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Nicolás
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They are using continuous cohomology, so that $$ H^n(G,M) = \varinjlim H^n(G/H,M^H) $$ if $G$ is topological and $M$ is continuous $G$-module (the limit runs over open compact subgroups $H$ of $G$). Look in p. 38 for the definition.

Nicolás
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