Timeline for Why, if the geometric realisation of a simplicial map $p$ is a (topological) covering map, must $p$ be a (simplicial) covering map?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 24, 2022 at 19:27 | comment | added | FShrike | By using the fibrewise property of the homeomorphism, and counting CW cells, we know $Y$ can only at most one pair of that form. So exactly one pair exists since $f$ surject. I think injectivity of $f$ follows from this. I used similar reasoning to avoid appealing to invariance of domain for other some points in your post (hopefully it makes sense) | |
| Dec 24, 2022 at 19:26 | comment | added | FShrike | Nice, thanks again. I think for injectivity you can do this: fix $v\in B_n$. For injectivity, we want the associated pullback $Y:=\Delta_n\times_{B}X$ to be isomorphic to $\Delta^n$ (over $\Delta^n$). Taking realisations we know for compactness reasons and the exactness of $|\cdot|$ that $|Y|\cong|\Delta^n|\times_{|B|}|X|$ over $|\Delta^n|$, and the fact $|f|$ is a homeomorphism tells us the latter pullback (can be taken to be) just a copy of $|\Delta^n|$, hence $|Y|\cong|\Delta^n|$ over the natural maps. If $Y$ contains more than one pair of the form $(id_n,x)$, (cont.) | |
| Dec 24, 2022 at 18:55 | comment | added | Maxime Ramzi | Yes ! that was the "by hand" argument I had in mind :) | |
| Dec 24, 2022 at 18:52 | comment | added | FShrike | Ok. Do you agree with my invariance of domain argument? | |
| Dec 24, 2022 at 18:29 | comment | added | Maxime Ramzi | No worries, I just hadn't had the time to come back to them yet :) So, for reflection of isomorphisms, let's see. If $|f|$ is surjective, then the pushout $* \coprod_{|A|}|B|$ is a point, so $|*\coprod_A B|$ is a point. It follows that $*\coprod_A B$ is a point, and therefore that $A\to B$ is surjective. For injectivity, I might have spoken too fast, and maybe you do need something like invariance of domain. | |
| Dec 24, 2022 at 16:31 | comment | added | FShrike | I apologise if you’ve felt badgered by all the comments I’ve posted and subsequently deleted. I think I understand the extra details now, so I’m just curious about a proof about reflecting isomorphisms that doesn’t use invariance of domain | |
| Dec 24, 2022 at 16:30 | vote | accept | FShrike | ||
| Dec 24, 2022 at 0:39 | comment | added | FShrike | For the reflectance of isomorphisms by $|\cdot|$, I’m not sure why exactness of $|\cdot|$ is related (it would be cool to know a slick categorical proof). I do have a thought on the lines you suggested: since the map $|f|$ is acts by identifying simplices in the colimit cocone, and $f$ maps $n$-simplices to $n$-simplices, because of invariance of domain we know the homeomorphism maps the $n$-cells (~ nondegenerate simplices) to $n$-cells so $f$ restricts to an isomorphism $E_n^{\rm{nd}}\cong X_n^{\rm{nd}}$ for each $n$. It should follow from this $f$ is a simplicial isomorphism. | |
| Dec 23, 2022 at 22:59 | comment | added | FShrike | Thank you so much, I will think on this. +1 for now, I may have to ask for some further information | |
| Dec 23, 2022 at 22:09 | history | answered | Maxime Ramzi | CC BY-SA 4.0 |