Is there any extension theorem about $n$-sphere? Let $A$ be a closed subset of $n$-sphere, $f$ a continuous surjective map from $A$ to $n$-sphere, could it be extended to the whole $n$-sphere? If not, what conditions should be added to make it true?
Since there exists surjective map from $[0,\infty)$ to $\mathbb{R}^n$, thus to $S^n$, I want to make use of it and find a proof similar with Tietze extension theorem, but I haven't thought of anything concrete.
 A: For metric spaces $X$ the following is true: suppose $f: A \rightarrow S^n$ is continuous and where $A \subseteq X$ is closed. Then there is an open subset $U$ of $X$ that contains $A$ such that $f$ can be continuously extended to $U$. This is saying that $S^n$ is an ANR (=ANE) for metric spaces. (ANR =absolute neighbourhood retract, ANE = absolute neighbourhood extender; an AE (Absolute extender) =essentially any space $X$ we can use as $[0,1]$ like in Tietze). As you cannot extend the identity embedding from $S^1 \subset B^2$ to all of $B^2$ (the non-retraction theorem), this is the best one can hope for. But the dimension $\operatorname{Ind}(B^2)= 2 > 1$ turns out to be the problem:
$X$ has large inductive dimension (see here e.g.) $\operatorname{Ind}(X) \le n$ iff for every closed subset $A \subseteq X$, every continuous $f:A \rightarrow S^n$ can be continuously extended to $X$.
(A proof of this theorem, and probably better references can be found in Jan van Mill's book, "infinite dimensional topology, introduction and prerequisites", and probably in Engelking's "theory of dimensions, finite and infinite")
This last theorem shows that this indeed holds for $X = S^n$ as asked for.
