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From a previous post, the following fact has been brought to my attention: For any two continuous functions $f, g: \mathbb{R} \rightarrow \mathbb{R}$, the following function is an isotopy of $\mathbb{R}^{2}$ that takes the graph of $y = f(x)$ to the graph of $y = g(x)$:

$$H(x, y, t) = (x, t(g(x) - f(x)) + y)$$

Specifically, $$H(x, f(x), 0) = (x, f(x))$$

And

$$H(x, f(x), 1) = (x, g(x))$$

However, the story becomes very different if $f(x)$ and $g(x)$ are to be switched, i.e. if it is necessary for both $H(x, f(x), 1) = g(x)$ and $H(x, g(x), 1) = f(x)$, then I believe an isotopy should not exist in general. However, I am a little confused on how to prove this, even for the case where $g(x) = - f(x)$ (a reflection about the $y$-axis). One idea would be to specify a set of vertical line segments between the graph of $y = f(x)$ and the graph of $y = -f(x)$ and consider their image under the isotopy. If this image results two points that have same image for some $t$-value, then the homotopy will fail to be an isotopy for this $t$-value and the contradiction will be complete. However, I really do not know how to fill in the details. Any assistance with this would be greatly appreciated!

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The answer is that there is no such isotopy for each $f\neq g$. The problem changes if one asks only for isotopy between the supports of the graphs without asking for the final $x$-coordinates to correspond. See comments and other answers for this. In fact there is no reason to suspect such a property to be determined just by their intersection points. What one can prove is however that when $f, g$ do not touch then there is such a "weaker" isotopy. Anyway, coming back to OP problem.

Let us consider arbitrary $f, g$ and let $[a, b]\subset \{x| f(x)\neq g(x)\}$ with $a\neq b$ and assume $f>g$ here. Let us call $R$ the path connected component of $x=((a+b)/2, (f((a+b)/2)+g((a+b)/2))/2)$ in the complement of the union of the graphs.

Now consider the loop $\gamma: (a, f(a))\stackrel{f}{\rightarrow} (b, f(b))\stackrel{\text{down}}{\rightarrow} (b, g(b))\stackrel{g}{\rightarrow}(a, g(a))\stackrel{\text{up}}{\rightarrow}(a, f(a))$. This is a simple closed curve and it has winds just one time clockwise around $x$. This is easily seen to be the case because it is homotopic to a rectangle by simply projecting the graph of $f$ and of $g$ on sufficiently close but disjoint horizontal lines near $x$. Suppose a wanted isotopy $H((\bullet, \bullet), t)$ swapping "vertically" the graphs of $f, g$ exists.

Consider the image $H(\gamma, t)$ of this loop as an element of the first homology group of the region $\mathbb{R}^2\setminus H(x, t)$.This is always well defined because $H((\bullet, \bullet), t)$ is bijective for each $t$.

Now compare the loop $\gamma$ and $H(\gamma, 1)+x-H(x, 1)$.

These two loops are homotopic (not necessarily in a pointed sense) as loops in $\mathbb{R}^2\setminus x$ (via $H$, or to be more precise via a translated version of $H$ by the vector $x-H(x, t)$).

However, it can be seen that the the loop $H(\gamma, 1)+x-H(x, 1)$ rotates counterclockwise around $x$.

One way to see this is by using Green's Theorem on an appropriate "winding function around $x$". and realizing that the integrals of the arcs on the graphs of $f, g$ get their $y-$position relative to $x$ swapped.

Anyway, the winding number represents one of the $3$ classes that a simple loop can have in the real homology of a punctured plane. In particular this new loop is not homologous to the starting $\gamma$ and so is not even free homotopic to $\gamma$, which gives us the desired contradiction (again, if $H$ existed it would have witnessed such a free homotopy).

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  • $\begingroup$ Where do you see that the homotopy must preserve $x$-coordinate? It is easy to swap $f$ and $g$ if their graphs have no common point. $\endgroup$ Commented Aug 29 at 15:25
  • $\begingroup$ How do you do it? In the problem it seems to me that the OP is asking $H(x, f(x), 1)=(x, g(x))$ and $H(x, g(x), 1)=f(X)$. So that at the end one obtains swapped graph only vertically. Am I misintepreting the problem? @ChristopheBoilley $\endgroup$ Commented Aug 29 at 15:31
  • $\begingroup$ Take e.g. $f(x)=1$ and $g(x)=-1$ for all $x\in\mathbb R$, and just rotate the whole plane around $(0,0)$ until both graphs are swapped globally (but individual points are moved along $x$-axis). $\endgroup$ Commented Aug 29 at 15:50
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    $\begingroup$ You are right. I acknowledge I built an answer to a more general question. $\endgroup$ Commented Aug 29 at 15:54
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    $\begingroup$ The winding number of $H(\gamma, 1)$ about $x$ is 1. Since $H(\gamma, 0) = \gamma$ by assumption and $H(*, *, 1)$ is a supposed to be a homeomorphism, this will lead to a contradiction because it will take a curve with winding number -1 to a curve with winding number 1. $\endgroup$ Commented Sep 8 at 16:11
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Let $f(x)=x^3-x$ and $g(x)=-f(x)$. Then $f$ goes above $g$ at the points $-1$ and $1$, but goes below only at $0$. You can see the graph of these functions between $-1$ and $1$ as a multigraph with three vertices.

If you had an isotopy switching the graphs of $f$ and $g$, then it would map the multigraph onto itself, and especially it would fix $(0,0)$. But it would swap the other vertices and keep upside and downside, or the other way around, therefore it would be orientation reversing on $(0,0)$, and that is not possible for an isotopy of $\mathbb R^2$.

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  • $\begingroup$ I do not understand. Why should it a priori be orientation reversing? The isotopy can just "open" the acute angle in $(0, 0)$ till it is obtuse and then rotate sufficiently so as to obtain the effect of swapping $\endgroup$ Commented Aug 29 at 15:39
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    $\begingroup$ No, because the edges of this acute angle must join again on the next vertex. Your idea works when there is only one common point (e.g. with two lines) but not with 3. $\endgroup$ Commented Aug 29 at 15:52

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