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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?

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    $\begingroup$ 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)? $\endgroup$ – Mariano Suárez-Álvarez May 10 '15 at 20:24
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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 $f$.

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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.

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