# Show that a continuously differentiable function is not 1-1

1. Oct 2, 2018

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Let $f:ℝ^2\rightarrow ℝ$ be a continuously differentiable function. Show that $f$ is not one-to-one."

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

3. 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.

Last edited: Oct 2, 2018
2. Oct 2, 2018

### 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).

3. Oct 2, 2018

### 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 $f(x,y)=z_0$ ......... :)

4. Oct 2, 2018

### PeroK

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

5. Oct 2, 2018

### Ray Vickson

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.

6. Oct 2, 2018

### PeroK

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

7. Oct 3, 2018

### Eclair_de_XII

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 $\lim_{(0,y)\rightarrow (0,0)} f(0,y)$?

8. Oct 3, 2018

### PeroK

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.

9. Oct 3, 2018

### Eclair_de_XII

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.

10. Oct 3, 2018

### PeroK

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?

11. Oct 3, 2018

### Eclair_de_XII

Is it the pre-image of its inverse?

12. Oct 3, 2018

### PeroK

It's an open interval.

13. Oct 3, 2018

### 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?

14. Oct 3, 2018

### Ray Vickson

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!

Last edited: Oct 4, 2018
15. Oct 4, 2018

### Eclair_de_XII

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 $(x,y_1)\neq (x,y_2)$.

16. Oct 4, 2018

### 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.

Last edited: Oct 4, 2018
17. Oct 5, 2018

### 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?

18. Oct 6, 2018

### WWGD

I think it can be done with Sard's theorem

19. Oct 8, 2018

### 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?

20. Oct 8, 2018

### WWGD

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.

Last edited: Oct 8, 2018