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Let $A$ be a subset of $\mathbb{R}$ and let $f: A \rightarrow \mathbb{R}$ be a continuous function. If $f(x) = mx$ for some positive number $m > 0$, then the function $H: \mathbb{R}^{2} \times [0, 1] \rightarrow \mathbb{R}^{2}$ defined by

$$H(x, y, t) = (x\cos(\theta \cdot t) + y \sin (\theta \cdot t), y \cos(\theta \cdot t) - x \sin(\theta \cdot t))$$

where $\theta = \cos^{-1}(1 - m^{2})$

is an ambient isotopy of $\mathbb{R}^{2}$ that takes the graph of $y = f(x) = mx$ to the graph of $y = -f(x) = -mx$, i.e $H(x, f(x), 0) = (x, f(x))$ and $H(x, f(x), 1) = (x, -f(x), 1)$, while $H(x, y, t)$ remains a homeomorphism of $\mathbb{R}^{2}$ for each $t$ (a rotation).

If the graph of $y = f(x)$, not a line, I believe that there is no ambient isotopy of $\mathbb{R}^{2}$ that takes the graph of $y = f(x)$ to the graph of $y = -f(x)$.

My intuition is that an orientation can be assigned to $\mathbb{R}^{2}$ as a two-form. When this two-form is fed two linearly independent vectors formed from two distinct tangent lines to the graph of $y = f(x)$, a positive number should pop out. When these two vectors are reflected about the $x$-axis, and become tangent vectors to the graph of $y = -f(x)$ (associated to two distinct tangent lines of the graph of $y = -f(x)$) and are again fed into this orientation two form, they should produce a negative number. From a continuity argument, it should then be argued that for some $t$ the isotopy should take these two linearly independent vectors to two vectors whose value when fed into the orientation two-form is zero. This would then imply that thr two vectors would become linearly dependent. But this would contradict the fact that $H(x, y, t)$ is a howeomorphism of $\mathbb{R}^{2}$ since it's pushforward would send two linearly independent vectors to two dependent ones.

However, I am missing significant details in this argument and this argument presumes that $f(x)$ and each $H(x, y, t)$ are not just continuous by at least $C^{1}$ differentiable. Any assistance with filling in the details and generalizing this argument so that functions are only required to be continuous $C^{0}$ would be greatly appreciated. I feel like my initial argument requires not only differentiablitly, but also a connection on $\mathbb{R}^{2}$ (which could just be chosen to be the Levi-Cevita connection induced by the Euclidean metric). Is there way to keep the arguments primarily topological (with just continuous/$C^{0}$ functions or do we need to approximate these functions by differentiable functions and use a connection to translate between tangent lines/tangent spaces?

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Let me address a special case, namely where the subset $A \subset \mathbb R$ is closed. In this case, if $f : A \to \mathbb R$ is any continuous function, then there is indeed an ambient isotopy of $\mathbb R^2$ taking the graph of $y=f(x)$ to the graph of $y=-f(x)$.

To construct it, the first step is to apply the Tietze extension theorem to extend $f$ to a continuous function $F : \mathbb R \to \mathbb R$. It follows that any ambient isotopy which takes the graph of $y=F(x)$ to the graph of $y=-F(x)$ also takes the graph of $y=f(x)$ to the graph of $y=-f(x)$. So for the case that $A$ is closed, this argument reduces the construction to the special case that $A=\mathbb R$.

The idea for constructing the ambient isotopy for $F$ is to consider each vertical line, with some fixed value of the $x$-coordinate which I'll just denote $x$, and then as time $t$ advances from $t=0$ to $t=1$ that line is translated downward by the amount that starts from $0$ and advances to $2 F(x)$.

Here is a formula for an ambient isotopy which does that: $$H_t(x,y) = \bigl(x,y - 2tF(x)\bigr) $$


Regarding your thoughts about orientations etc., they just don't seem applicable to this problem. In particular, the construction I gave works not just for $F(x)=mx$, but for $F(x)=$... well... anything continuous whatsoever.

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  • $\begingroup$ Thank you for the clarification! I believe I had this confused with the case where the graph is a unit circle and there is no ambient isotopy that sends the unit circle to its reflection across the $x$-axis. In this case, the isotopy you described would not work because $(x, \sqrt{1-x^{2}})$ must get sent to $(x, -\sqrt{1-x^{2}})$ AND $(x, -\sqrt{1-x^{2})$ must get sent to $(x, \sqrt{1-x^{2}})$, so creating an istopy that simply slides down along vertical lines won't be sufficient. $\endgroup$ Commented Aug 6 at 7:43

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