Extending a continuous function from $S^1$ to $S^2$ to a continuous function from $D$ to $S^2$.

Question

Let $f\colon S^1\to S^2$ be a continuous function which is not onto. Show that $f$ extends to a continuous function $F$ from the closed unit disk $D$ in the plane to $S^2$ in the sense that the restriction of $F$ to $S^1$ is $f$.

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My attempt:

Since $\mathbb R^2$ is a metric space and $S^1$ is a closed subset of $\mathbb R^2$, we can extend $f$ to a continuous function $F$ on $\mathbb R^2\supset D$ by Tietze extension theorem. Now we need to restrict the image of $F$ in $S^2$ when the domain is $D$. And I have the intuition that we can project the image of $F$ to $S^2$ by the function $p\colon\vec v\mapsto\frac{\vec v}{|\vec v|}$. That being so, we can compose $F$ with $p$, such that $p\circ F$ satisfies our condition.

Is that correct? Thank you.

Answer 1

Here's a version that mimics your proposed solution.

1. Assume that $f$ misses the north pole, $n = (0,0,1)$.

2. Define the "stereographic projection" $$ h:S^2 \to \Bbb R^2 : p \mapsto \frac{p \cdot n}{p \cdot n -1 } n - \frac{1}{p \cdot n -1 }p $$ which projects $S^2 - \{n\}$ homeomorphically onto the plane.

3. Now let $g = h \circ f$. That's a continuous map $S^1 \to \Bbb R^2$; Tietze (following your argument) says you can extend to a map $G:D \to \Bbb R^2$.

4. Let $F:D \to S^2: u \mapsto h^{-1}(G(d))$,i.e., $F = h^{-1} \circ G$, and you're done.

Answer 2

Consider the cone on $S^1$, given by $CS^1=S^1 \times I/\sim$ where $(x,1) \sim (y,1)$. Then $CS^1 \cong D^2$. More importantly, we have that

$f:S^1 \to S^2$ restricts down to $f:S^1 \to S^2 \setminus \{pt\} \cong \mathbb R^2$. In $\mathbb R^2$, define $F:S^1 \times I\to \mathbb R^2$ by $F(x,t)=t+(1-t)f(x)$. Note that $F(x,0)=f(x)$ and $F(x,1)=0$, so by the universal property of the quotient map, we get an induced map $\tilde{F}:CS^1 \to \mathbb R^2 \cong S^2 \setminus pt \subset S^2$.

This maps extends $f$ to the entire disk.

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if you want a more "explicit" model, consider $CS^1=\{(x,1-\|x\|) \in D^2 \times \mathbb R\}$ while noting that the projection onto the first factor is a homeomorphism to the disk, and repeat the argument I had above, where you replace $S^1 \times I$ with $CS^1 \setminus \{(0,1)\} \cong S^1 \times I$.