1
$\begingroup$

Let $M$ be a locally compact (Hausdorff) space, and $g:M\to M$ an isomorphism (think of an action of a finite cyclic group). By some generalities one can show that the "obvious" map $(M^g)^+\to(M^+)^g$ is continuous. Here $M^+$ is the one-point-compactification, and $M^g$ are the fixed points with the subspace topology. (One extends $g$ to a pointed map on $M^+$, that is $g(+)=+$).

Edit: if $M$ happens to be compact, $M^+=M\coprod +$ is the disjoint union with a point.

Is it true that this is an isomorphism? A (weak and/or equivariant) homotopy equivalence? Is an appropriatly modified statement true in a "convenient" category of topological spaces?

$\endgroup$
4
  • $\begingroup$ I guess that if $M^g$ is already compact then you still add a point to it in $(M^g)^+$? $\endgroup$ – Alessandro Codenotti Feb 2 at 16:00
  • $\begingroup$ Exactly! Also only then is the "obvious" map a bijection. $\endgroup$ – Leonard Feb 2 at 16:17
  • 2
    $\begingroup$ I think it's pretty straightforward. What have you tried and where are you stuck? $\endgroup$ – YCor Feb 2 at 16:48
  • $\begingroup$ Oh you are right, i feel a bit foolish now. It is a continuous bijection between compact Hausdorff spaces, thus an isomorphism. $\endgroup$ – Leonard Feb 2 at 23:31
2
$\begingroup$

$M^+$ is still Hausdorff, so also $(M^+)^g$ is. Now we observe that the "obvious" map is a continuous bijection from a compact to a Hausdorff space, thus by standard textbook contents, an isomorphism.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.