Let $G$ be an abelian profinite group and $G=\varprojlim G_i$ (all $G_i$ are finite).
Then why $\hom_{\text{cont}}(\varprojlim G_i,\mathbb{Q}/\mathbb{Z}) = \varinjlim\hom_{\text{cont}}(G_i,\mathbb{Q}/\mathbb{Z})$ as topological groups.
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That's not true even as groups, you probably confused $\varinjlim$ and $\varprojlim$ – Maxime Ramzi Jul 21 '19 at 16:37
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I may be wrong. Can you check profinite groups by Luis Ribes(Lemma 2.9.3)? – math Jul 21 '19 at 19:31
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check the answer(https://math.stackexchange.com/a/127291/326275). – math Jul 21 '19 at 19:58
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2I assume that by $\operatorname{hom}$, you mean the group of continuous homomorphism, isn't it ? Then look for the compact subgroups of $\mathbb{Q/Z}$. Or directly, show that a continuous homomorphism $G\to\mathbb{Q/Z}$ has an open kernel. – Roland Jul 21 '19 at 20:00
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yes, a group of continuous homomorphism. – math Jul 21 '19 at 20:07
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How does topological equality follow from your statement? @Roland – math Jul 21 '19 at 20:16
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math : I'm sorry I may have spoken too quickly, I was thinking about the general situation and not the specific one – Maxime Ramzi Jul 21 '19 at 20:59