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 $f$.
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.
Best Answer
Here's a version that mimics your proposed solution.
Assume that $f$ misses the north pole, $n = (0,0,1)$.
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.
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 $G:D \to \Bbb R^2$.
Let $F:D \to S^2: u \mapsto h^{-1}(G(d))$,i.e., $F = h^{-1} \circ G$, and you're done.