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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.

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    $\begingroup$ 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 $X \rightarrow S^n$ $\endgroup$ Commented Mar 24, 2020 at 17:26
  • $\begingroup$ @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! $\endgroup$
    – robothead
    Commented Mar 25, 2020 at 0:36

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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 $f$.

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

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