Define the circle group $U_1 = \{z \in \mathbb{C} \mid |z| =1 \} \subset \mathbb{C}^*$. Prove: $\mathbb{C}^* \cong U_1 \times \mathbb{R}_{>0}^*$.
I would like to know if I did it correct. I believe I still have to say that my function is well-defined, but I don't know how to explain this, and for injectivity I would like to use that a function is injective if $\text{Ker}(f) = {0}$. I hope somebody can help me with this.
My idea: Define $f: \mathbb{R}_{>0}^* \times U_1 \rightarrow \mathbb{C}^* : (r, e^{2\pi ix}) \mapsto re^{2\pi i x}$. It's a function from a group mapping to a group.
- $f$ is surjective, as every element in $\mathbb{C}^*$ can be written as $re^{2\pi i x}$ with $r > 0$ and $x \in \mathbb{R}$. The element is mapped to $(r, e^{2\pi i x}).$
- $f$ is a homomorphism: $f((r_1, e^{2\pi i x_1})(r_2, e^{2\pi i x_2})) = f(r_1r_2, e^{2\pi i(x_1 + x_2)}) = r_1r_2e^{2\pi(x_1 + x_2})$ and $f(r_1, e^{2\pi i x_1})f(r_2, e^{2\pi ix_2}) = r_1e^{2\pi i x_1}r_2e^{2\pi i x_2} = r_1r_2e^{2\pi i (x_1 + x_2)}$.
- $f$ is injective when Ker$(f) = {e}$; $\mathbb{C}^*$ has 2 elements that satisfy $x^2 = e$, namely$x = 1$ and $x = -1$.
$\mathbb{R}_{>0}^* \times U_{1}$ also has two, namely $(1, 1)$ and $(1, -1)$.
Thus follows $\mathbb{C}^* \cong U_1 \times \mathbb{R}_{>0}^*$.
You proof is correct, but it can be simplified.
There is no reason to write the elements of $U_1$ in the form $e^{2\pi i x}$. Simply define $$f(r,z) = r \cdot z .$$
That $f$ is a homomorphism follows from the commutativy of complex multiplication: $$f((r_1,z_1) \cdot (r_2,z_2) = f(r_1r_2, z_1z_2) = r_1r_2z_1z_2 = r_1z_1r_2z_2 = f(r_1,z_1) \cdot f(r_2,z_2) .$$
To see that $f$ is a bijection, consider $$g : \mathbb C^* \to \mathbb R _{>0} \times U_1, g(w) = \left(\lvert w \rvert, \frac {w}{\lvert w \rvert} \right).$$ It is then easy to check that $g \circ f = id$ and $f \circ g = id$.