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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?

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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.

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  • $\begingroup$ 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? $\endgroup$
    – John
    Sep 30 '19 at 13:19
  • $\begingroup$ 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. $\endgroup$
    – Paul Frost
    Sep 30 '19 at 13:24
  • $\begingroup$ 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 $\phi \circ f = g$. $\endgroup$
    – John
    Sep 30 '19 at 13:35
  • $\begingroup$ 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. $\endgroup$
    – Paul Frost
    Sep 30 '19 at 13:39
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    $\begingroup$ 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). $\endgroup$
    – Paul Frost
    Sep 30 '19 at 16:15
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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.

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