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I am a beginner in Topology. I was going through Munkres book where I came across this example. The mapping $[0,1)\to S^1$ (unit circle) is bijective and continuous, but $f^{-1}$ is not continuous. The function $f(t)=(\cos 2\pi t, \sin 2\pi t)$ and $S^1$ is a subspace of the plane $\mathbb{R}^2$. I don't understand how the inverse is not continuous. Can someone please explain this simple thing to me?

Thanks a lot

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    $\begingroup$ $t \to e^{i2\pi t}$ is the mapping presumably. How does the inverse function behave at $(1,0)?$ $\endgroup$
    – zhw.
    May 20, 2015 at 19:54
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    $\begingroup$ $[0,1)$ is not compact. $\endgroup$ May 20, 2015 at 21:14
  • $\begingroup$ Its important to realize that the boundary on the interval does not map to a boundary on the circle. $\endgroup$
    – user795305
    May 20, 2015 at 23:41
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    $\begingroup$ How does the inverse mapping behave near the circle point corresponding to 0? $\endgroup$
    – user253751
    May 21, 2015 at 0:41
  • $\begingroup$ This is an incredibly important counterexample and I strongly suggest you commit it to memory. The fact this map isn't open isn't obvious despite the spaces in the domain and range being fairly simple in structure and familiar. $\endgroup$ May 21, 2015 at 3:28

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We need $(f^{-1})^{-1}(U) \subset S^1$ to be open for any open $U \subset [0,1)$, i.e. $f(U)$ needs to be open. Any interval of the form $[0,\epsilon)$ for $0<\epsilon<1$ is open in $[0,1)$, but $f([0,\epsilon))$ is not open in $S^1$.

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  • $\begingroup$ Actually that is the thing troubling me. How is $f([0,\epsilon))$ not open in $S^1$? I am not able to understand that. I know its a very basic thing. But I am not able to understand this concept. $\endgroup$
    – Khushboo
    May 20, 2015 at 20:02
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    $\begingroup$ As zhw commented, it would help to know the function $f$ you have in mind. Is it $f: [0,1) \mapsto S^1 \subset \mathbb{R}^2$ given by $t \mapsto (\cos 2\pi t, \sin 2\pi t)$? Is it $[0,1)\mapsto [0,1]/\sim$ where $0\sim1$? Regardless, the point is that the image of $[0,\epsilon)$ will "look like" a half-open interval in the middle of $S^1$. Depending on how you're defining $S^1$ (again, as a subset of $\mathbb{R}^2$ or as a quotient space), you should try to convince yourself that open sets in $S^1$ can't "look like" half-open intervals. $\endgroup$
    – Kyle
    May 20, 2015 at 20:06
  • $\begingroup$ I see your point. So because at $t=0$, $f(0)$ is closed, it is not continuous? If I consider any other interval of the form $[a,\epsilon)$, where $a>0$, the inverse map would be continuous? $\endgroup$
    – Khushboo
    May 20, 2015 at 20:14
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    $\begingroup$ The issue isn't quite that $f(0)$ is closed. (Because $f(0)$ is a point, which will be a closed set in any $T_1$ topological space.) The problem is that $f(0)=(1,0)$ does not have a neighborhood contained in $f([0,\epsilon))$, so $f([0,\epsilon))$ is not open. Another way to see this is to observe that the complement $S^1 \setminus f([0,\epsilon))$ is not closed, since it contains a sequence of points that converges to $f(0)$ but $f(0) \not \in S^1 \setminus f([0,\epsilon) )$. $\endgroup$
    – Kyle
    May 20, 2015 at 20:20
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What is the image of $[0,1/3)$? It's open in $[0,1)$ but its image is not open in $S_1$. Let $g=f^{-1}$. If $g$ were continuous then $g^{-1}(\text{open})=\text{open}$. But $g^{-1}([0,1/3))$ is not open. So $g$ is not continuous.

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Name $P$ the point $f(0)$. What are the inverse images of points close to $P$ under $f$?

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The intuition is that a homeomorphism captures the idea of two spaces being the same. Since an interval and a circle do not have the same shape, you should expect this map to not be a homeomorphism.

Formally, you should investigate an open neighborhood of 0 in $[0,1)$. If $f^{-1}:S^1 \to [0,1)$ were continuous, then the preimage of this open neighborhood will be an open set in $S^1$.

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Intuitively the inverse of f is ripping open the circle. So consider a pair of points on the circle that are on opposite sides of (1,0). Make them very close. Now consider the inverse of f. It sends each of the pair close to a different end of [0,1). How can this be a continuous map?

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