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Let $\widehat{\mathbb{R}/\mathbb{Z}}$ be the set of all homomorphisms from $\mathbb{R}/\mathbb{Z}$ to $\mathbb{C}$. Is $\widehat{\mathbb{R}/\mathbb{Z}} = \mathbb{Z}$? I think that $\mathbb{R}/\mathbb{Z} = \{t + \mathbb{Z}: t \in \mathbb{R}\}$. We can define $\varphi: \mathbb{R}/\mathbb{Z} \to \mathbb{C}$ by letting $\varphi(t+\mathbb{Z}) = e^{2\pi i n t}$, $n \in \mathbb{Z}$. It is easy to see that $\varphi$ is a homomorphism. Is it true that every element in $\widehat{\mathbb{R}/\mathbb{Z}}$ is of this form? Thank you very much.

LJR
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    You need continuous homomorphisms. – Andrea Mori Mar 19 '14 at 13:11
  • Are you using $\mathbb C$ as a multiplicative group? If so then $\mathbb C^\times$ is more usual (which doesn't include $0$) – Ben Millwood Mar 19 '14 at 13:25
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    Maybe these posts may help: http://math.stackexchange.com/questions/148155/why-is-that-widehat-mathbbr-mathbbz-cong-mathbbz and http://math.stackexchange.com/questions/123588/varphi-in-operatornamehoms1-s1-are-of-the-form-zn/ – Martin Sleziak Mar 19 '14 at 13:55

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