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My apologies for this rather elementary question, but here goes: I didn't have any trouble figuring out how to prove that the 'standard' homotopy/isotopy definitions give rise to an equivalence relation, but I've been struggling trying doing the same for ambient isotopy.

I'll include the definition I'm using for good measure:

Two kots $K_1$ and $K_2$ are ambient isotopic iff there exists a continuous map $H: \mathbb{R}^3\times[0,1]\to\mathbb{R}^3: (x,t)\mapsto h_t(x)$ such that $H(x,0)=x$ for all $x\in \mathbb{R}^3$, $h_1(K_1)=K_2$ and every $h_t$ is a homeomorphism.

To prove that this is symmetric, my first instinct was to use the 'reverse time'-trick, but I don't see how that works because $H_1 \neq \operatorname{Id}$, right? The next thing I thought of was just taking the inverse function of every $h_t$, but I'm having trouble showing that $(x,t)\mapsto h_t^{-1}(x)$ is continuous.

I'd be happy if someone could provide me with a little guidance.

mval
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  • You want to use the inverse function. And the continuity works for the same reason that the non-inverse works. – N. Owad Feb 18 '14 at 16:35

2 Answers2

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As I pointed out in the comments, the answer of Mark B is incorrect. Here there is a valid proof, but first I need to clarify the following

Definition: Let $K_1$ and $K_2$ be subsets of $\mathbb{R}^3$. We say that $K_1$ is ambient isotopic to $K_2$ (denote $K_1\sim K_2$) if there is a continuous function $H:\mathbb{R}^3\times [0,1]\to \mathbb{R}^3$ such that the functions of the form $H_t:\mathbb{R}^3\to \mathbb{R}^3$, $H_t(y):=H(y,t)$ for $t\in[0,1]$, satisfy the following:

  1. $H_t$ is a homeomorphism for all $t\in[0,1]$.
  2. $H_0=id_{\mathbb{R}^3}$ and $H_1(K_1)=K_2$.

The map $H$ is called an ambient isotopy from $K_1$ to $K_2$.

Proof of symmetry of $\sim$: Suppose that $K_1\sim K_2$ and let $H$ an ambient isotopy from $K_1$ to $K_2$. Consider the function $G:\mathbb{R}^3\times [0,1]\to \mathbb{R}^3$, $G(x,t)=H_t^{-1}(x)$. We can check directly that $G$ satisfies the conditions 1 and 2 of an ambient isotopy from $K_2$ to $K_1$. To prove that $K_2\sim K_1$ it is enough to prove that $G$ is continuous. We will do it using the Invariance of Domain Theorem.

Consider the function $Z:\mathbb{R}^4\to \mathbb{R}^4$, defined by

\begin{equation*} Z(x,t)= \begin{cases} \displaystyle (x,t)& \mbox{ if }(x,t)\in\mathbb{R}^3\times(-\infty,0]\\ (H_t(x),t) &\mbox{ if }(x,t)\in\mathbb{R}^3\times[0,1]\\ \displaystyle (H_1(x),t) &\mbox{ if }(x,t)\in\mathbb{R}^3\times[1,+\infty). \end{cases} \end{equation*} Using the Pasting Lemma we can prove that $Z$ is continuous. Also we can prove the injectivity of $Z$ without tricks. Now because of the Invariance of Domain Theorem, we conclude that $Z$ is an open mapping. Therefore the restriction $\hat{H}:\mathbb{R}^3\times [0,1]\to \mathbb{R}^3\times[0,1]$, $\hat{H}(x,t)=(H_t(x),t)$ is continuous, open and injective. Notice that $\hat{H}$ is also onto. Thus its inverse, the function $\hat{G}:\mathbb{R}^3\times [0,1]\to \mathbb{R}^3\times[0,1]$, $\hat{G}(x,t)=(H_t^{-1}(x),t)$, is continuous.

Hence our function $G=\pi_1\circ\hat{G}$ is composition of continuous functions, and therefore continuous as well, where $\pi_1:\mathbb{R}^3\times [0,1]\to \mathbb{R}^3$ is defined by $\pi_1(x,t)=x$.

Notice that this proof is valid for $\mathbb{R}^n$ in general.

Chilote
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1

To show that the ambient isotopy relation (which I'm going to denote by $\sim$) is symmetric, you want to show that if $K_1 \sim K_2$, then $K_2 \sim K_1$. Start by assuming that $K_1 \sim K_2$. That is, there exists a continuous map $H: \mathbb{R}^3 \times [0,1] \to \mathbb{R}^3$ satisfying the required conditions.

To show that $K_2 \sim K_1$, we need to find a continuous map $G:\mathbb{R}^3 \times[0,1] \to \mathbb{R}^3$ which also satisfies the required conditions. You are correct that our map is given by $$G(x,t) = g_t(x) = h_t^{-1}(x)$$ It's clear that $G(x,0) = x$ and $g_1(K_2) = K_1$. It's also clear that each $g_t$ is a homeomorphism, so we only need to show that G is continuous.

Definition: A function is continuous iff the pre-image of every open set is open.

Proof that G is continuous: Let $U$ be an open set in $\mathbb{R}^3$. The pre-image under $G$ is $$G^{-1}[U] = \{\, (x,t) \in \mathbb{R}^3 \times [0,1] \,\,\, | \,\,\, G(x,t) \in U\,\}$$ That is, $$G^{-1}[U] = \{\, (x,t) \,\,\, | \,\,\, g_t(x) \in U\,\} $$ $$ = \bigcup_{t \in [0,1]} \{ \, x \in \mathbb{R}^3 \,\,\,| \,\,\, g_t(x) \in U \}$$ $$= \bigcup_{t \in [0,1]} g_t^{-1}[U]$$ But for homeomorphisms, the pre-image is the same as the image of the inverse function. That is, $g_t^{-1}[U] = \{g_t^{-1}(x) \,|\, x \in U\} = \{h_t(x) \,|\, x \in U\} = h_t(U)$. We finally get $$G^{-1}[U] = \bigcup_{t \in [0,1]} h_t(U)$$.

Each $h_t$ is a homeomorphism, and so maps open sets to open sets. It follows that $G^{-1}[U]$ is the union of open sets and is therefore open.

Mark B
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