# Bijection between isomorphism class of torsors and first cohomology set

Let $$G$$ be a profinite group and let $$A$$ be a $$G$$-group. A torsor over $$A$$ is a non-empty $$G$$-set $$P$$ endowed with a simply transitive right action $$\star$$ of $$A$$ which is compatible with the left action of $$G$$, i.e.

$$\sigma \cdot (x \star a) = (\sigma \cdot x) \star (\sigma \cdot a), \, \textrm{for all} \, \sigma \in G, a \in A, x \in P.$$

A homomorphism between torsors is a map that is equivariant with respect to the left action of $$G$$ and the right action of $$A$$. Any homomorphism between torsors is automatically an isomorphism.

What I want to show is that the isomorphism classes of torsors over $$A$$ are in bijective correspondence with $$H^{1}(G,A)$$.

I am able to fill in most of the details in the proof given by Serre in his book on Galois Cohomology but there is one thing I'm having trouble with.

Let $$\alpha \in \mathrm{Z}^{1}(G,A)$$ be be a 1-cocycle. We construct a torsor $$P_{\alpha}$$ over $$A$$ by endowing $$A$$ with a left action of $$G$$ given by $$(\sigma,x) \mapsto \alpha_{\sigma} (\sigma \cdot x)$$ and a right action of $$A$$ given by $$(x,a) \mapsto xa$$.

I then want to show that if two 1-cocycles $$\alpha$$ and $$\alpha'$$ are cohomologous, then $$P_{\alpha} \simeq P_{\alpha'}$$. I tried to do so by constructing a homomorphism of torsors $$f: P_{\alpha} \rightarrow P_{\alpha'}$$ (since any homomorphism of torsors is an isomorphism). I feel like the crucial part here is the fact that this homomorphism needs to be equivariant with respect to the left action $$(\sigma,x) \mapsto \alpha_{\sigma} (\sigma \cdot x)$$ of $$G$$. Writing down the equivariant condition unfortunately didn't give me any insight on how to define my homomorphism and I am stuck.

Any help and comments would be highly appreciated.

Edit: what I find is that we should have $$f(\alpha_{\sigma})= \alpha'_{\sigma}$$ (equivariance with respect to the left action) but this doesn't help me. If I am not mistaken, equivariance with respect to the right action would yield the identity which seems completely off.

• The first 2 sections of Chapter X of Silverman's Arithmetic of Elliptic Curves I gives quite a detailed and (I think) accessible introduction to this, filling in most of the details Sep 23, 2021 at 22:17
• I find the story much clearer in terms of homotopy theory in the case that $G$ is a discrete group, which could be generalized to profinite groups by passing to condensed mathematics. For the discrete case, see the nLab page: ncatlab.org/nlab/show/… Sep 23, 2021 at 22:26

Since $$\alpha$$ and $$\alpha'$$ are cohomologous, there exists $$a \in A$$ such that $$\alpha'_{\sigma} = a \alpha_{\sigma} (\sigma \cdot a^{-1})$$ for all $$\sigma \in G$$.
For $$f: P_{\alpha} \rightarrow P_{\alpha'}$$ to be a morphism of $$G$$-torsors, we need to have $$f(\sigma x) = \sigma f(x)$$, where $$\sigma x$$ is the left action of $$G$$ on $$A$$ defined by $$\sigma x= \alpha_{\sigma} (\sigma \cdot x)$$.
$$f(\sigma x) = f(\alpha_{\sigma} (\sigma \cdot x))$$ and $$\sigma f(x) = \alpha'_{\sigma} (\sigma \cdot f(x)) = (a \alpha_{\sigma} (\sigma \cdot a^{-1}))(\sigma \cdot f(x))$$.
We thus want to construct a map $$f$$ such that $$f(\alpha_{\sigma} (\sigma \cdot x))=(a \alpha_{\sigma} (\sigma \cdot a^{-1}))(\sigma \cdot f(x))$$. It is now easy to see that setting $$f(x) = ax$$ gives the desired equality. It is also easy to check that $$f$$ is invariant with regard to the right action $$(x,a) \mapsto xa$$.