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$f:[0,1) \to S$ where $ f(t) = e^{2\pi i t}$ is continuous. (Also it is called wrapping function.) Notice that the function is one to one and onto so $f^{-1}$ exists. Why is the Inverse (Unwrapping) Function not continuous?

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yes it was missing i just corrected it thanks –  da_elysian_fields Dec 8 '13 at 19:26
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5 Answers 5

up vote 3 down vote accepted

I'll try an give the intuition, as the details are already given by others:

A continuous function maps neighboring points to neighboring points. For example, $f(x)=x^2$ is continuous at $x=1$ (and, in fact, everywhere) because $f(1)=1$ and if $y$ is close to $1$ then $f(y)$ will be close to $1$ as well.

Your inverse map can't be continuous on all $S^1$ because it has to cut the circle somewhere. To be specific, $f(0)$ is the "zero angle" on the circle, and when $x \rightarrow1$, then $f(x)$ gets closer and closer to the zero angle from below. But this means that the neighborhood of the zero angle is torn apart by $f^{-1}$ into 2 far away sections - one around $0$ and another one near $1$.

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One is compact the other is not.

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but how does that relate to the unwrap function being not continuous? –  da_elysian_fields Dec 8 '13 at 7:55
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If the inverse function was continuous then the image of $S$ should be compact. –  user71352 Dec 8 '13 at 7:57
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If $V$ is a neighborhood of $0$ in $[0,1)$, then there is no neighborhood $U$ of $1$ in $\mathbb{S}$ satisfying $f^{-1}(U) \subset V$.

Roughly speaking, two points near to $1$ in $\mathbb{S}$ may be sent to far points in $[0,1)$: $f^{-1}$ cut the circle, the transformation cannot be continuous.

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It follows from that

If there exists a continuous mapping $f: X \to Y$ of a compact space $X$ onto a Hausdorff space $Y$, then $Y$ is a compact space. In other wrds, a continuous image of a compact space, provided it is a Hausdorff space.

Proof: Let $\{U_s\}_{s\in S}$ be an open cover of the space $Y$. The family $\{f^{-1}(U_s)\}_{s\in S}$ is an open cover of $X$; thus there exists a finite set $\{s_1,s_2,...s_k\} \subset S$ such that $$f^{-1}(U_{s_1})\cup f^{-1}(U_{s_2})\cup...\cup f^{-1}(U_{s_k})=X,$$ and this implies that $U_{s_1}\cup U_{s_2}\cup...\cup U_{s_k}=Y.$

Note that $S$ is compact and $[0,1)$ is not compact.

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Taking a small open (in the space $[0,1)$) neighbourhood$~U$ of $0$, its image $f(U)$ by the wrapping function$~f$ is not open in $S$ (and not a neighbourhood of $f(0)$). Since $f(U)$ is also the inverse image of$~U$ under$~f^{-1}$, that function is not continuous, by the basic definition of continuous.

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