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?

  • $\begingroup$ I guess that if $M^g$ is already compact then you still add a point to it in $(M^g)^+$? $\endgroup$ Feb 2, 2021 at 16:00
  • $\begingroup$ Exactly! Also only then is the "obvious" map a bijection. $\endgroup$
    – Leonard
    Feb 2, 2021 at 16:17
  • 2
    $\begingroup$ I think it's pretty straightforward. What have you tried and where are you stuck? $\endgroup$
    – YCor
    Feb 2, 2021 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, 2021 at 23:31

1 Answer 1


$M^+$ 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.


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