Proof $\mathbb{C}^* \cong \mathbb{C} / \mathbb{Z}$

Question

I am trying to prove the isomorphism between $\mathbb{C}^*$ and $\mathbb{C} / \mathbb{Z}$. I already established the way to do it:

find a surjective homomorphism $f: \mathbb{C} \to \mathbb{C}^*$, such that $Ker(f)=\mathbb{Z}$
take the homomorphism $\phi: \mathbb{C} \to \mathbb{C}/\mathbb{Z}$
Then there exists a homomorphism $g: \mathbb{C}/\mathbb{Z} \to \mathbb{C}^*$, and then we have to prove that $g$ is a isomorphism.

My problem mostly is in finding a surjective homomorphism $f$ such that the $Ker(f)=\mathbb{Z}$. Anyone that can help me out?

You probably mean $(\mathbb C^*, \times)$ and $(\mathbb C/\mathbb Z, +)$. It helps to make this clear. — Trebor, Jun 19 '20 at 7:12
In general, when looking for a homomorphism that goes from "addition" to "multiplication", a good first thing to try is an exponential function, i.e. something like $f(x) = e^x$. — Ben Grossmann, Jun 19 '20 at 7:17

HallaSurvivor · Accepted Answer · 2020-06-19 07:14:55Z

5

You have the right idea! To finish it up, remember the following facts:

$e^{z_1 + z_2} = e^{z_1} \cdot e^{z_2}$. But we know $e^{n 2\pi i} = 1$ for each $n \in \mathbb Z$...

Can you use your idea (the first isomorphism theorem) to finish the proof with this?

I hope this helps ^_^

answered Jun 19 '20 at 7:14

HallaSurvivor

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$\begingroup$ Ok yes I had something like that in mind as well. So the homomorphism would be: $f: z\mapsto e^{2i\pi z}$. However, is that surjective? These are only the answers for the unit circle, right? (so the solutions for $z=re^{ni2\pi}$, where $r=1$) $\endgroup$ – dikkemaatjes Jun 19 '20 at 7:22

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