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I have to prove that $\mathbb{C}/ \mathbb{Z} \cong \mathbb{C}-\{0\}$ holds.

I`m using this theorem: If $\phi: \mathbb{C} \rightarrow \mathbb{C}-{0}$ is a homomorphism and $H=Ker(\phi)$, with $H$ as a normal subgroup for $\mathbb{C}$. Then, $\mathbb{C}/ \mathbb{Z} \cong \mathbb{C}-{0}$ if $\phi$ is a surjective homomorphism.

My attempt:

I try to find an homomorphism $\phi: \mathbb{C} \rightarrow \mathbb{C}-{0}$, but I can`t seem to find it. Any help would be really welcome!

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Try $$\phi(z)=e^{2 \pi i z} .$$

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  • $\begingroup$ but that is only for the unit circle in $\mathbb{C}-{0}$, I need to get values mapped to the whole of $\mathbb{C}-{0}$. $\endgroup$ – Pim Jun 8 '14 at 12:04
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    $\begingroup$ @Pimziengs If $z$ is complex $e^{2 \pi i z}$ is not necessarily on the unit circle. Try setting $z=-i$ for instance. $\endgroup$ – user1337 Jun 8 '14 at 12:06
  • $\begingroup$ Aw, I see thanks! $\endgroup$ – Pim Jun 8 '14 at 13:57

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