y$$. As $$f$$ is $$G$$ equivariant, we have $$y=f(x.g)$$. Thus, $$f$$ is surjective. So, $$f$$ is bijection.

Observe that $$f$$ maps $$X_x$$ to $$Y_{f(x)}$$ where $$X_x=\{g.x:g\in G\}$$. But, $$X_x\cong G$$ and $$Y_{f(x)}\cong G$$. Thus, $$X_x\rightarrow Y_{f(x)}$$ is an diffeomorphism. Checking locally, we conclude, $$f:X\rightarrow Y$$ is a diffeomorphism.

So, any morphism of $$G$$-torsors is an isomorphism. What ever may be your definition of $$G$$-torsor, I am sure this proof works without much changes.