# Fixed points of one-point-compactification

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?

• I guess that if $M^g$ is already compact then you still add a point to it in $(M^g)^+$? Feb 2, 2021 at 16:00
• Exactly! Also only then is the "obvious" map a bijection. Feb 2, 2021 at 16:17
• I think it's pretty straightforward. What have you tried and where are you stuck?
– YCor
Feb 2, 2021 at 16:48
• Oh you are right, i feel a bit foolish now. It is a continuous bijection between compact Hausdorff spaces, thus an isomorphism. Feb 2, 2021 at 23:31

$$M^+$$ 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.