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Consider the map $C^{0}\left(\mathbb{R}^{2} \setminus\{(0,0)\}\right) \rightarrow C^{0}\left(S^{1}\right):\left.f \mapsto f\right|_{S^{1}}$. I am trying to prove this map is surjective. I know I need to show for every continuous function on $S^1$ there is a continuous function on $\mathbb{R}^{2} \setminus\{(0,0)\}$, however, I am having trouble on constructing a formal demonstration.

user432987423
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  • I’m a bit confused, the restriction of a generic mapping $f$ to $\Bbb S^1$ may take values outside the unit circle. For instance the mapping $(x,y)\mapsto(2x,2y)$. – InsideOut Sep 28 '21 at 20:09
  • @InsideOut It's conventional at least in functional analysis that, when writing $C^0(X)$ for some topological space $X$, the codomain of the functions in $C^0(X)$ are assumed to be the scalar field ($\Bbb{R}$ or $\Bbb{C}$ typically). In this case, $\Bbb{R}$ seems the natural choice. – Theo Bendit Sep 28 '21 at 20:14
  • In the future, try to give at least some of the work you've done on a problem in your writeup. "Having trouble" is really broad. – user3716267 Sep 28 '21 at 20:16
  • I would edit this posting to read as follows: $${}$$Consider the map $C^0\left(\mathbb{R}^{2} \setminus{(0,0)}\right) \rightarrow C^0(S^1):\left.f \mapsto f\right|_{S^1}$. I am trying to prove this map is surjective. I know I need to show for every continuous function on $S^1$ there is a continuous function on $\mathbb{R}^{2} \setminus {(0,0)}$, however, I am having trouble on constructing a formal demonstration. – Michael Hardy Sep 28 '21 at 20:17
  • Here$\quad\uparrow\quad$you see the difference between \setminus and \backslash. The former has horizontal spacing appropriate to a binary operation symbol. However, the moderators of this site prefer to have people use MathJax and LaTeX badly, so I won't edit the question. – Michael Hardy Sep 28 '21 at 20:19
  • Here is a dual question about radial extension, possibly from a classmate, asking about some intuition, if you're interested. – Theo Bendit Sep 30 '21 at 22:49

2 Answers2

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Let $f$ be a continuous function on $S^1$. We extend this radially to a function on $\mathbb{R}^2 \setminus (0, 0)$, such that along each radius our function is constant (note that the removal of the origin allows such a construction to remain well-defined).

It remains to show that this function is continuous on $\mathbb{R}^2 \setminus (0, 0)$, but this is not very hard.

user3716267
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I think the following works: Take some $g \in C^0(S^1)$. Consider $\mathbb{R}^2\setminus \{(0,0)\}$ as $\mathbb{C}\setminus \{(0,0)\}$. Elements in this set look like $z=re^{it}$ with $r>0$ and $t \in [0,2\pi)$. Then, define $$ f(z) = f(re^{it}) := g(e^{it}), $$ i.e. $f$ is constant along the ray with angle $t$ with respect to the $x$-axis. This works, because we exclude the origin $(0,0)$. Then we see that $f|_{S^1}=g$, as desired.

I leave it to you to fill in the details. I hope this helped!

jasnee
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