$\mathbb{C}/\mathbb{Z} \cong \mathbb{C}-{0}$

Question

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!

score 5 · Accepted Answer · answered Jun 08 '14 at 11:50

5

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

answered Jun 08 '14 at 11:50

user1337

24,381

but that is only for the unit circle in $\mathbb{C}-{0}$, I need to get values mapped to the whole of $\mathbb{C}-{0}$. – Pim Jun 08 '14 at 12:04
1

@Pimziengs If $z$ is complex $e^{2 \pi i z}$ is not necessarily on the unit circle. Try setting $z=-i$ for instance. – user1337 Jun 08 '14 at 12:06
Aw, I see thanks! – Pim Jun 08 '14 at 13:57

$\mathbb{C}/\mathbb{Z} \cong \mathbb{C}-{0}$

1 Answers1