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I've taken $f(t)=(\cos 2\pi t,\sin 2\pi t)$,then as $t $ varies from $0$ to $1$,$(\cos2\pi t,\sin2\pi t)$ covers up the whole $S^{1}$ once,which means $f:[0,1)\to S^{1}$ is a bijection.Also,as $\cos2\pi t$ & $\sin2\pi t$ are both continuous functions which means $f$ is continuous also.

Now,in order to show $f$ is not a homeomorphism we have to show that $f^{-1}$ is not continuous,which can be done by showing that $[0,1)$ has atleast one open set whose image under $(f^{-1})^{-1}$ is not open in $S^1$.

How to do this,i don't know?What is the form of open sets in $S^1$?

Disclaimer:I cannot use Compactness & Connectedness.

(Using these concepts this problem already has answers- 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?)

We can use only above highlighted notion.

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Styles
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2 Answers2

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HINT: $\left[0,\frac12\right)$ is open in $[0,1)$.

But you’ve misunderstood what you need to do to show that $f^{-1}$ is not continuous: you need to find an open set $U$ in $[0,1)$, the codomain of $f^{-1}$, such that $(f^{-1})^{-1}[U]$ is not open in $S^1$.

Brian M. Scott
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  • M.Scott:What is the form of open sets in $S^1$? – Styles Aug 15 '16 at 21:15
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    @PKStyles: It’s the subspace topology on $S^1$: the intersections of open sets in $\Bbb R^2$ with the set $S^1$. If you think about that for a bit, you’ll see that open intervals around $S^1$ are a base for its topology. – Brian M. Scott Aug 15 '16 at 21:23
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You can use two facts: if $f: X \to Y$ is a homeomorphism then so is $g: X \setminus\{x_0\} \to Y \setminus \{f(x_0)\}$ is also a homeomorphism where $g$ is just the restriction of $f$. Now recall that homeomorphic spaces have the same number of connected components.

Faraad Armwood
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  • Why was this answer down voted? It seems right to me.. – PtF Aug 15 '16 at 22:40
  • @PtF: There are a lot of things on this site that get down voted out of personal bias. These two facts, if used correctly, answers the problem in one line. – Faraad Armwood Aug 16 '16 at 04:35
  • @FaraadArmwood:you're downvoted may be because use of connectedness is not allowed.But,i upvoted you,because of the way you represent $g$. –  Aug 16 '16 at 13:24