Artin exercise 2.5.4: Let $f:\Bbb R^+\to\Bbb C^\times$ be the map $f(x)=e^{ix}$. Prove that $f$ is a homomorphism, and determine its kernel and image.

I am trying to solve:

Let $$f: \mathbb{R}^{+} \to \mathbb{C}^{\times}$$ be the map $$f(x) = e^{ix}$$. Prove that $$f$$ is a homomorphism, and determine its kernel and image.

Here is my attempt.

Given $$x,y \in \mathbb{R}$$, we have \begin{align*} f(x+y) = e^{i(x+y)} = e^{ix + iy} = e^{ix} e^{iy} = f(x) f(y), \end{align*} so $$f$$ is a homomorphism. I claim that $$\mathrm{ker}(\varphi) = \{2\pi k \mid k \in \mathbb{Z}\}$$. Indeed, we have: \begin{align*} x \in \mathrm{ker}(f) & \iff f(x) = 1 \\ & \iff e^{ix} = 1 \\ & \iff \cos(x) + i \sin x = 1 \\ & \iff \cos x = 1, \; \sin x = 0 \\ & \iff x = 2\pi k, \; k \in \mathbb{Z} \end{align*} Finally, I claim that $$\mathrm{Im}(f) = S^1 = \{z \in \mathbb{C} \mid |z| = 1\}$$. We have: \begin{align*} z \in \mathrm{Im}(f) & \iff \exists x \in \mathbb{R}, \; f(x) = e^{ix} = z \\ & \iff |z| = 1 \\ & \iff z \in S^1 \end{align*}

How does this look?

• Pretty good, but the definition of $S^1$ needs an equal sign, not a less than. Commented Apr 9, 2021 at 19:32
• Thank you, I fixed it. Commented Apr 9, 2021 at 19:35
• This works well. Commented Apr 9, 2021 at 19:35

Some might say that relying too much on the symbol "$$\iff$$" is bad form, but I disagree, along with many others.
• I appreciate the feedback, thank you very much. I originally proved the inclusions separately but saw $\iff$ as the "more rigorous approach." It's very good to know that there are different opinions on this. Commented Apr 9, 2021 at 19:43