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:

1. find a surjective homomorphism $f: \mathbb{C} \to \mathbb{C}^*$, such that $Ker(f)=\mathbb{Z}$
2. take the homomorphism $\phi: \mathbb{C} \to \mathbb{C}/\mathbb{Z}$
3. 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?

Answer 1

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 ^_^