Show Circle Group $ \mathbb{T} $ isomorphic to $\mathbb { C } ^ { * } / \mathbb { R } ^ { * } $

Question

I'm trying to think of an extension to this question, which asks to show whether $\mathbb { C } ^ { * } / \mathbb { R } ^ { + } \simeq \mathbb { T } $. They do it using the First Isomorphism Theorem.

I think my postulate should be true because $ \mathbb { R } ^ { + } \simeq { R } ^ { * } $, (the positive reals under multiplication and the reals under multiplication ), but I'm having difficulty coming up with a homomorphism

$$ \phi : \mathbb { C } ^ { * } \mapsto \mathbb{T} $$

Any suggestions?

Answer 1

Just use the same homomorphism, and then double the angle. Dividing by negatives means that we identify numbers on opposite sides of the origin. Double the angle, and that's not a problem anymore. After all, we can map the circle 2-1 to the circle.

(Or, as Tsemo Aristide's answer put it, going from the projective line to the circle is an angle-doubling)

Answer 2

$\mathbb{R}^*$ is not isomorphic to $\mathbb{R}^+$ as topological groups because the first set is not connected and the second is connected. But the quotient of $\mathbb{C}^*$ by $\mathbb{R}^*$ is the projective line which is the circle.