$ \mathbb{R}^+ \to \mathbb{C}^{\ast}$ with $x \mapsto e^{2\pi i x}$.
To prove that this is a group homomorphism, we prove $f(a+b) = f(a) \cdot f(b)$.
$$f(a+b) = e^{2\pi i (a+b)} = e^{2\pi i a + 2 \pi i b} = e^{2\pi i a} \cdot e^{2\pi i b} = f(a) \cdot f(b)$$
Thus group homomorphism is proven.
Next we show $f$ is injective.
To prove this we show that $\forall x,y \in \mathbb{R}^+$ s.t. if $f(x) = f(y)$ then $x=y$.
$$f(x) = f(y)$$
$$e^{2\pi i x} = e^{2\pi i y}$$
$$2\pi i x = 2\pi i y$$
$$ix = iy$$
$$x = y$$
Thus injectivity is proven.
Now my problem is to show that the map is (not) surjective.
The definition for surjectivity is $\forall x \in \mathbb{C}^{\ast}, \exists y \in \mathbb{R}^+$ s.t. $f(y) = x$.
Is showing that
$$y = e^{2\pi i x}$$
$$ln(y) = 2 \pi i x$$
$$\frac{ln(y)}{2 \pi i} = x$$
enough to conclude that the map is surjective, since the inverse exists $\forall x \in \mathbb{C}^{\ast} $?
ps. I use the fact that I know that the map is injective to check if a bijection exists (bijective iff. surjective and injective). I am unsure how to prove that the map is surjective without using this fact.
For surjectivity: note that $f$ maps to only those complex numbers that lie on the unit circle since $|e^{2\pi i x}|=1$. So choose any complex number outside the unit circle to show it is not surjective.
Your proof for injectivity also needs to be checked: Note that $e^{2\pi i x}=e^{2\pi i y} \implies 2\pi i x = 2\pi i y$ is not true. Think about $e^{2\pi i(1)}=e^{2\pi i (2)}=1$.
So the given $f$ is a homomorphism but it is not one-one and not onto.