Question

prove that h is continuous and bijective, but it is not a homeomorphism

Answer 1

Let S denote the unit circle in the plane S={\[{(x,y)\in R^{2} : x^{2}+y^{2}=1}\]} define h: [0,2pi] rightarrow S by \[h \left( \theta \right) =( \cos \theta , \sin \theta )\]

Answer 2

Are you sure its supposed to be from [0,2pi]? Its not injective (hence not bijective) in that case, since:\[h(0)=h(2\pi)\]

Answer 3

So let's assume you have a function:\[h:[0,2\pi)\rightarrow S\]\[h(\theta)=(\cos\theta,\sin\theta)\]Which part are you having trouble with? Showing that its continuous and bijective? Or showing that its not a homeomorphism?

Answer 4

showing that its not a homeomorphism

Answer 5

Ah ok. You need to use the fact that if h is a homeomorphism, then the pre-image of a connect set should still be a connected set. Can you think of a subset of S thats connected, such that when you look at the pre-image of that set under h:\[h^{-1}(S)\]its not connected anymore?

Answer 6

and bijective, sorry my network is bad

Answer 7

thats the problem, i cant com up wit it

Answer 8

Alright, first we show its bijective. Lets do injectivity (one-to-one) first. We are going to assume that h(x)=h(y), and we will prove x=y. So assume x and y are both in the interval [0,2pi), and that:\[h(x)=h(y)\Longrightarrow (\cos x,\sin x)=(\cos y, \sin y).\]Then it follows that x = y +(2pi)*k for some integer k. But since x and y are both in the interval [0,2pi), it follows that x = y.

Now we need surjectivity (onto).

Answer 9

ok cool, how to show onto?

Answer 10

Is it clear that the image of [0,2pi) under h is all of S to you?\[h\left([0,2\pi)\right)=S\]

Answer 11

Is it clear that as theta travels from 0 to 2pi, that we will sweep the entire unit circle?

Answer 12

yes that is clear

Answer 13

As t goes from 0 to 2pi, we get the whole circle. Alright, I'm glad thats clear, because otherwise the details would kill us lol. Thats the surjectivity.