$\mathbb{R}^n$ contains no subspace homeomorphic to $S^n$ (Borsuk-Ulam)

I want to prove that $\mathbb{R}^n$ contains no subspace homeomorphic to $S^n$, as per the wikipedia article

I'm trying to set up a function $g:S^n \to \mathbb{R}^n$ that would be injective, which cannot occur by Borsuk-Ulam.

If I have a subspace, $A \subset \mathbb{R}^n$, homeomorphic to $S^n$ then I can have a map $h:S^n \to A$. Can I compose this with the inclusion $i:\hookrightarrow \mathbb{R}^n$ to give an (injective) map $S^n \to \mathbb{R}^n$, and am done?

(As an aside, I am sure there are other point-set topology ways to prove this problem, maybe using invariance of domain)

Edit: Since the above works, to make this a little more interesting - can anyone provide any other proofs without using algebraic topology? (Although this is probably debatable, because some other methods to use it are probably based on algebraic topology)

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Yes. ${}{}{}{}$ – Jonas Meyer Jun 1 '11 at 4:15
@Jonas - ok, thanks. I guess I'll keep the question here in case anyone else ever needs it – Juan S Jun 1 '11 at 4:17
@Qwirk: OK, but you shouldn't take my word for it!  I'm honestly not sure what part you have a question about, but you outlined how the existence of $A$ would imply the existence of an injective continuous map $S^n\to\mathbb{R}^n$, which is impossible by Borsuk-Ulam. – Jonas Meyer Jun 1 '11 at 4:20
@Jonas - I guess there is no question! Sometimes I have trouble working out if what I did is correct (a lack of faith perhaps?). Also writing it out properly (not just my scribbles on a pad) probably made me write it out better – Juan S Jun 1 '11 at 4:27
@Jonas - well I edited it, since it was, as stands, not a very interesting question! I would be interested to see other ways to prove this then – Juan S Jun 1 '11 at 4:30
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2 Answers

Another way to see this is to use dimension theory (part of general topology, see e.g. Engelking's book "Theory of dimensions, finite and infinite", or others). One important theorem is that (for separable metrisable spaces, say, where the theory is nice and well-understood) $\dim(X) \le n$ is equivalent to "every continuous function $f$ from a closed subset $A$ of $X$ to $\mathbb{S}^n$ can be continuously extended over $X$". So Tietze's extension theorem with the $n$-sphere as the image.

This theorem follows from Brouwer's fixed point theorem (eventually) and can be used to give a nice proof of the invariance of domain, that you already mentioned. In fact, Borsuk-Ulam, the sphere-extension theorem above, invariance of domain, the no retract theorem ($\mathbb{S}^n$ is not retract of $\mathbb{B}^{n+1}$) and Brouwer's fixed point theorem are all closely related, and having one we can relatively easily prove the others.

If now a subset $A$ of $\mathbb{R}^n$ were homeomorphic to $\mathbb{S}^n$, then by compactness of the latter, $A$ would be closed in $\mathbb{R}^n$, which has dimension $n$, so we would have an extension of the homeomorphism to a continuous function, and this means we have a retraction (composing with the inverse of the homeomorphism) of $\mathbb{R}^n$ onto $A$, which is supposedly homeomorphic to $\mathbb{S}^n$, and this cannot be.

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Thank you for this very nice answer – Juan S Jun 1 '11 at 12:04

When you compose $h$ with the inclusion, you're using the fact that the inclusion is continuous due to the definition of the subspace topology: The preimage of any open set $U$ of $\mathbb R^n$ under the inclusion is the intersection of $U$ with $A$, and that's by definition an open set of the subspace. You could also use this to show directly (i. e. without appealing to the fact that composition preserves continuity) that $h$ considered as a map to $\mathbb R^n$ is continuous, since the preimage of any open set $U$ of $\mathbb R^n$ under $h$ is the preimage of its intersection with $A$, which is open.

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