1
$\begingroup$

I was trying to prove this statement, and multiple sources say that it is clear from the 'Alexander's trick'. However, in the proof that I thought of, I only used the fact that a homeomorphism $f:S^{n-1} \to S^{n-1}$ can be extended to a homeomorphism of $D^n$. Now Wikipedia mentions that some authors indeed call this the Alexander's trick, but I just wanted to make sure that my proof is correct.

What I did was to consider both my manifold $M$ and $S^3$ as $D^3 \cup_{s}D^3$, take any homeomorphism $f:S^2 \to S^2$ of the splitting surface and extend it to homeomorphisms of $D^3$ on 'both sides'. These local homeomorphisms patch up to give a homeomorphism from $S^3$ to $M$.

$\endgroup$
1
  • $\begingroup$ This proof works well. $\endgroup$ – bjn Sep 24 '17 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.