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$X=[0,2\pi)$ with the relative topology determined by the usual topology on $\mathbb R$, and the unit circle $S=\{(x_1,x_2)\mid x_1^2+x_2^2=1\}$, with the relative topology determined by the usual topology on $\mathbb R^2$. Define $f:X\to S$ by $f(t)=(\cos(t),\sin(t))$. Prove $f$ is continuous and prove $f$ inverse is not continuous.

I keep looking at this and for some reason solve it. Please help me

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3 Answers 3

HINT: First show that the inverse image of an open interval in $S$ is either an open interval in $X$ or a set of the form $[0,x)\cup(y,2\pi)$, which is also open in $X$; from this you can conclude that $f$ is continuous. Now let $g=f^{-1}$. $[0,1)$ is an open set in $X$; is $g^{-1}\big[[0,1)\big]$ open in $S$?

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thank you.....do you know an easy way to prove f is one-to-one and onto?? –  moe Mar 18 '13 at 14:54
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@moe: $f$ just sends $t\in[0,2\pi)$ to the point whose polar coordinates are $\langle 1,t\rangle$, so it clearly traverses the circle $S$ anticlockwise; I doubt that you really need more proof than that. –  Brian M. Scott Mar 18 '13 at 15:02
    
I appreciate your help on everything –  moe Mar 18 '13 at 15:05
    
@moe: My pleasure. –  Brian M. Scott Mar 18 '13 at 15:06

Draw pictures of both sets. Draw an epsilon interval around $0$ on the line segment $X$. Now draw the smallest possible delta interval around the point $(1,0)$ on the unit circle. Does $f^{-1}$ map all of the points in this delta interval into the epsilon interval on the line segment?

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f^(-1) is not even a function as df is not 1-1.

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$f$ certainly is 1-to-1! Note that it includes 0 but not $2\pi$. –  Twiceler Mar 18 '13 at 0:58

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