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For a given topological space $X$ (one can assume a simplicial complex if required), define it's homological dimension $\operatorname{hdim}(X)$ as the largest integer $n$ such that $H_n(X,A)\ne 0$ for some closed subspace $A$ of $X$.

I have been stuck on the following problem for a while now. It is clear that if $X$ and $Y$ are homeomorphic then their homological dimensions are the same. But what happens in the following case:

Suppose there exists a surjective, local homeomorphism $f:X\to Y$. Then can we say that $\operatorname{hdim}(X)=\operatorname{hdim}(Y)$ (or at least $\operatorname{hdim}(X)\ge \operatorname{hdim}(Y)$)?

I am kind of at a lost what to even attempt here. I can intuitively understand that since locally $X$ and $Y$ are homeomorphic, their dimensions are the same locally. My naive attempt was to consider the closed subspace $A_Y$ we get for $Y$ and pull it back to $X$ getting $f^{-1}(A_Y)$ which is also closed. But then I am not sure how to proceed.

Any help is appreciated!

Rainy
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    I don't know anything about relative homology but I can tell you that this statement is badly false if you replace relative homology with ordinary homology. In fact you can find covering maps $f : X \to Y$ where $X$ is contractible but $Y$ has infinite homological dimension, for example by taking $X = S^{\infty}, Y = \mathbb{RP}^{\infty}$. – Qiaochu Yuan Jul 04 '24 at 10:02
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    For reasonably nice spaces (say, Hausdorff, paracompact) the answer is positive (at least, if one works with cohomological dimension and Chech cohomology; I would have to think about homological dimension). For simplicial complexes, all notions of dimension are equivalent and dimension equals the supremum of dimensions of simplices. – Moishe Kohan Jul 04 '24 at 16:02
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    @QiaochuYuan: Yes, our ancestors defined (co)homological dimension using relative/sheaf (co)homology for a reason, since they wanted dimension of $\mathbb R^n$ to be $n$, not $0$. – Moishe Kohan Jul 04 '24 at 16:05
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    This is a fairly interesting notion of dimension, and as Moishe mentions, it is true for cohomological dimension of nice spaces. Do you think there is some more structure on the space $X$ for your question? Also separately, it might be interesting to look at the relation between this notion of dimension and cohomological dimension as a separate question. +1 – HackR Jul 07 '24 at 15:31
  • If local homological dimensions coincide with homological dimensions of $X$ and $Y$, they are equal. – Bob Dobbs Jul 14 '24 at 13:27
  • @BobDobbs What do you mean by "local homological dimensions?" Are you saying that if $f\colon X \to Y$ is a surjective local homeomorphism then $\max {n \mid H_n(X) \neq 0} = \max {n \mid H_n(Y) \neq 0}$? ...because that is false. – Ben Steffan Jul 14 '24 at 13:36
  • https://encyclopediaofmath.org/wiki/Local_homology @BenSteffan There is a paragraph starting with homological dimension... – Bob Dobbs Jul 14 '24 at 17:14
  • @MoisheKohan Do you there use the following definition (for locally compact Hausdorff spaces) $\dim X$ is the least integer $n$ such that $\mathscr{H}^{n+1}_C(X;\mathscr{F})=0$ for all Abelian sheaves $\mathscr{F}$ on $X$? – FShrike Aug 13 '24 at 22:02
  • @FShrike: Of course. – Moishe Kohan Aug 14 '24 at 01:56
  • @MoisheKohan In which case, using Cech cohomology and looking the cofinal family of covers of $X$ by open sets homeomorphic with opens in $Y$, I think I can show $\mathscr{H}^\bullet(Y;\mathscr{F})\to\mathscr{H}^\bullet(X;f^\ast\mathscr{F})$ is always injective, which would imply the desired inequality. Is that what you figured as well? – FShrike Aug 14 '24 at 10:25

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