Let $G$ be an abelian profinite groups. Then we have the Sylow group decomposition $$G\cong \prod_p G_p.$$ In the case of finite groups, we have $ \prod_p G_p\cong \bigoplus_p G_p$ and thus $$\text{Hom}\left( \prod_p G_p,T\right)\cong \text{Hom}\left(\bigoplus_p G_p, T\right)\cong \prod_p \text{Hom}\left(G_p,T\right).$$ In the profinite case, we have $ \prod_p G_p\ncong\bigoplus_p G_p$ and thus in general we don't know the maps out of $\prod_p G_p.$ My question is if there exists some nice class of topological groups $T$ such that $$\text{Hom}\left( \prod_p G_p,T\right)\overset{?}{\cong} \prod_p \text{Hom}\left(G_p,T\right).$$ For instance, using the injective map $\bigoplus_p G_p\rightarrow \prod_p G_p$, for injective topological groups we have a surjection $\text{Hom}\left( \prod_p G_p,T\right)\rightarrow \text{Hom}\left(\bigoplus_p G_p, T\right).$ Is there a good additional condition making this map injective?
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$\begingroup$ If $T$ is a compact abelian group and Hom refers to continuous group homomorphisms, this is an isomorphism. $\endgroup$– YCorCommented Nov 30, 2020 at 23:03
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$\begingroup$ @YCor What is the argument in that case? $\endgroup$– curious math guyCommented Nov 30, 2020 at 23:13
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3$\begingroup$ @YCor: really? That doesn't seem right. If, say, $T = S^1$ then what I'd expect (using that Pontryagin duality is an equivalence of categories) is $\text{Hom}(\prod G_p, S^1) \cong \oplus_p \text{Hom}(G_p, S^1)$. $\endgroup$– Qiaochu YuanCommented Dec 1, 2020 at 3:31
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3$\begingroup$ @QiaochuYuan thanks, I was not careful. I should have said, for $T$ a profinite abelian group. $\endgroup$– YCorCommented Dec 1, 2020 at 3:35
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