- #1

- 757

- 0

^{n}to R

^{m}for n>m?

I'm thinking in the context of geometry, where you want to map a bounded region from a higher space to a lower space.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Curl
- Start date

- #1

- 757

- 0

I'm thinking in the context of geometry, where you want to map a bounded region from a higher space to a lower space.

- #2

chiro

Science Advisor

- 4,790

- 132

^{n}to R^{m}for n>m?

I'm thinking in the context of geometry, where you want to map a bounded region from a higher space to a lower space.

This is an interesting question.

Intuitively since this is bounded, you would think that it is possible, but given the nature of real numbers, I'm not so sure it is.

It is definitely not possible if m = 0. You can not map an interval onto a point: it is impossible to go from dimension > 0 to 0 and back again.

My guess is if you wanted to prove this, you should use an inductive argument that shows that you can't go from m to m-1 in terms of one-to-one properties, and then if that is correct, you're finished.

- #3

Deveno

Science Advisor

- 906

- 6

imagine you have the peano space-filling curve (well, actually the hilbert one works better)

(http://en.wikipedia.org/wiki/Space-filling_curve)

now this map may not be injective, but it IS surjective, so we can define a right-inverse from R

i don't think you can do this with differentiable curves, though.

- #4

disregardthat

Science Advisor

- 1,861

- 34

- #5

Deveno

Science Advisor

- 906

- 6

- #6

disregardthat

Science Advisor

- 1,861

- 34

^{2}, the injection we are seeking will map to two disconnected sets, but R^{2}-{x_{0},y_{0}} is connected.

An injection isn't necessarily an homeomorphism. Also, such an injective function need not necessarily be surjective. But, as it turns out, injections from R^n to R^n have open images and are homeomorphisms onto their image by the invarience of domain theorem.

http://en.wikipedia.org/wiki/Invariance_of_domain

As R^m can be seen as a subspace of R^n, such an injective function f would be homeomorphic to an open subset of R^m. This set would then be open and

- #7

Deveno

Science Advisor

- 906

- 6

An injection isn't necessarily an homeomorphism. Also, such an injective function need not necessarily be surjective. But, as it turns out, injections from R^n to R^n have open images and are homeomorphisms onto their image by the invarience of domain theorem.

http://en.wikipedia.org/wiki/Invariance_of_domain

As R^m can be seen as a subspace of R^n, such an injective function f would be homeomorphic to an open subset of R^m. This set would then be open andconnectedin R^m (since R^n is connected), thus generating an homeomorphism from R^n to R^m. But this is impossible due to different fundamental groups.

yes, i know that a continuous injection need not be a homeomorphism (even a continuous bijection need not be, the inverse also has to be continuous). but, as you point out, we can easily translate our proposed injection into a homeomorphism, so the connectedness should run both ways (i seem to recall, dimly, this (the open mapping property of continuous injection in R

i don't understand your remark about fundamental groups, because π

- #8

disregardthat

Science Advisor

- 1,861

- 34

i don't understand your remark about fundamental groups, because π_{1}(R^{n}) is trivial.

You're right, faulty train of thought there. I was thinking about fundamental groups as one remove a certain subset from R^m. It works for m = 1 by removing a point, m = 2 by removing a line I guess, but I don't see how to formalize it for general m. It can actually be seen easily by the invariance of domain theorem however. R^m can be seen as a closed subspace of R^n, so a homeomorphism would imply an injective function from R^n to R^n with image R^m which must be open by the theorem.

- #9

- 431

- 0

However, they obviously have the same cardinality (ignoring the n=0 case) so there will be bijections between then.

- #10

- 431

- 0

But each of R^n and R^m with a point removed are homotopy equivalent to S^{n-1} and S^{m-1} respectively, which aren't homeomorphic for n=/=m (just look at homology groups).

- #11

lavinia

Science Advisor

Gold Member

- 3,239

- 625

But each of R^n and R^m with a point removed are homotopy equivalent to S^{n-1} and S^{m-1} respectively, which aren't homeomorphic for n=/=m (just look at homology groups).

You can make this argument even stronger. R[itex]^{m}[/itex] is a subspace of R[itex]^{n}[/itex] so if there were a continuous map that was 1-1 (or even only 1-1 on R[itex]^{m}[/itex]) from R[itex]^{n}[/itex] onto R[itex]^{m}[/itex] then S[itex]^{m-1}[/itex] would be a retract of S[itex]^{n-1}[/itex]. But your homology argument shows that this is impossible.

- #12

disregardthat

Science Advisor

- 1,861

- 34

Regarding the terminology one-to-one and 1-1, does this mean bijective or simply injective?

- #13

- 431

- 0

1-1 = one- to- one = injective

onto = surjective

onto = surjective

- #14

- 431

- 0

You can make this argument even stronger. R[itex]^{m}[/itex] is a subspace of R[itex]^{n}[/itex] so if there were a continuous map that was 1-1 (or even only 1-1 on R[itex]^{m}[/itex]) from R[itex]^{n}[/itex] onto R[itex]^{m}[/itex] then S[itex]^{m-1}[/itex] would be a retract of S[itex]^{n-1}[/itex]. But your homology argument shows that this is impossible.

Sorry Lavinia, could you explain this a little more? For a retract S^{n-1} to S^{m-1}, wouldn't we need the points of S^{m-1} contained in S^{n-1} to be invariant under the induced map? How do we know that the subspace R^m is left invariant under such a homeomorphism?

- #15

lavinia

Science Advisor

Gold Member

- 3,239

- 625

Regarding the terminology one-to-one and 1-1, does this mean bijective or simply injective?

1-1 means injective to me but I also said onto - I hope

- #16

lavinia

Science Advisor

Gold Member

- 3,239

- 625

Sorry Lavinia, could you explain this a little more? For a retract S^{n-1} to S^{m-1}, wouldn't we need the points of S^{m-1} contained in S^{n-1} to be invariant under the induced map? How do we know that the subspace R^m is left invariant under such a homeomorphism?

My idea was to remove the origin so that the inclusion of R[itex]^{m}[/itex] into R[itex]^{n}[/itex] followed by the 1-1 map from R[itex]^{m}[/itex] onto R[itex]^{m}[/itex] induces an isomorphism on the top homology of R[itex]^{m}[/itex] minus a point - so I was a little hasty because this is not exactly a retract - sorry. But the idea is the same.

I guess once could say that this map when restricted to the m-1 sphere is a continuous bijection and therefore since the sphere is compact, is a homeomorphism onto its image. This shows that the map must induce an isomorphism on homology from the m-1 sphere onto its its image which is itself homeomorphic to an m-1 sphere. But this map is factored through n space minus a point and its m-1st homology is zero.

Do you think this is wrong? I suddenly got worried. But I think its OK.

Last edited:

- #17

- 757

- 0

Can anyone show how to map the unit disk to an interval, or the unit square to an interval on R?

Or how about a cube or sphere mapped to R^2?

- #18

Deveno

Science Advisor

- 906

- 6

this gives us a surjection. using the axiom of choice, pick a point in the preimage of any point in the plane region, this gives a bijection between a subset of a real interval and a region of the plane. some possible things to look at:

Hilbert curve

Peano curve

Moore curve

Dragon curve

Terdragon curve

Gosper curve

Lesbegue curve (this one is "almost differentiable", which is kinda scary)

one thing i did not know until recently, is that these types of things are actually used to convert higher-dimensional information such as maps or IP addresses, into linear sequences.

- #19

lavinia

Science Advisor

Gold Member

- 3,239

- 625

Can anyone show how to map the unit disk to an interval, or the unit square to an interval on R?

Or how about a cube or sphere mapped to R^2?

As my previous post showed, there is no continuous bijection from R^m to R^m with m <n. The same type of arguments apply for the unit disk or a square or a cube or whatever.

There are continous space filling maps as was previously posted but they are not 1-1.

There are bijection between any n dimensional region and any m dimensional region, m>0, but these can not be continuous.

- #20

- 431

- 0

There are no homeomorphisms between different dimensional spaces.

For n<m there are continuous maps R^n onto R^m but they will always fail to be one to one.

In general, of course there are non-continuous bijections between them too.

But the original question was can there exist a transformation R^n into R^m which is one to one for n>m. I'll assume we mean continuous here. We don't seem to have answered this question- space filling is mapping lower dimensional things to cover higher ones.

The answer, I think, is clearly that you can't have a continuous map from R^n into R^m for n>m which is one to one. Is there an easy way to see this?

- #21

Deveno

Science Advisor

- 906

- 6

There are no homeomorphisms between different dimensional spaces.

For n<m there are continuous maps R^n onto R^m but they will always fail to be one to one.

In general, of course there are non-continuous bijections between them too.

But the original question was can there exist a transformation R^n into R^m which is one to one for n>m. I'll assume we mean continuous here. We don't seem to have answered this question- space filling is mapping lower dimensional things to cover higher ones.

The answer, I think, is clearly that you can't have a continuous map from R^n into R^m for n>m which is one to one. Is there an easy way to see this?

suppose f:R

choose any x in (a,b), and consider the set A = R

this is still connected, but f(A) is not, contradicting the fact is f is continuous. therefore, there can be no such continuous injection.

to generalize to m dimensions, remove a set of m-1 dimensions from f(R

- #22

lavinia

Science Advisor

Gold Member

- 3,239

- 625

suppose f:R^{2}→R is a continuous injection. then f(R^{2}) is connected (since R^{2}is), so we may assume it contains an interval (a,b).

choose any x in (a,b), and consider the set A = R^{2}- f^{-1}(x).

this is still connected, but f(A) is not, contradicting the fact is f is continuous. therefore, there can be no such continuous injection.

to generalize to m dimensions, remove a set of m-1 dimensions from f(R^{n}) (a line from a region of R^{2}, a plane from a region of R^{3}...in general, a hyperplane of R^{m}), such that f(R^{n}) is disconnected after removing the hyperplane. R^{n}is still connected, even after removing the pre-image of the hyperplane.

good argument. The homology argument also works. But yours reveals the idea simply.

Last edited:

- #23

- 431

- 0

Indeed, we have already shown that there exist odd space filling maps from lower dimensional spaces to higher ones- how do we know that removing a circle in the plane doesn't remove a sphere from the preimage in 3-space, for example? (the preimage isn't even necessarily continuous).

Also, how do we know we can remove these sorts of sets- I don't see an obvious generalisation here. What if the higher dimensional space mapped to the plane which leaves a fractal like image, for which we can't find circles or hyperplanes to remove?

Your argument seems like the right one, but with perhaps some annoying tweaks missing.

- #24

lavinia

Science Advisor

Gold Member

- 3,239

- 625

Indeed, we have already shown that there exist odd space filling maps from lower dimensional spaces to higher ones- how do we know that removing a circle in the plane doesn't remove a sphere from the preimage in 3-space, for example? (the preimage isn't even necessarily continuous).

Also, how do we know we can remove these sorts of sets- I don't see an obvious generalisation here. What if the higher dimensional space mapped to the plane which leaves a fractal like image, for which we can't find circles or hyperplanes to remove?

Your argument seems like the right one, but with perhaps some annoying tweaks missing.

So you are saying that the map on the removed hyperplane could be space filling. Hmmm That seems like a good point. We are back to the homology argument.

Share: