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We have a unit circle with subspace topology induced from $\mathbb{R}^2$

How do I show that $f: [0,1) \to S^1$, $f(t) = (\cos(2\pi t), \sin(2\pi t))$ is not a homeomorphism?

So we have two topological spaces, $\mathbb{R}$ with the standard topology and unit circle with subspace topology induced from $\mathbb{R}^2$.

I know that a homeomorphism is bijective, continuous, and its inverse is continuous.

It looks to me that f is continuous, and bijective. So then the inverse is not continuous. Im just not sure what the inverse of f is?

Mhenni Benghorbal
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sarah
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2 Answers2

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Hint You should be able to use the fact that if you remove any point from $S^1$, the remaining space is connected. The same is not true for $[0,1)$.

If you want to use the inverse, note that $$\lim_{t \to 0^+} f(t)=\lim_{t \to 1^-}f(t)=f(0)=(1,0) \,. $$

Can you deduce from here that $f^{-1}$ cannot be continuous at $(1,0)$?

Martin Sleziak
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N. S.
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    what is the formula for the inverse of f? – sarah Nov 11 '13 at 05:10
  • To explicitely write down the inverse is a bit tedious and not necessary. It suffices to know that it exists. – Hagen Knaf Nov 11 '13 at 07:18
  • @sarah: it suffices to note that the $\lim_{t\to (1,0)}f^{-1}(t)$ does not exist (because its two side limits differ). Thus $f^{-1}$ cannot be continuous at $(1,0)$. – KonKan Feb 07 '17 at 15:21
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$S^1$ is compact. so $[0,1)$ cannot be continuous image of that.

GA316
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