Show $\widehat{\mathbb{Z}}$ is isomorphic to $S^1$ I am trying to solve a problem from the book "Fourier series and integrals" by McKean. :)

Problem:
Prove that the group $\widehat{\mathbb{Z}}$ (the dual to $\mathbb{Z}$) is isomorphic to $S^1$.

I have been given that $e_n\mapsto n$ gives an isomorphism between $(S^1)^{\wedge}$ and $\mathbb{Z}$. I tried to kinda mimic the idea and I ended up constructing the function:
$$\begin{cases}f:S^1\to\mathbb{Z}^{\wedge}\\
\alpha\mapsto \chi_{\alpha}.
\end{cases}$$
In the book, they defined $S^1$ as "The circle $[0,1)$ under addition modulo $1$". When i read the sentence, I interpret it as the operation is addition on the unit circle, that is, if $\alpha,\beta\in S^1$ then $\alpha+\beta\in S^1$. But I don't think this is the correct interpretation of it, since $\alpha=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\beta=1$ are both on the unit circle, but $\alpha+\beta=\frac{3}{2}+\frac{\sqrt{3}}{2}$ and $|\alpha+\beta|=\sqrt{3}$ which shows us that $\alpha+\beta\not\in S^1$. Since I know multiplication of complex numbers on the unit circle gives us a group, I ended up thinking that's what they mean. Even though I don't know how I could interpret the above sentence as multiplication.

What we now have to do is to prove $f$ is a homomorphism, bijective, continuous and that its inverse is continuous.
Homomorphism:
For the homomorphism part, we have $f(\alpha\beta)=\chi_{\alpha\beta}(z)=(\alpha\beta)^z=\alpha^z\beta^z=f(\alpha)f(\beta)$.
$\text{ }$
Bijection:
I have two suggestions for the injectivity. Either we consider the equation $f(\alpha)=f(\beta)$ and try to conclude $\alpha=\beta$ or we try to prove kernel is trivial.
For the first suggestion, we have that $f(\alpha)=f(\beta)\iff\alpha^z=\beta^z\iff \alpha^z-\beta^z=0$ and I know a factor to $\alpha^z-\beta^z$ will be $\alpha-\beta$. Perhaps we can use it to conclude $\alpha=\beta$?
For my second suggestion, we have that $f(1)=\chi_1$ and $\chi_1(z)=1^z=1$, this shows us $1\in S^1$ gets mapped to the identity in $\mathbb{Z}^{\wedge}$. Perhaps it is trivial that this is the only element which gets mapped to $1\in\mathbb{Z}^{\wedge}$ and this is the better approach?
I have actually no suggestions or idea how to prove the map is surjective.
$\text{ }$
Continuity:
We have to check $f(\alpha)=\alpha^{z}$ is continuous for all $z$. But since $f(\alpha)=\alpha^z$ is a function from $S^1$ to $S^1$. Since it is a restriction of the map $\mathbb{C}\to\mathbb{C}$ defined by $\alpha\mapsto\alpha^z$, which is continuous, the map $f(\alpha)=\alpha^z$ is also continuous.
$\text{ }$
Continuity - Inverse function:
Unfortunately, I do not even know where to begin here. How Should I construct an inverse function? Can I "just" define it as $\chi_{\alpha}\mapsto\alpha?$ How do I prove it is continuous?

My questions are:

*

*Is my interpretation of the group structure correct? If not, can you guide me in the right direction to help me understand this problem and be able to solve it?

*Is my solution for the homomorphism part and the continuity correct?

*What do you think about the proof of the injection? Do you think you
can help me conclude it is, indeed, injective?

*Can you help me with the surjection and the inverse map?

I appreciate any help, thanks!
 A: "Addition modulo $1$" is not the same as addition of real numbers. Also, where you say $\frac 1 2 + \frac{\sqrt3} 2$ is on the unit circle, I suspect you meant $\frac 1 2 + i \frac{\sqrt3} 2.$ Not notice that $\frac 1 2 + i\frac{\sqrt3} 2 = \cos\frac \pi 3 + i\sin\frac\pi 3,$ and $\frac \pi 3 = \frac{\text{whole circle}} 6.$ The number within $[0,1)$ to which $\frac 1 2 + i\frac{\sqrt 3} 2$ corresponds is therefore $\frac 1 6 = \frac{\pi/3}{\text{whole circle}}. \vphantom{\dfrac{\displaystyle\sum}1}$ Suppose you have $\alpha,\beta\in[0,1).$ Then
\begin{align}
& \Big(\cos(\alpha\times\text{whole circle)} + i\sin(\alpha\times \text{whole circle}) \Big) \\
\times {} & \Big( \cos(\beta\times\text{whole circle)} + i\sin(\beta\times \text{whole circle}) \Big) \\[10pt]
= {} & \cos((\alpha+\beta)\times\text{whole circle)} + i \sin((\alpha+\beta)\times\text{whole circle}),
\end{align}
where $\alpha+\beta$ is the sum modulo $1$ of $\alpha$ and $\beta.$ For example, if $\alpha= 0.8$ and $\beta = 0.9,$ then the ordinary sum of $\alpha$ and $\beta$ is $1.7,$ so the sum modulo $1$ of $\alpha$ and $\beta$ is $0.7.$
