Show that a continuously differentiable function is not 1-1

[Eclair_de_XII]

Homework Statement

"Let $f:ℝ^2\rightarrow ℝ$ be a continuously differentiable function. Show that $f$ is not one-to-one."

Homework Equations

A function $f:ℝ^n\rightarrow ℝ^m$ is continuously differentiable if all the partial derivatives of all the components of $f$ exist and are continuous.

Hint: If, for example, $D_1f(x,y)\neq 0$ for all $(x,y)\in A\subset ℝ^2$, consider $g:A\rightarrow ℝ^2$ defined by $g(x,y)=(f(x,y),y)$.

Notation: $D_1f(x,y)=\frac{d}{dx}f(x,y)$

The Attempt at a Solution

I'm at a loss as to what to do here... I've tried finding $Dg(x,y)$ and then $D(fg)(x,y)$, but I cannot seem to figure out a way to show that whenever $(x,y_1)\neq (x,y_2)$, then $f(x,y_1)= f(x,y_2)$. In order to do that, I plan on showing that $D_2f(x,y)=0$. So thus far, I have:

$Dg(x,y)=\begin{pmatrix} D_1f(x,y) & D_2f(x,y) \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix}$

So: $D(fg)(x,y)=Df(g(x,y))Dg(x,y)=\begin{pmatrix} D_1f(f(x,y),y) & D_2f(f(x,y),y) \\ \end{pmatrix} \begin{pmatrix} D_1f(x,y) & D_2f(x,y) \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix}= xD_1f(x,y)D_1f(f(x,y),y)+y[D_1f(f(x,y),y)D_2f(x,y)+D_2f(f(x,y),y)]$

So I either must show that $D_2f(x,y)=0$, or I'm approaching this problem incorrectly. Any insight on this will be very much appreciated. This is my most difficult class this semester, and it takes a while for me to pick up a new topic and learn it.

 

[Dick]

I'm not sure I see how the hint is supposed to work, but if you want an alternative hint, think about what the function looks like on the unit circle (or any other closed path).

 

[wrobel]

So you have a function $z=f(x,y)$. If $f=const$ then the assertion is trivial. In other case there is a point $(x_0,y_0)$ such that
$$\frac{\partial f}{\partial x}(x_0,y_0)\ne 0$$ or $$\frac{\partial f}{\partial y}(x_0,y_0)\ne 0.$$ Let for definiteness be
$$\frac{\partial f}{\partial x}(x_0,y_0)\ne 0,\quad z_0=f(x_0,y_0).$$
From the implicit function theorem we know that the equation $f(x,y)=z_0$ ... :)

 

[PeroK]

Any insight on this will be very much appreciated.

Do you think simple continuity of $f$ might be sufficient to exclude one-to-one-ness?

 

[Ray Vickson]

Homework Statement

"Let $f:ℝ^2\rightarrow ℝ$ be a continuously differentiable function. Show that $f$ is not one-to-one."

Homework Equations

A function $f:ℝ^n\rightarrow ℝ^m$ is continuously differentiable if all the partial derivatives of all the components of $f$ exist and are continuous.

Hint: If, for example, $D_1f(x,y)\neq 0$ for all $(x,y)\in A\subset ℝ^2$, consider $g:A\rightarrow ℝ^2$ defined by $g(x,y)=(f(x,y),y)$.

Notation: $D_1f(x,y)=\frac{d}{dx}f(x,y)$

The Attempt at a Solution

I'm at a loss as to what to do here... I've tried finding $Dg(x,y)$ and then $D(fg)(x,y)$, but I cannot seem to figure out a way to show that whenever $(x,y_1)\neq (x,y_2)$, then $f(x,y_1)= f(x,y_2)$. In order to do that, I plan on showing that $D_2f(x,y)=0$. So thus far, I have:

$Dg(x,y)=\begin{pmatrix} D_1f(x,y) & D_2f(x,y) \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix}$

So: $D(fg)(x,y)=Df(g(x,y))Dg(x,y)=\begin{pmatrix} D_1f(f(x,y),y) & D_2f(f(x,y),y) \\ \end{pmatrix} \begin{pmatrix} D_1f(x,y) & D_2f(x,y) \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix}= xD_1f(x,y)D_1f(f(x,y),y)+y[D_1f(f(x,y),y)D_2f(x,y)+D_2f(f(x,y),y)]$

So I either must show that $D_2f(x,y)=0$, or I'm approaching this problem incorrectly. Any insight on this will be very much appreciated. This is my most difficult class this semester, and it takes a while for me to pick up a new topic and learn it.

You have explained your understanding of the concept of differentiability. What is your understanding of the concept of "one-to-one" or not?

What theorems are already available to you? Have you had the implicit function theorem or the inverse function theorem?

Try to jetission some of the symbols and fussy formalism; try to concentrate instead on meaning and intuitive understanding, so you know what you are aiming for. Formalism and rigor can come later, when come to translate your intuition into a firm, logical argument.

 

[PeroK]

I'm not sure I see how the hint is supposed to work, but if you want an alternative hint, think about what the function looks like on the unit circle (or any other closed path).

@Eclair_de_XII Another alternative is to think about how $f$ maps the $x-$ and $y-$ axes. What happens around the point $(0,0)$?

 

[Eclair_de_XII]

Do you think simple continuity of $f$ might be sufficient to exclude one-to-one-ness?

It shouldn't be enough; a function can be continuous and not be one-to-one.

What is your understanding of the concept of "one-to-one" or not?

A function $f:ℝ\rightarrow ℝ$ is one-to-one if whenever $x\neq y$, then $f(x)\neq f(y)$. But my book showed me another condition for one-to-one-ness for higher dimensions of Euclidean space... If a function $f:ℝ^n\rightarrow ℝ^n$ is continuously differentiable at $a\inℝ^n$, then if $det[f'(a)]\neq 0$, then there exists an open set $A\in ℝ^n$ containing $a$ such that for all $x\in A$, $det[f'(x)]\neq 0$, and $f$ is one-to-one on $A$.

Have you had the implicit function theorem or the inverse function theorem?

I have had the inverse function theorem explained to me, but not the implicit function theorem. Though I had much trouble following the former theorem's proof. Let me see:

"The inverse function theorem states that if $f:ℝ^n\rightarrow ℝ^n$ is continuously differentiable at $a\in ℝ^n$, with $det f'(a) \neq 0$, then there exists an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that there exists a continuous inverse $f^{-1}:W\rightarrow V$ that is continuously differentiable, with $(f^{-1})'(y)=f'(f^{-1}(y))^{-1}$"

Another alternative is to think about how $f$ maps the $x-$ and $y-$ axes. What happens around the point $(0,0)$?

Are you asking about the values of $\lim_{(x,0)\rightarrow (0,0)} f(x,0)$ and $\lim_{(0,y)\rightarrow (0,0)} f(0,y)$?

 

[PeroK]

It shouldn't be enough; a function can be continuous and not be one-to-one.

Are you asking about the values of $\lim_{(x,0)\rightarrow (0,0)} f(x,0)$ and $\lim_{(0,y)\rightarrow (0,0)} f(0,y)$?

And a function can be continuously differentiable and not one to one. So, your logic does not hold.

For your information, there are no continuous one-to-one functions from $\mathbb{R^2}$ to $\mathbb{R}$. You can find that easily on the Internet.

I'm not talking about limits at all. Assume $f(0,0) = a$. Where could $f$ map the x-axis? And where could $f$ map the y-axis? Assuming $f$ is continuous.

 

[Eclair_de_XII]

Where could$f$ map the x-axis?

If I'm understanding you correctly, then I guess that $f$ can map the x-axis to $f(x,0)=a+sin(x)$, and I guess I could throw in a bunch of polynomials with degree greater than or equal to one.

 

[PeroK]

If I'm understanding you correctly, then I guess that $f$ can map the x-axis to $f(x,0)=a+sin(x)$, and I guess I could throw in a bunch of polynomials with degree greater than or equal to one.

I don't follow that at all. If $f$ is continuous and one-to-one, then its restriction to the x-axis is also continuous and one-to-one.

Can you see what the image of a continuous and one-to-one function from R to R must be?

 

[Eclair_de_XII]

Can you see what the image of a continuous and one-to-one function from R to R must be?

Is it the pre-image of its inverse?

 

[PeroK]

Is it the pre-image of its inverse?

It's an open interval.

 

[Eclair_de_XII]

So I'm going to hazard a guess here, and say that the image of the x-axis under $f$ is an open interval containing $a$. Am I wrong?

 

[Ray Vickson]

It shouldn't be enough; a function can be continuous and not be one-to-one.
A function $f:ℝ\rightarrow ℝ$ is one-to-one if whenever $x\neq y$, then $f(x)\neq f(y)$. But my book showed me another condition for one-to-one-ness for higher dimensions of Euclidean space... If a function $f:ℝ^n\rightarrow ℝ^n$ is continuously differentiable at $a\inℝ^n$, then if $det[f'(a)]\neq 0$, then there exists an open set $A\in ℝ^n$ containing $a$ such that for all $x\in A$, $det[f'(x)]\neq 0$, and $f$ is one-to-one on $A$.
I have had the inverse function theorem explained to me, but not the implicit function theorem. Though I had much trouble following the former theorem's proof. Let me see:

"The inverse function theorem states that if $f:ℝ^n\rightarrow ℝ^n$ is continuously differentiable at $a\in ℝ^n$, with $det f'(a) \neq 0$, then there exists an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that there exists a continuous inverse $f^{-1}:W\rightarrow V$ that is continuously differentiable, with $(f^{-1})'(y)=f'(f^{-1}(y))^{-1}$"

Are you asking about the values of $\lim_{(x,0)\rightarrow (0,0)} f(x,0)$ and $\lim_{(0,y)\rightarrow (0,0)} f(0,y)$?

You have presented an explanation of 1:1-ness (or non 1:1-ness) "formally", but what is you intuitive understanding of the concept? Can you explain it without a lot of symbols, etc?

If I were looking at it informally, I would just note that the surface $z = f(x,y)$ is smooth under the hypotheses of the question, and for most values of $z_0$ there are either NO values of $(x,y)$ giving $f(x,y) = z_0$ (for example, if $z_0$ is above the maximum height of the surface) or else there will be many values of $(x,y)$ giving $f(x,y) = z_0$. In other words, most of the time when $(x_0,y_0,z_0)$ is a point on the surface, there will be a curve passing through $(x_0,y_0)$ what solves the equation $f(x,y)= z_0$ in a non-trivial neighbourhood of $(x_0,y_0)$. Of course, for some particular values of $z_0$ there might be a single root of the equation $f(x,y) = z_0$ --- for example, right at the maximum on the surface $z = f(x,y)$ --- but the question is asking you to prove that there are at least some values of $z_0$ for which the equation $f(x,y) = z_0$ has more than one solution---a curve, in fact.

When it comes time to get formal, you can cite the appropriate theorems that guarantee this. The exception would be for $f(x,y) \equiv c$, where $c$ is a constant, and in that case the mapping $f(x,y)$ is certainly not 1:1, so you are done anyway!

 

[Eclair_de_XII]

Can you explain it without a lot of symbols, etc?

Basically, in $ℝ^2$, if you take a horizontal line and move it across the y-axis, that line will intersect a one-to-one function no more than once for a given $y$. In any case, I've already figured it out. Basically, with $g(x,y)=(f(x,y),y)$, we have that by the inverse function theorem, that $g$ is invertible. So let $g^{-1}(x,y)=(\phi^1(x,y),\phi^2(x,y))$ for some generic functions $\phi(x,y)$. Then it must be true that $gg^{-1}(x,y)=g(\phi^1(x,y),\phi^2(x,y))=(f(\phi^1(x,y),\phi^2(x,y)),\phi^2(x,y)=(x,y)$. so $\phi^2(x,y)=y$ and $f(\phi^1(x,y),y)=x$, so $f$ is independent of the second variable, and so it fails to be one-to-one for $(x,y_1)\neq (x,y_2)$.

 

[mathwonk]

If you understand the two fundamental theorems on one variable calc, the max/min value theorem, and the intermediate value theorem, then you can use the hints of Dick and Perok to finish.

e.g. the restriction of f to the unit circle has a max at some point p and a min at some point q. Then both arcs of the circle between p and q must map onto the same interval. do you see why?

the hint seems to be trying to get you to use the inverse function theorem. I.e. that theorem would allow you to conclude that g is a local diffeomorphism on some open disc, and then to conclude that the function f can be factored as a local diffeomorphism followed by a projection, which cannot be injective.

this is a really tortured proof of non injectivity though and uses wayyy too much machinery. So your professor seems to have been bending over backward to test the inverse function theorem by this problem which is solved far easier by hints of Dick and Perok.

 

[mathwonk]

By the way, I thought I would jack up the level by suggesting you try to prove a smooth function from R^3 to R^2 cannot be injective. Unfortunately the only proof I can think of again imitates the ideas of Dick and Perok, and uses the Jordan curve theorem, and only continuity of the map. I was hoping for a more elementary proof using smoothness and inverse function theorem, but have not been able to think of one. anyone?

 

[WWGD]

By the way, I thought I would jack up the level by suggesting you try to prove a smooth function from R^3 to R^2 cannot be injective. Unfortunately the only proof I can think of again imitates the ideas of Dick and Perok, and uses the Jordan curve theorem, and only continuity of the map. I was hoping for a more elementary proof using smoothness and inverse function theorem, but have not been able to think of one. anyone?

I think it can be done with Sard's theorem

 

[mathwonk]

Seems like a nice idea. Assuming we allow such a non elementary theorem, what would be the approach? Can one show the image of an injective function would contain an open set. e.g?

 

[WWGD]

Homework Statement

"Let $f:ℝ^2\rightarrow ℝ$ be a continuously differentiable function. Show that $f$ is not one-to-one."

Homework Equations

A function $f:ℝ^n\rightarrow ℝ^m$ is continuously differentiable if all the partial derivatives of all the components of $f$ exist and are continuous.

Hint: If, for example, $D_1f(x,y)\neq 0$ for all $(x,y)\in A\subset ℝ^2$, consider $g:A\rightarrow ℝ^2$ defined by $g(x,y)=(f(x,y),y)$.

Notation: $D_1f(x,y)=\frac{d}{dx}f(x,y)$

The Attempt at a Solution

I'm at a loss as to what to do here... I've tried finding $Dg(x,y)$ and then $D(fg)(x,y)$, but I cannot seem to figure out a way to show that whenever $(x,y_1)\neq (x,y_2)$, then $f(x,y_1)= f(x,y_2)$. In order to do that, I plan on showing that $D_2f(x,y)=0$. So thus far, I have:

$Dg(x,y)=\begin{pmatrix} D_1f(x,y) & D_2f(x,y) \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix}$

So: $D(fg)(x,y)=Df(g(x,y))Dg(x,y)=\begin{pmatrix} D_1f(f(x,y),y) & D_2f(f(x,y),y) \\ \end{pmatrix} \begin{pmatrix} D_1f(x,y) & D_2f(x,y) \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix}= xD_1f(x,y)D_1f(f(x,y),y)+y[D_1f(f(x,y),y)D_2f(x,y)+D_2f(f(x,y),y)]$

So I either must show that $D_2f(x,y)=0$, or I'm approaching this problem incorrectly. Any insight on this will be very much appreciated. This is my most difficult class this semester, and it takes a while for me to pick up a new topic and learn it.

I think you may restrict your supposed map to a compact subset, say a closed ball. . Then the restriction is acontinuous injection between compact and Hausdorff, meaning it is a homeomorphism, which cannot happen by dimension reasons. An open 2-ball cannot be homeomorphic to an interval . You can use similar for different n,m.
EDIT: I may have repeated Dick's argument, sorry if this was the case.

 

[WWGD]

Seems like a nice idea. Assuming we allow such a non elementary theorem, what would be the approach? Can one show the image of an injective function would contain an open set. e.g?

OR, like you said, something simpler, using same argument of continuous bijection between compact and Hausdorff is a homeo, selecting a closed 3-ball $B^3$and considering $f:B^3 \rightarrow f(B^3)$ as a homeo. This is a pretty straight forward result from pointset topology. More annoying than difficult to prove.

 

[mathwonk]

but it is not clear that such a homeo must have as image a set containing an open set. I.e. it is "invariance of domain" that an open set in R^3 is not homeo to any set in R^2. Am I missing the idea? If we could get diffeo, then we could use linear algebra for a contradiction. I.e. I don't know what we can conclude after saying we have a homeo from a ball in R^3 to its image R^2. ??

 

[WWGD]

but it is not clear that such a homeo must have as image a set containing an open set. I.e. it is "invariance of domain" that an open set in R^3 is not homeo to any set in R^2. Am I missing the idea? If we could get diffeo, then we could use linear algebra for a contradiction. I.e. I don't know what we can conclude after saying we have a homeo from a ball in R^3 to its image R^2. ??

Well, this is still maybe using a tank to kill a fly, but we can use third Homology . Both are orientable, so third Homology over integers is the integers for the 3 ball and 0 for the 2- ball. Or that the homeo restricts to s homeo between the boundariess S^2 ,S^1 respectively, one has trivial fundamental group, other does not. But over the top, let me look for something simpler.

 

[mathwonk]

my problem is we do not know anything about the homeomorphic image of the 3 ball, i.e. not that it is a 2 ball, or indeed anything about it except that it is a connected subset of R^2. are you saying the 3rd homology of the boundary of the image should be zero because it is a subspace of R^2? That seems promising but again it uses a great deal more than the OP seems to have available. If we are going to use homology then we can perhaps also use Jordan curve theorem. I wonder if there is an argument that actually illustrates the power of the implicit/inverse function theorem. So far I see no reason for his prof to have assigned this problem with the hypothesis of smoothness. ?

 

[WWGD]

my problem is we do not know anything about the homeomorphic image of the 3 ball, i.e. not that it is a 2 ball, or indeed anything about it except that it is a connected subset of R^2. are you saying the 3rd homology of the boundary of the image should be zero because it is a subspace of R^2? That seems promising but again it uses a great deal more than the OP seems to have available. If we are going to use homology then we can perhaps also use Jordan curve theorem. I wonder if there is an argument that actually illustrates the power of the implicit/inverse function theorem. So far I see no reason for his prof to have assigned this problem with the hypothesis of smoothness. ?

But the 3-ball has non-empty interior and homeos take interior points to interior points, so image is nn-empty and I think has non-empty interior. But, yes, still too heavy machinery. I will go back to trying with sard's or something more basic if I can find it.

 

[StoneTemplePython]

how about Borsuk-Ulam?

 

[WWGD]

I don't really understand the objection. Sure, it is not as simple as could be expected but ultimately the 3 ball will have 3-Homology equal to the coefficient ring, since it is orientable. Its image is a subset of the plane so its 3-homology will be 0. So you end up with a homeomorphism that does not preserve homology (when just homotopy equivalence will do) which is a clear contradiction.

 

[mathwonk]

i guess we should use the 2nd homology of the 2 sphere? i.e. the (solid) 3 ball is contractible so all higher homology is zero. the problem i don't know is why a subset of the plane has to have 2nd homology zero. what does that follow from? it seems obvious but why couldn't some weird subset of the plane have homology even though the plane itself does not? I assume it is false but i don't know the proof. of course no submanifold of the plane has 2nd homology but why is the image of an injective map an embedded submanifold?

I wonder if someone can think of an elementary topological proof say that there is no continuous injection from the 2 sphere to the plane. yes borsuk ulam will do it, but what is the proof of that theorem? what does it use? I have not studied topology for a long time and have forgotten these standard proofs.

If one considers the compactness of the 2 sphere and the minimum disc it maps into in the plane, one obtains locally, an open disc mapping continuously into say the (closed) upper half plane, with the center of the disc mapping to the origin. Why is such a map non injective? It seems obvious, but what is a proof? and does assuming smoothness help in any way to make the argument easier?

Of course we can look at an injective continuous map of a 2 sphere into the plane and consider where the equator goes. then if we have the Jordan curve theorem, we know both hemispheres of the sphere must map onto the unique bounded component of the complement of the image of the equator. But this is not very elementary, and i don't know the proof off hand without looking it up.

apologies if i am just being obtuse, to recall shawshank redemption.

 

[StoneTemplePython]

I wonder if someone can think of an elementary topological proof say that there is no continuous injection from the 2 sphere to the plane. yes borsuk ulam will do it, but what is the proof of that theorem? what does it use? I have not studied topology for a long time and have forgotten these standard proofs.

Well they prove exactly this (3 dimensional case of Borsuk-Ulam) on pages 116 - 119 of Chinn and Steenrod's First Concepts of Topology. It needs a concept of a winding number, which typically gets blank stares. Even so, it is self contained in this is book, which is nominally targeted at high school students...

- - - - -
The underlying idea to me (ignoring a lot of topology) for this entire thread is of a defective stereographic projection --i.e. defective because someone comes and tries to add the single point at the north pole back... this inevitably causes a collision somewhere.

 

[WWGD]

i guess we should use the 2nd homology of the 2 sphere? i.e. the (solid) 3 ball is contractible so all higher homology is zero. the problem i don't know is why a subset of the plane has to have 2nd homology zero. what does that follow from? it seems obvious but why couldn't some weird subset of the plane have homology even though the plane itself does not? I assume it is false but i don't know the proof. of course no submanifold of the plane has 2nd homology but why is the image of an injective map an embedded submanifold?

I wonder if someone can think of an elementary topological proof say that there is no continuous injection from the 2 sphere to the plane. yes borsuk ulam will do it, but what is the proof of that theorem? what does it use? I have not studied topology for a long time and have forgotten these standard proofs.

If one considers the compactness of the 2 sphere and the minimum disc it maps into in the plane, one obtains locally, an open disc mapping continuously into say the (closed) upper half plane, with the center of the disc mapping to the origin. Why is such a map non injective? It seems obvious, but what is a proof? and does assuming smoothness help in any way to make the argument easier?

Of course we can look at an injective continuous map of a 2 sphere into the plane and consider where the equator goes. then if we have the Jordan curve theorem, we know both hemispheres of the sphere must map onto the unique bounded component of the complement of the image of the equator. But this is not very elementary, and i don't know the proof off hand without looking it up.

apologies if i am just being obtuse, to recall shawshank redemption.

No man, the third Homology of the 3-ball ( which is nonzero , since it is orientable) with the third of the image. A homeo ( homotopy equivalence) gives rise to isomorphism on each Homology.

 

[mathwonk]

@WWGD, do you refer to the closed unit ball in R^3, i.e. the set of x such that |x| ≤ 1 as the 3ball? If so, surely this is a contractible space, hence homotopy equivalent to a point, hence has zero homology everywhere except at degree zero. What am I missing? Are you referring to the fact that a compact orientable 3 manifold (without boundary) has non zero 3rd homology? the problem here seems to be that the 3 ball has a boundary. Since it is homeomorphic to its image, the image also has no 3rd homology. What do you think?

@StoneTemplePython. Thanks for the relatively elementary reference to Chiin and Steenrod. I am somewhat reluctant to read it there however since I feel any elementary argument should suggest itself to me just by reflecting on it. And I consider winding numbers relatively elementary. The difficulty for me is how to prove that if we have an injective continuous map from the circle to the plane, that there is some point of the plane that the image curve does actually wind around in a non zero way, i.e. why does the complement of the image curve have a bounded component? I don't quite get it from your remark on projections.

Here is a simple statement I would like a simple proof of: a smooth, or even just continuous, injection from R^2 to R^2, is an open mapping, hence a homeomorphism onto an open subset of R^2. This seems to follow from Jordan curve theorem, but since the hypothesis is stronger, might be easier to prove. Notice in dimension one this is just the intermediate value theorem hence pretty elementary, although admittedly as deep as any foundational result in one variable calc.

 

[mathwonk]

A quick course in winding numbers:

Consider a continuous map f from the unit interval [0,1] to the unit circle in the plane such that f(0) = f(1). This is essentially a continuous map from the circle to the circle. We want to measure how many times f “winds around” the circle.

method 1) We subdivide the interval into small subintervals so that the image of each subinterval under f is short (say shorter than 1, just to make sure it does not contain any two antipodal points, so that the image is contained in an angle of less than 180 degrees). Then we note the angle difference between the images of the two ends of each subinterval, and we add up those angle differences. The answer must be 2πn for some n, and n is the number of times f winds around the circle counterclockwise. This must be shown to be independent of the subdivision into small subintervals.

method 2) Consider the exponential map taking t of R to the point e^it in the (complex) plane. (Thus t is the angle measure of the corresponding arc on the circle from e^i.0 to e^it.) Then given a continuous map f:[0,1]—> unit circle, there is a unique continuous map g from [0,1]—>R taking t to 0, and such that for all t in [0,1], we have e^ig(t) = f(t). I.e. g is the log of f. Then the image point g(1) is some number of form 2πn, and we define n as the winding number of f around the unit circle.

These methods are the same, since g is defined essentially the way we measured the angle differences in the images of f, using a subdivision of [0,1].

Now if f is any map from [0,1] to the plane with f(0) = f(1), i.e. a closed curve (not necessarily simple), but missing the origin, we define the winding number of f about the origin to be the winding number of the map taking t to f(t)/|f(t)|, i.e. the map to the unit circle obtained by projecting f(t) radially out to the unit circle.

If p is any point missed by f, we define the winding number of f about p, to be the winding number of the map taking t to (f(t)-p)/|f(t)-p|, i.e. the same but after translating by p to make p into the origin.This concept of winding number has the following basic properties:

1) If f, considered as a map from the circle to the plane, extends to a continuous map F of the closed disc to the plane, still missing the point p, then the winding number of f about p is zero.
2) If there is a continuous map F of a cylinder to the plane, still missing p, which restricts to f on one boundary circle of the cylinder, then the winding number of the restriction of F to the other boundary circle is the same as that of f. I.e. winding number is constant under continuous deformation.
3) If f misses both p and q, and p and q are in the same connected component of the complement of the image of f, then f has the same winding number about p and q. (At least this seems true.)Cor: (contrapositive of 1): If f is a continuous map of the circle to the plane, missing p, but with winding number non zero about p, and if F is any continuous extension of f to the closed disc, then F does not miss p, i.e. for some point q inside the unit disc we must have F(q) = p.

This corollary is the 2 dimensional analog of the intermediate value theorem.Now somehow we want to use these ideas to prove that if F is any continuous map of the 2 sphere into the plane, there are 2 points of the sphere that map to the same point. (weak Borsuk - Ulam).

E.g. if we could prove that for some disc embedded in the sphere, centered say at q, that the boundary of the disc misses F(q) but does wind around F(q) a non zero number of times, then the restriction of F to the other disc in the sphere with the same boundary circle, would have the same (or negative of) winding number number, so would have to also hit F(q). This would give us two points with the same image.

But how to prove that the image of some circle in the sphere, does wind around some point of the plane, a non zero number of times? I.e. how do we know that it cannot happen that the image of every circle fails to wind around any point at all?

help? (Chinn and Steenrod spend 3 pages on this theorem, so some cleverness may be used somewhere.)

Rmk: The fundamental theorem of algebra is relatively easy using these tools, since one can directly calculate that the polynomial map z^n has winding number n about the origin, and that any polynomial of degree n has the same winding number as z^n on some large circle.

But here we don't have a concrete map like z^n to compute with, just the continuity hypothesis.

heres an idea: if you think of a simple closed curve in the plane and one point inside it and another point ouside it, then an arc going from one point to the other should have to cross the closed curve once, or an odd number of times. maybe one can prove that if an arc intersects a simple closed curve once (transversely?) then its endpoints are on opposite sides of the closed curve, which would enable us to know? they have different winding numbers wrt the curve?

So if we have an injection from a 2 sphere to the plane, then a circle close to the south pole does not wind around the image of the north pole, since the constant map at the spouth pole does not. now an arc running from the north pole to the south pole crosses that little circle near the south pole exactly once, and its image joins the images of the south and north poles. So its winding number should be different for the two poles, hence non zero for the south pole. Then by the argument above, the map should actually hit the south pole again at some point on the north pole side of the little circle, a contradiction.

but this does not seem airtight at all to me, since I have not proved the key assumption nor made precise the hypothesis of transversality. But I believe it anyway. Maybe Chinn and Steenrod do something like this.

reactions?

 

[WWGD]

Habout this? We have a smooth map by assumption
.and ythe map restricts to a homeo on the ball. Don't we then have a local diffeomorphism?

 

[mathwonk]

i think the subtlety is the difference between a homeo from the ball to its image, which could be awful, and a homeo from the ball to nice subset. I.e. we have no idea what structure the image has. Unbelievably, even though the image is homeomorphic to a ball, it "could", until proven otherwise, be a strange subset of R^3, say with no interior.

As crazy as it sounds, it is not at all clear theoretically, even though obvious intuitively, that under a continuous or even smooth injective map from the disc to the plane, the image of a disc has a non empty interior. I.e. using the jordan curve theorem we know that the image of the circular boundary of the disc, is a simple closed curve, which separates the plane into two disjoint open components, and that the interior of the disc maps homeomorphically onto the bounded component, an open set in the plane homeomorphic to the disc.

But we do not know without the jordan curve theorem, that the disc might not map homeomorphically onto some strange subset of the plane with no interior at all.

At the moment I cannot even prove that, under an injective continuous map from the disc to the plane, taking 0 to 0, that the image of the boundary circle does actually wind once around the origin.

Well I have an idea, based perhaps on old memories of Chinn and Steenrod, that it may follow from some vector field results. Assume that we have a smooth map from the sphere to the plane such that no two antipodal p[oints map to the same image point. Use that to define a non zero vector field on the disc as follows: to each point of the upper hemisphere of the sphere, you have two distinct points of the plane, namely the image of the given point and the image of its antipodal point. Since thiose are by assumption different, by joining them you get a non zero oriented segment in the plane hence a non zero vector, which you can attach to the original given point of the uppr hemisphere, which is a disc. This gives you then a nerver zero vector field on the disc, which takes oppositely directed vectors at opposite points of the equator.

Now I claim that a non zero vector field on the equator, which sends opposite points to opposite vectors, must define a map from the equator to the circle with non zero winding number. hence it cannot extend to a map from the disc to the circle. i.e. by results above, any extension to the disc must map some point to zero, which does not happen if the points are all projectible onto the circle.I think this works, but it is a bit sophisticated to have to factor the argument through a vector field argument. I don't understand why it has to be so sophisticated.
almost bet this is the kind of proof in Chinn and Steenrod, indeed a quick look at their contents shows the 3 previous sections are about vector fields. So sorry, this is the kind of magic that some beautiful topology results often seem to involve.

the best I can say is that somehow this vector argument let's us use the winding number theory without picking a specific point to wind around, i.e. somehow the vector field includes winding around all points at once in terms of the twisting of the vector field...

you may recall the fixed point arguments for maps of a disc to itself also use a retraction argument from the disc to its boundary defined by the vector field going from each point to its image, assumibng there is no fixed point.

 

[WWGD]

I think the image of the ball would have non-empty interior. Homeos map interior to interior . Or just take an open subset of the ball. Since homeos are open maps, it's image would be open in R^2.