# Prove that : every continuous map $f : A \to S^n$ can be extended over some neighborhood of A

That is one of my topology exercises.

I want to prove that if A is a close subset of metric space, then the continuous map can be extended over the whole space.

Maybe i can extend it over $$\mathbb{R}^{n-1}$$ first, then use the Homeomorphism between $$\mathbb{R}^{n-1}$$ and $$S^n$$ to complete it?

Any hint would be helpful to me!

Edit

I want to prove that if A is a close subset of metric space, then the continuous map can be extended over the whole space.

That should be

I want to prove that if A is a close subset of metric space, then the continuous map can be extended over the neighborhood of $$A$$.

Sorry to make you confused.

• There is no homeomorphism between $\mathbb R^{n-1}$ and $S^n$. There is a homeomorphism between $S^n - \{p\}$ and $\mathbb R^n$ however. Also it is not clear what metric space $A$ is a closed subset of. It is not true in general that if $A$ is a closed subset of a metric space $X$ then $f$ extends to a map $X \rightarrow S^n$ Commented Mar 24, 2020 at 17:26
• @Noel Lundström I'm sorry that i made a mistake on the detail part : ' the whole space ' should be ' neighborhood of A '. I will correct it. Thanks for your comment! Commented Mar 25, 2020 at 0:36

I want to prove that if A is a close subset of metric space, then the continuous map can be extended over the whole space.

This is not true. Consider $$X=\mathbb{R}^{n+1}$$, $$A=S^n$$ and $$f:A\to S^n$$ the identity and note that $$f$$ cannot be extended to whole $$X$$ because that would mean that $$S^n$$ is a retract of $$\mathbb{R}^{n+1}$$. And this cannot happen (by comparing $$n$$-th homology groups for example).

Maybe i can extend it over $$\mathbb{R}^{n-1}$$ first, then use the Homeomorphism between $$\mathbb{R}^{n-1}$$ and $$S^n$$ to complete it?

There is no such homeomorphism. Obviously dimensions don't agree. But it doesn't matter, there is no homeomorphism between $$\mathbb{R}^n$$ and $$S^m$$ regardless of $$n,m$$. You seem to think about homeomorphism between $$\mathbb{R}^n$$ and $$S^n\backslash\{p\}$$ (the stereographic projection). But I don't see how we can utilize it here.

Anyway indeed, every $$f:A\to S^n$$ can be extended to some open neighbourhood $$U$$ of $$A$$ in $$X$$ if $$X$$ is metrizable (or more generally normal). You do it as follows: first define

$$g:A\to\mathbb{R}^{n+1}$$ $$g(x)=f(x)$$

i.e. $$g$$ is the composition of $$f$$ and the inclusion $$i:S^n\to\mathbb{R}^{n+1}$$. Then we apply Tietze extension theorem (here the assumption about $$X$$ being normal is important) and obtain the extension

$$G:X\to\mathbb{R}^{n+1}$$ $$G(x)=g(x)\text{ if }x\in A$$

So how to pass from $$\mathbb{R}^{n+1}$$ back to $$S^n$$? Well, we consider $$U=G^{-1}(\mathbb{R}^{n+1}\backslash\{0\})$$, which is open in $$X$$, and the restriction

$$G':U\to\mathbb{R}^{n+1}\backslash\{0\}$$ $$G'=G_{|U}$$

The crucial observation is that $$A\subseteq U$$. We are very close now. The final step is to take the retraction

$$r:\mathbb{R}^{n+1}\backslash\{0\}\to S^n$$ $$r(v)=\frac{v}{\lVert v\rVert}$$

and define $$F:U\to S^n$$ by $$F=r\circ G'$$. Note that $$F$$ extends our initial $$f$$.

• That is very helpful! I think i made a mistake on the detail part, i will correct it. Commented Mar 25, 2020 at 0:38