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Regarding Gregory Berhuy's book "An Introduction to Galois Cohomology and its Applications":

The book defined a cohomology sets for non-abelian $G$-groups. Let $A$ be a $G$-group, we define a 1-cocycle as follows: it is a map $$ \alpha : G \to A \quad , \quad \sigma \mapsto \alpha_\sigma$$ such that for all $\alpha \in A $: $\alpha_1 = 1$ (1 is the unit element of each group) and it satisfies the following relation for all $\sigma , \tau \in G$: $$ \alpha_{\sigma \tau} = \alpha_\sigma \sigma \cdot \alpha_\tau \ . $$

In page 52, Berhuy's uses that fact to deduce that for an exact sequence of groups $$ 1 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 1 $$ where $f(A)$ is a central subgroup of $b$, that $$ g( \beta_\sigma (\sigma \cdot \beta_\tau) \beta_{\sigma \tau}^{-1}) = \gamma_\sigma \sigma \cdot \gamma_\tau \gamma_{\sigma \tau}^{-1} = 1 $$ however, in page 53, he writes $$ f(\alpha_{\sigma, \tau}) = \beta_\sigma (\sigma \cdot \beta_\tau) \beta_{\sigma \tau}^{-1} $$ (notice the brackets on $\sigma \cdot \beta_\tau$) but he doesn't say it is equal to 1, why? If we use the 1-cocycle definition (the equation with the $\alpha$'s) then $$\beta_\sigma (\sigma \cdot \beta_\tau) = \beta_{\sigma \tau} $$

Where is the difference and why?

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It seems to me like he's using the fact that $g(\beta_\sigma(\sigma \cdot \beta_\tau)\beta_{\sigma\tau}^{-1}) =1$ to say that \beta_\sigma(\sigma \cdot \beta_\tau)\beta_{\sigma\tau}^{-1} is in the image of $f$, which is of course true by the assumption that we have a short exact sequence of $G$-groups. –  stankewicz Jun 24 '12 at 11:36
    
I'm not sure. He first states that $$g( \beta_\sigma (\sigma \cdot \beta_\tau ) \beta_{\sigma \tau}^{-1}) = \gamma_\sigma \sigma \cdot \gamma_\tau \gamma_{\sigma \tau}^{-1} = 1 $$ and then deduces from that, that $$ \beta_\sigma (\sigma \cdot \beta_\tau ) \beta_{\sigma \tau}^{-1} = f( \alpha_{\sigma , \tau})$$ (I'm quoting: " = 1, so $\beta ... = f( \alpha_{\sigma , \tau})$ for some unique $\alpha_{\sigma , \tau} \in A$). However, what about the general case? Why it is not =1 regardless the $f,g$ homomorphisms? –  Zachi Evenor Jun 24 '12 at 12:05
    
@stankewicz See my last comment, it may help. –  Zachi Evenor Jun 24 '12 at 17:15
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In this book on page 52, $\beta_\sigma$ is SOME preimage of $\gamma_\sigma$. Although $\gamma$ is a cocycle, $\beta$ in general is NOT a cocycle, and this is the point! It is a cochain, whose coboundary lives in $f(A)$ (because $\gamma$ is a cocycle), which gives us a 2-cocycle in $A$. The cohomology class of this cocycle is, by definition, $\delta^1(\gamma)$. –  Mikhail Borovoi Jun 24 '12 at 20:23
    
Thanks. Now that this understood, I got the rest (I recall the since the sequence is exact $g \circ f = 1$ the trivial map. –  Zachi Evenor Jun 25 '12 at 10:00
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1 Answer 1

Without answering your question in particular, you might like to look at the paper

R. Brown, ``Fibrations of groupoids'', J. Algebra 15 (1970) 103-132.

which uses semidirect products to turn derivations (or crossed morphisms, as the term is used there) into sections and also uses exact sequences of a fibration of groupoids to deal with exact sequences: see Theorem 5.7 of that paper.

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