The cohomological-dimension tag has no usage guidance.

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### Does the singular cohomology for a metric space of finite topological dimension vanish in high dimensions?

It is known that by applying the universal coefficient theorem, the singular cohomology of closed manifold with coefficient $\mathbb{Z}_2$ vanishes in high dimensions. But for a metric space $M$ with ...

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### what's the cohomological dimension of a Stein space?

I want to know the "cohomological dimension" of a Stein manifold.
I know that:
for $X$ differential manifold and for every sheaf $F$ of abelian
groups, I have $H_c^j(X,F)=H^j(X,F)=0$ for ...

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### What is the intuition for $\mathbb{Q}^{ab}$ having cohomological dimension $1$?

I frequently talk to people who think of finite fields as arithmetic analogs of punctured discs. This makes some sense: the absolute Galois group of a finite field is the profinite completion of ...

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### Continued fractions and projective resolutions

Hello,
This question might be vague and not thought-through enough.
If we have a real positive number $x$, we can start to write it as a continued fraction:
$x = a_0 + \frac{1}{x_1} , \ldots , ...

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### cohomological dimension for coarser/finer topologies

Given a sheaf $\mathcal{F}$ with respect to some Grothendieck topology, is the cohomological dimension for this sheaf less than or equal to the cohomological dimension of a finer topology?
Example: ...

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### Cohomological dimension of $\mathcal{B}_n$

What is the cohomological dimension of the braid group $\mathcal{B}_n$ on n-strands ?
A reference would be appreciated.

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### cohomological dimension of the push-forward functor

Let $\ell$ be a prime number and let $f:X \to Y$ be a morphism of schemes of finite type over the complex numbers (or a regular scheme of dimension at most 1, in which $\ell$ is invertible). How to ...

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### Do the homological dimension and cohomological dimension for a group agree?

Or equivalently, if $G$ is a group, do the projective and injective dimension of $Z$ (viewed as a $ZG$-module) agree?
Thanks!