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?
Best Answer
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.