Hello Veeky Forums...

Hello Veeky Forums, how can i prove that [math]f: \mathbb{C} _{\neq 0} \rightarrow \mathbb{S}^1 \times \mathbb{R}_{>0} [/math] defined as [math]f(z):=(\frac{z}{|z|},|z|)[/math] is an isomorphism?

*defined as [math]f(z):=(\frac{z}{|z|},|z|)[/math]

fuuuck [math] f(z):= ( \frac{z}{|z|, |z| ) [/math]

by checking that it's injective and surjective

the classic way? i.e. Let [math]z_1 , z_2 \in \mathbb{C}_{\neq 0} [/math] [math]z_1 \neq z_2[/math] and i gotta show that [math]f(z_1) \neq f(z_2)[/math] . But how do i show that its surjective?

>But how do i show that its surjective?
show that every point on the cylinder has a preimage

how? for example let [math]z \in \mathbb{S}^1 \times \mathbb{R} _{>0} [/math]. i have to construct [math] f^{-1} (z)[/math]?

>i have to construct f−1(z)?
yes

Well fuck. Have no idea how I can find an inverse of a vector function.

any help?

this is something you know how to do. stop thinking of the symbols and trying to move them around to get the answers. think of the picture. what is the function drawing? how would you draw the image and the range? what happens when you fix a value in one coordinate and move the other?

(hint: polar coordinates)

lemme try
[math] \displaystyle
f(z):= \left ( \frac{z}{|z|},|z| \right )
[/math]

Let [math]z\in \mathbb{C}_{\neq 0}[/math]. every complex number can be written as [math]z=\frac{z}{|z|} \cdot |z|[/math] and also [math]\mathbb{S}^1 \cap \mathbb{R}_{>0}={1}[/math] so, \mathbb{C}_{\neq 0} = \mathbb{S}^1 \oplus \mathbb{R}_{>0}[/math].

then let [math]z=r e^{i \theta}[/math] so we can construct the map [math]g:\mathbb{S}^1 \times \mathbb{R}_{>0} \rightarrow \mathbb{C}_{\neq 0}[/math] defined as [math](r,e^{i \theta}) \mapsto r\cdot e^{i \theta}[/math]. Is this [math]f^{-1}(z)[/math]?

It's just polar coordinates you brainlet.

* [math]\mathbb{C}_{\neq 0} = \mathbb{S}^1 \oplus \mathbb{R}_{>0}[/math]

yeah

???

okay guyz ive got it thanks

>How do I show that it's injective
First check that f is a homomorphism and then verify that its kernel is trivial.

>But how do i show that its surjective?
You take a generic point on the cylinder ((x,y),z) and a point on the complex unit circle (x+iy) and show that f maps z(x+iy) to ((x,y),z)

Is this not okay for showing [math]f[/math] is injective? also for showing its bijective i just use [math]g[/math] as stated above dont i?

it's okay, there's more than one way to show this. e.g. if you have an explicit formula for the inverse, you can just show that fg=id and gf=id by a straightforward computation.

i hear you

Yes. Basically there are a number of equivalent ways to show that a function is an ismorphism.

1) check that it is a homomorphism, has trivial kernel and is surjective.
2) check that it is a homomorphism, then demonstrate an explicit inverse as in 3) check that it is a homomorphism, then show that it is injective and surjective by using the definitions of injectivity and surjectivity.

Just count to 6