Continuous Orbit Equivalence on Self-Similar Graph Actions

Round 1

Reviewer 1 Report

In this paper author was interested in self-similar graph actions. Author shown that isomorphic inverse semigroups of self-similar graph actions are a complete invariant of the topological orbit equivalence of inverse semigroup actions. Also author was interested in relations between group actions and inverse semigroup actions and shown that the transformation groupoid of the group action is a subgroupoid of the groupoid of germs of inverse semigroup actions.  

In my opinion, this is interesting result and it gives space for future steps. Background results and new content is clearly presented although I would like future steps to be more clearly. That aside, there are minor grammatical and spelling errors. For example in the list of keywords it is written semigroupd.

Author Response

Dear Editor and Reviewer,

Thank you for the fast reply and encouraging suggestions and comments.

The following is my reply to the Reviewer's comments.

"I would like future steps to be more clearly. "

Reply: I almost finish to write another paper expanding results of this paper.

I would be very happy if the reviewer could review the second paper.

"there are minor grammatical and spelling errors. For example in the list of keywords it is written semigroupd."

Reply: I corrected the type.

Best regards,

Inhyeop Yi.

Reviewer 2 Report

First remark: even if I found the main new claim very interesting, it took me some time to find it. As I understand it, you start from Theorem 1 (line 70), and improve it by adding a fourth equivalence, "4. The semigroups S_{G,E} and S_{H,F} are isomorphic."

(One can debate terminology; I would rather say "are conjugate", since proving that they're abstractly isomorphic is, at least in appearance, different to proving that they are isomorphic as partial permutation inverse semigroups)

Then Proposition 1 proves 4=>2, and Proposition 2 proves 1=>4.

It would definitely help to state it that way.

Now, about the maths. As I see it, the real meat is all concentrated in Lemma 2; but I can't follow its proof.

When, line 102+3, you write that f(Z(v)) is covered by finitely many cylinders, it's entirely possible to take x=x_1=emptysequence, and likewise y=y_1=emptysequence. Then (line 103-1) we get U=E^omega, but (line 105+1) we are supposed to magically get that a is a constant map on U, namely on all of E^omega! There must be a problem somewhere along the way.

As a first step, is it possible to claim that f(Z(v)) is exactly a disjoint union of finitely many cylinders, and not just contained in one? That would already point to the right place to look for a constant part of a.

Comments for author File: Comments.pdf

Author Response

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Author Response File: Author Response.pdf

Reviewer 3 Report

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Comments for author File: Comments.pdf

Author Response

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Author Response File: Author Response.pdf

Round 2

Reviewer 2 Report

I am entirely satisfied with the author's changes.