Prove $S^3$ and $\mathbb{R}^3$ are not homeomorphic.

I've encountered this question on a PhD exam in topology. This is at a level where we are expected to understand cohomology already, so there are already a lot of obvious one line proofs I could give (e.g. they don't have the same homology groups). But this appears among the "give a detailed answer" questions, as opposed to the more computational questions in the latter half. (The heading says to show all work and support all statements to the best of my ability.)

So, I am confused about the level of detail I would have to include here. Is there an obvious choice for how to prove this directly without relying on any one liners that assume higher level stuff?

I realize this question is a little opinion based, but maybe the answer will be unambiguous to those with more experience in topology. How would you answer this question?

  • $\begingroup$ You have stated the problem to be answered only in the title. Please include the problem statement in the body of the Question as well, usually with additional details (that may not be needed or entailed here). $\endgroup$
    – hardmath
    Aug 12, 2016 at 0:47
  • $\begingroup$ You get a close vote from me: you (more or less) know the answer and are concerned with how to get the exam points - that's not in scope for MSE. $\endgroup$
    – Rob Arthan
    Aug 12, 2016 at 0:56
  • 2
    $\begingroup$ Pretty annoying way to state the expectations... Is it OK to just say $S^3$ is compact? Do you need to say it's closed and bounded? (Do you need to prove it is closed and bounded?) Is that enough, or do you need to prove that closed and bounded implies compact? Where do you draw the line? If I was the grader, paf's answer would be a perfect score. $\endgroup$
    – user325968
    Aug 12, 2016 at 0:57
  • $\begingroup$ I agree @mathguy, and if you'll excuse the whine, the main reason I have trouble studying for this exam is that many of the other questions are stated with this sort of ambiguity. $\endgroup$ Aug 12, 2016 at 1:35
  • $\begingroup$ @RobArthan That is not true, and I resent the implication that I am grade grabbing. I am asking whether those more experienced than me understand (what I perceive to be) an ambiguously worded question. Even setting aside from the material content of the question, interpretation of mathematical writing is well within the scope of MSE. $\endgroup$ Aug 12, 2016 at 1:39

2 Answers 2


$S^3$ is compact whereas $\Bbb R^3$ is not. Thus they cannot be isomorphic.


$S^3$ is compact, while $\mathbb{R}^3$ is not. Since any continuous function $f:S^3\rightarrow \mathbb{R}^3$ maps compact subsets of $S^3$ to compact subsets of $\mathbb{R}^3$, it can't be surjective (or else $f(S^3)=\mathbb{R}^3$ is also compact).

  • $\begingroup$ Aren't you assuming the conclusion there? $\endgroup$ Aug 12, 2016 at 3:24
  • $\begingroup$ No. I am just using that homeomorphism preserves the topological properties $\endgroup$ Aug 13, 2016 at 5:11
  • $\begingroup$ You want to prove that S^3 is compact and R^3 is not compact. Then you claim that a continuous map can't be surjective, because otherwise R^3 would be compact. But that argument relies on both the assumption that S^3 is compact and R^3 isnt, which is what you sought to prove $\endgroup$ Aug 13, 2016 at 5:26
  • $\begingroup$ No, I don't want to prove that $S^3$ is compact and $\mathbb{R}^3$. It is a known fact. If you want to prove this just use closed and bounded subset of $\mathbb{R}^3$ is compact. $\endgroup$ Aug 13, 2016 at 5:48
  • $\begingroup$ Then I don't see how your second sentence is relevant... Weren't you done in the first? $\endgroup$ Aug 17, 2016 at 4:03

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