# Create a surjective function from [0,1]^n→S^n

• I
• laurabon
In summary: Next stretch the latitude lines in the Southern Hemisphere northward from the equator until they touch the North Pole. The equator will be mapped to the pole by this stretching and the rest of the southern hemisphere will be mapped to the sphere minus the North Pole.A Topologist might consider this visualization to be a proof even though no specific mapping is written down.In summary, there are two proposed proofs for finding a continuous surjective function from [0,1]^n to S^n. The first method involves using the quotient map and composing it with a continuous surjective function from [0,1] to [0,1]^n, such as the Peano curve. The second method involves using a space

#### laurabon

the first method is this : I think I can create a surjective function f:[0,1]^n→S^n in this way : [0,1]^n is omeomorphic to D^n and D^n/S^1 is omeomorphic to S^n
so finding a surjective map f is equal to finding a surjective map f':D^n →D^n/S^n and that is quotient map.
Now if I take now a continuous surjective function [0,1]→[0,1]^n e.g Peano curve and I compose the continuous surjective function [0,1]^n→S^n above I should have the proof .

the second one I founnd on internet is this A non-empty Haussdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected,locally connected andsecond countabke space , S^n has all this properties so there exist a continuous and surjective function from [0,1]^n→S^n

Are the 2 proofs correct?

laurabon said:
the first method is this : I think I can create a surjective function f:[0,1]^n→S^n in this way : [0,1]^n is omeomorphic to D^n and D^n/S^1 is omeomorphic to S^n
so finding a surjective map f is equal to finding a surjective map f':D^n →D^n/S^n and that is quotient map.
Now if I take now a continuous surjective function [0,1]→[0,1]^n e.g Peano curve and I compose the continuous surjective function [0,1]^n→S^n above I should have the proof .

the second one I founnd on internet is this A non-empty Haussdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected,locally connected andsecond countabke space , S^n has all this properties so there exist a continuous and surjective function from [0,1]^n→S^n

Are the 2 proofs correct?

"The unit interval" $[0,1]$ is not the same thing as $[0,1]^n$, which is a unit hypercube. Which are you trying to show? A surjection from $[0,1]$ or a surjection from $[0,1]^n$? And do you need it to be continuous?

So it seems this may just be a parametrization you're looking for?
I can think for ##n=1##, i.e. using ##[0,1]##, you can adjust the map ##f(t)=( cost, sint)## maybe as
##f(t)=( cos(2\pi t), sin( 2 \pi t))##
And then use a more general parametrization for other ##n##.

Set $f_1(t_1) = \cos(2\pi t_1)e_1 + \sin(2\pi t_1)e_2$ and construct $f_n$ recursively as $$f_n(t_1, \dots, t_n) = f_{n-1}(t_1, \dots, t_{n-1})\cos(2\pi t_n) + \sin(2\pi t_n)e_{n+1}$$ where $e_1, \dots, e_{n+1}$ are the standard basis vectors of $\mathbb{R}^{n+1}$. Then in the euclidean inner product $$\|f_n(t_1,\dots,t_n)\|^2 = \|f_{n-1}(t_1, \dots, t_{n-1})\|^2\cos^2(2\pi t_n) + \sin^2(2\pi t_n) = 1.$$

Office_Shredder
Btw, if you accept [0,1] is connected, this shows ## S^n ## is both compact and connected, as the continuous image of a connected , compact set ##[0,1]^n ##.

laurabon said:
the first method is this : I think I can create a surjective function f:[0,1]^n→S^n in this way : [0,1]^n is omeomorphic to D^n and D^n/S^1 is omeomorphic to S^n
so finding a surjective map f is equal to finding a surjective map f':D^n →D^n/S^n and that is quotient map.
Now if I take now a continuous surjective function [0,1]→[0,1]^n e.g Peano curve and I compose the continuous surjective function [0,1]^n→S^n above I should have the proof .

the second one I founnd on internet is this A non-empty Haussdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected,locally connected andsecond countabke space , S^n has all this properties so there exist a continuous and surjective function from [0,1]^n→S^n

Are the 2 proofs correct?
@laurabon Your idea for the second proof is correct but you have stated it in a confusing way. It is true that there is a space filling curve that covers the n-sphere. One gets a continuous surjective map of the n-cube onto the sphere by first projecting the n-cube onto the unit interval then following the projection with the space filling curve.

A visualization of a quotient mapping might be to first inflate the n-cube into an n-ball then project the n-ball onto the southern hemisphere of the n-sphere. Next stretch the latitude lines in the Southern Hemisphere northward from the equator until they touch the North Pole. The equator will be mapped to the pole by this stretching and the rest of the southern hemisphere will be mapped to the sphere minus the North Pole.

A Topologist might consider this visualization to be a proof even though no specific mapping is written down.

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lavinia said:
Your idea for the second proof is correct but you have stated it in a confusing way. It is true that there is a space filling curve that covers the n-sphere. One gets a continuous surjective map of the n-cube onto the sphere by first projecting the n-cube onto the unit interval then following the projection with the space filling curve.

A visualization of a quotient mapping might be to first inflate the n-cube into an n-ball then project the n-ball onto the southern hemisphere of the n-sphere. Next stretch the latitude lines until they touch the North Pole. The equator will be mapped to the pole by this stretching and the rest of the southern hemisphere will be mapped to the sphere minus the North Pole.

A Topologist might consider this visualization to be a proof even though no specific mapping is written down.
I think there's something wrong there if I understood you correctly. There can't be such bijection between [0,1]^n and S^n, as for, e.g., n=1, it would be a continuous bijection between compact and Hausdorff , which implies a homeomorphism, which is impossible here.

WWGD said:
I think there's something wrong there if I understood you correctly. There can't be such bijection between [0,1]^n and S^n, as for, e.g., n=1, it would be a continuous bijection between compact and Hausdorff , which implies a homeomorphism, which is impossible here.
Right. The map is not a bijection, only a surjection. A space filling curve is many to 1 as is the projection of the cube onto the interval.

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WWGD
@WWGD A weird thought. What if the sphere is given the quotient topology under the space filling curve,

Since the space filling curve is continuous all of the standard open sets on the sphere are open in the quotient topology. But how does one characterize the others?

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