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Let $G$ be a profinite group. $A$ and $B$ are discrete G-modules. If $A^U=A$ for some open subgroup $U \subseteq G,$ then why Hom(A,B) is a discrete G-module. $\big( g(\phi)(a) = g(\phi(g^{-1}(a))) \big)$

math
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    If you mean $Hom$ of abelian groups then $Hom(A,B)$ is a $G$-module and since $G/U$ is finite the $G$-action on $Hom(A,B)$ is continuous if the $G$-action on $B$ is continuous – reuns Jul 23 '19 at 20:57
  • sorry for the late response. first, I thought I did the prove but now I think the above statement is not correct. Please check the errata https://www.mathi.uni-heidelberg.de/~schmidt/papers/errata-nsw2e.pdf which I got from the web page of Alexander Schmidt. if you have a proof, can you please provide the details? @reuns – math Nov 10 '19 at 03:32

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