Explain me, please
Show that $\phi (t) = cis t $is a group homomorphism of the additive group $\mathbb R$ onto the multiplicative group T:={z:|z|=1}

Question

loser66 · Answer

The usual way is easy to prove and my prof gave us other way but I don't understand it. Please, explain me
$ker \phi = \{t \in \mathbb R ~~~|~~~ \phi (t) =1\}$ . How? it should be $\phi(t) =0$, right?
$\phi (t) =1 $ iff $t = 2\pi n, ~~n\in \mathbb Z$
$ker \phi = \{2\pi n: n\in \mathbb Z\}\cong \mathbb Z$ 
hence $\phi $ is onto. Again, how? 
Hence $S^1 =\{z:|z|=1\} \cong \mathbb R /\mathbb Z$

loser66 · Answer

@Michele_Laino

anonymous · Answer

$$ker \phi = \{t \in \mathbb R ~~~|~~~ \phi (t) =1\}$$because the identity in the multiplicative group is 1

loser66 · Answer

Got it, thanks, how about the next "how"? :)

michele_laino · Answer

no, since the unit element of T is 1

loser66 · Answer

@Michele_Laino  same idea!!, he puts it under the quotient group concept

michele_laino · Answer

second part:
we have:
$$\Large {e^{i2\pi n}} = 1,\quad n \in \mathbb{Z}$$

loser66 · Answer

Got it too. Thanks you guys

michele_laino · Answer

:)