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$1. $ Is there exist a continuous onto function $f : S^1 \rightarrow S^2 $?

$2.$ Is there exist a continuous onto function $f : S^2 \rightarrow S^1$?

My thinking : I thinks in both cases $1$ and $2$ it is not possible because $S^2$ minus two points is connected while $S^1$ minus two points isn‘t.

balddraz
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jasmine
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1 Answers1

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For $t\in [0,\pi]$ let $f_1(\cos t,\sin t)=t/\pi.$ For $t\in [\pi,2\pi]$ let $f_1(\cos t, \sin t)=(2\pi-t)/\pi.$

There exists a continuous surjection $f_2:[0,1]\to [0,1]^2.$ See "Peano Curve" in Wikipedia.

For $(x,y)\in [0,1]^2$ let $f_3(x,y)=(\cos 2\pi x, \sin 2\pi x \cos \pi y, \sin 2\pi x \sin \pi y).$

Now $f=f_3f_2f_1$ is a continuous surjection from $S^1$ to $S^2.$

DanielWainfleet
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  • @PaulFrost $[0,1]^2$ IS the range of $f_2$. It's a SURjection. – DanielWainfleet Aug 13 '19 at 07:58
  • @PaulFrost . Thank you. That was a typo. Typo is my native language. – DanielWainfleet Aug 13 '19 at 08:02
  • Perhaps you have another typo, do you mean $f=f_3\circ (f_2\times f_1)$ instead of $f=f_3f_2f_1$? Or, possibly $f=f_3\circ (f_1\times f_2)$. Also, you edited so now $f_3$ is a function of two variables (if you look at the domain), but also of three variables (if you look at the definition). – Mirko Aug 13 '19 at 08:07
  • @Mirko . Yes, another typo. In the 1st line $f$ should have been $ f_1$.....$f_1$ maps $S^1$ to $[0,1]$ and $f_2$ maps $ [0,1] $ to $[0,1]^2$ and $f_3$ maps $[0,1]^2$ to $S^2.$ – DanielWainfleet Aug 13 '19 at 14:54