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

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
  • #1
laurabon
16
0
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?
 
Physics news on Phys.org
  • #2
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" [itex][0,1][/itex] is not the same thing as [itex][0,1]^n[/itex], which is a unit hypercube. Which are you trying to show? A surjection from [itex][0,1][/itex] or a surjection from [itex][0,1]^n[/itex]? And do you need it to be continuous?
 
  • #3
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##.
 
  • #4
Set [itex]f_1(t_1) = \cos(2\pi t_1)e_1 + \sin(2\pi t_1)e_2[/itex] and construct [itex]f_n[/itex] recursively as [tex]
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}[/tex] where [itex]e_1, \dots, e_{n+1}[/itex] are the standard basis vectors of [itex]\mathbb{R}^{n+1}[/itex]. Then in the euclidean inner product [tex]
\|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.[/tex]
 
  • Like
Likes Office_Shredder
  • #5
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 ##.
 
  • #6
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.
 
Last edited:
  • #7
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.
 
  • #8
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.
 
Last edited:
  • Like
Likes WWGD
  • #9
@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?
 
Last edited:

1. What is a surjective function?

A surjective function, also known as a surjection, is a function in which every element in the range is mapped to by at least one element in the domain. In other words, every element in the output has a corresponding input value.

2. What is the significance of [0,1]^n and S^n in this function?

[0,1]^n and S^n are mathematical notations for the unit n-dimensional cube and the n-dimensional sphere, respectively. In this function, [0,1]^n represents the domain and S^n represents the range. This means that the function will map elements from the unit n-dimensional cube to the n-dimensional sphere.

3. How can I create a surjective function from [0,1]^n to S^n?

To create a surjective function from [0,1]^n to S^n, you can use a mapping that projects the points from the unit n-dimensional cube onto the surface of the n-dimensional sphere. This can be achieved by using a transformation such as a rotation or a projection.

4. Can a surjective function from [0,1]^n to S^n be invertible?

No, a surjective function from [0,1]^n to S^n cannot be invertible. This is because an invertible function must be both injective (or one-to-one) and surjective. However, in this case, the function is only surjective, meaning that multiple elements in the domain can map to the same element in the range.

5. What are some real-world applications of a surjective function from [0,1]^n to S^n?

A surjective function from [0,1]^n to S^n can be used in various applications, such as computer graphics, computer vision, and image processing. It can also be used in physics and engineering to map points in a three-dimensional space onto a two-dimensional surface, such as a sphere, for visualization and analysis.

Similar threads

Replies
15
Views
1K
Replies
11
Views
4K
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
  • Topology and Analysis
Replies
10
Views
2K
  • Sticky
  • Topology and Analysis
Replies
9
Views
5K
  • Topology and Analysis
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
523
  • Topology and Analysis
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
490
  • Topology and Analysis
Replies
8
Views
1K
Back
Top