# Ambient isotopy of the plane carrying unit circle minus North Pole onto the $x$-axis

On the plane $$\mathbb{R}^{2}$$ consider the unit circle $$S^{1}$$ and let $$N$$ denote the North Pole $$(0,1)$$. The stereographic projection is a homeomorphism of $$S^{1}-N$$ onto the $$x$$-axis. However, is there an ambient homeomorphism of $$\mathbb{R}^2$$ onto itself which carries $$S^{1}-N$$ onto the $$x$$-axis?

Are $$S^{1}-N$$ and the $$x$$-axis ambient-isotopic in $$\mathbb{R}^{2}$$? How to construct such map?

It is impossible. For any map $$h : \mathbb R^2 \to \mathbb R^2$$ the set $$h(S^1 \setminus \{ N\})$$ is contained in $$h(S^1)$$ which is compact and thus cannot contain the $$x$$-axis.

Edited :

What you mean seems to be this: Let $$s : S^1 \setminus \{ N\} \to \mathbb R$$ be stereographic projection. It is a homeomorphism and $$\lvert s(p) \rvert \to \infty$$ as $$p \to N$$. You have two embeddings $$i, f : S^1 \setminus \{ N\} \to \mathbb R^2$$: One is $$i(p) = p$$, the other is $$f(p) = (s(p),0)$$.

Then the map $$E : (S^1 \setminus \{ N\})\times [0,1] \to \mathbb R^2, E(p,t) = (1-t)p + tf(p)$$ is an isotopy from $$i$$ to $$f$$. To see this, it remains to show that $$E_t = E(-_,t) : S^1 \setminus \{ N\} \to \mathbb R^2$$ is an embedding for all $$0 < t < 1$$. Clearly each $$E_t$$ is injective: If $$E(x_1,x_2,t) = E(x'_1,x'_2,t)$$, then by considering the second coordinate of $$E_t$$ we see that $$(1-t)x_2 = (1-t)x'_2$$, i.e. $$x_2 = x'_2$$, and can then conclude by considering the first coordinate of $$E_t$$ that $$x_1 = x'_1$$. Let us next show that $$E_t$$ is a closed map for $$0 < t < 1$$ (this will imply that $$E_t$$ is an embedding). So let $$A \subset S^1 \setminus \{ N\}$$ be closed in the subspace topology. Let $$(q_n)$$ be a sequence in $$E_t(A)$$ converging to some $$q \in \mathbb R^2$$. We habe to show that $$q \in E_t(A)$$. Let $$p_n = E_t^{-1}(q_n) \in A$$. $$(p_n)$$ has a subsequence converging to some $$p \in S^1$$. W.l.o.g. we may assume that $$p_n \to p$$. If $$p \ne N$$, then necessarily $$p \in A$$ since $$A$$ is closed in $$S^1 \setminus \{ N\}$$. Hence $$q_n = E_t(p_n) \to E_t(p)$$ and we conclude $$q = E_t(p) \in E_t(A)$$. Now assume that $$p = N$$. Then $$y_n = E_t(p_n)$$ is unbounded since $$(1-t)p_n \to (1-t)N$$ and $$\lVert tf(p_n) \rVert = t\lvert s(p_n) \rvert \to \infty$$. Hence $$(q_n)$$ does not converge which is a contradiction. Therefore $$p = N$$ cannot occur.

• You are correct, but if I get a piece of string and place it straight on a table, I can bend it into the shape of a circle (minus a point). The process can be reversed. Is not this a description of an ambient isotopy of the plane? Is it because the piece of string is compact why this works?
– John
Sep 30 '19 at 13:19
• What you mean is something else: It is not an ambient isotopy which deforms the whole space $\mathbb R^2$, but an isotopy of two embeddings $e_i : \mathbb R \to \mathbb R^2$ which is map $E : \mathbb R \times [0,1] \to \mathbb R^2$ such that $E(x,i) = e_i(x)$ and all $E(-,t)$ being embeddings. Such an isotopy exists. Sep 30 '19 at 13:24
• I see, is the notion of isotopy of two embeddings the same as the notion of two embeddings being equivalent? By equivalent I mean the following: two embeddings $f$,$g: X\rightarrow Y$ are equivalent if there exists a (topological) homeomorphism $\phi$ of $Y$ onto itself such that $\phi \circ f = g$.
– John
Sep 30 '19 at 13:35
• No, equivalence as you define is much weaker than isotopy. As a trivial example consider $X= \{0\}, Y = \{0,1\}$ and $f(0) = 0, g(0) = 1$. $f$ and $g$ are equivalent, but not even homotopic. Sep 30 '19 at 13:39
• We have ambient isotopy $\Rightarrow$ isotopy and ambient isotopy $\Rightarrow$ equivalence. But isotopy does not imply equivalence. See my above edit. A reference is Rushing, T. Benny. Topological embeddings. Vol. 52. Academic Press, 1973. However, its focus are very special classes of spaces (polyhedra, manifolds). Sep 30 '19 at 16:15

There is no ambient homeomorphism of $$\mathbb{R}^{2}$$ onto itself which carries $$S^{1}−N$$ onto the $$x$$-axis. If such an ambient homeomorphism existed then the complements of $$S^{1}-N$$ and the $$x$$-axis in $$\mathbb{R}^{2}$$ would be homeomorphic. However, this is impossible as the former space is connected while the latter is not.