This should be really well known but I don't seem to find a statement about it nor a question in MO answering this.

Consider a Compact Hausdorff topological space $X$. The cohomological dimension of $X$ is the first natural number where the cohomology vanishes. The topological dimension is as defined in wikipedia (see also Hurewicz Wallman's book).

For CW complexes, it seems easy to prove that finiteness of topological dimension implies finiteness of the other. However, for general spaces I don´t figure out how to prove (or construct a counterexample) to

Q) Does finite topological dimension of $X$ implies finite cohomological dimension?

Maybe there are well known inequalities also (for example, it seems that for spaces homotopy equivalent to CW complexes one should have that the topological dimension is bigger than or equal to the cohomological one). Information about this kind of inequalities would be welcome.

  • 6
    $\begingroup$ Engelking's "General Topology" (or something like that) has an exhaustive discussion of topological dimension theory, and in particular proves for separable metrizable spaces that covering dimension coincides with the large & small inductive dimensions. The covering dimension clearly is an upper bound on non-vanishing for Cech cohomology, and on paracompact Hausdorff spaces Cech cohomology coincides with sheaf cohomology. So on those spaces, finite covering dimension implies finite cohomological dimension. And likewise on separable metrizable spaces the question has an affirmative answer. $\endgroup$
    – BCnrd
    Oct 24, 2010 at 17:45

3 Answers 3


Well, I think it depends on which dimension you mean and which cohomology. The best fit I think is covering dimension and Čech cohomology. The Čech cohomological dimension is indeed bounded (more or less by definition) by the covering dimension.

Addendum: Just to comment on BCnrd's comment. The usual definition of cohomological dimension is the vanishing of all sheaf cohomology above that dimension. I however got the impression that the OP interpreted it in the sense of all cohomology with constant coefficients vanishing. In any case the Čech cohomology of sheaves also vanish above the covering dimension so for that it doesn't matter what one means, unless sheaf cohomology is meant. However, a Hausdorff space (I am asuming that that is part of the OP's assumption) that either has finite covering dimension or is compact is paracompact and hence it doesn't matter after all.


First of all, "cohomological dimension" means two different things. The cohomological dimension of $X$ could be the largest $n$ such that $H^n(X)$ does not vanish. This definition is appropriate for defining the cohomological dimension of a group $G$ to be the cohomological dimension of its classifying space $BG$. It could also mean the supremum of $n$ for all closed subsets $C \subseteq X$ such that $H^n(X,C) \ne 0$. This second definition is more appropriate for geometry. By this second definition, an $n$-ball is $n$-dimensional, because you can let $C$ be its boundary. In this second, more geometric approach, it's standard and better to use Čech cohomology. So I should really write $\check{H}^n(X,C)$, but I won't bother. Also, you can ask for the cohomological dimension with various coefficients, and you might get different answers.

There is a very interesting review by Dranishnikov of cohomological dimension theory of compact metric spaces. It includes his famous construction of a compact metric space whose covering dimension is $\infty$ and whose cohomological dimension (over $\mathbb{Z}$) is 3.

Theorem: (elementary) If $X$ is any topological space and $A$ is an abelian group, then $\dim_A X \le \dim X$, where the right side is the covering dimension.

As Torsten Ekedahl points out, this inequality follows quickly from the definition of Čech cohomology and the fairly similar definition of covering dimension. In both cases you use the nerve of an open covering. In Čech cohomology, you take the limit of the cohomology of the nerve. In covering dimension, you take the lim inf of the dimension of the nerve, QED.

Theorem: (elementary) (1) $\dim_A X \le \dim_{\mathbb{Z}} X$; (2) For every $A$, $\dim_A X = 0$ iff $\dim X = 0$; (3) $\dim_\mathbb{Z} X = 1$ iff $\dim X = 1$; (4) For every $A$ and every $n$-dimensional compact simplicial complex $K$, $\dim_A K = n$.

I would guess that (4) also applies to compact CW complexes.

Theorem: (Alexandroff) If $X$ is a compact metric space and if its covering dimension $\dim X$ is finite, then $\dim X = \dim_{\mathbb{Z}} X$.

Theorem: (Pontryagin) For each prime $p$, there is a "surface" $X$ whose $\mathbb{Z}/p$-dimension is 2, and whose $\mathbb{Z}/q$-dimension is 1 for any prime $q \ne p$.

In light of Alexandroff's theorem, if you want cohomological dimension and covering dimension to differ for a compact metric space, the covering dimension has to be infinite. Whether this was possible was a long-standing open problem; Dranishnikov found the first example. He also describes an example of Dydak and Walsh of a compact metric space $X$ such that $\dim_{\mathbb{Z}} X = 2$ but $\dim_{\mathbb{Z}} X \times X = 3$.

  • $\begingroup$ Wow thanks! I need to catch up with quite lot of stuff from here. Thanks for the very complete answer, it helps on where to look at. $\endgroup$
    – rpotrie
    Oct 24, 2010 at 22:01
  • $\begingroup$ Do you know by a chance if this result is known mathoverflow.net/q/391839/121665 ? $\endgroup$ May 24, 2021 at 1:36
  • $\begingroup$ I am not expert in the matter, sorry. $\endgroup$ May 25, 2021 at 2:25

If you're interested in singular homology, I think the answer is no.

The $2$--dimensional analog of the Hawaiian earring should be a counterexample.

See Hatcher's answer to this question.

  • $\begingroup$ Thanks. I chose the other answer since it is more similar to what I expected, but this answer was very helpfull too, in my search I could not find the MO question you refer too. $\endgroup$
    – rpotrie
    Oct 24, 2010 at 20:14
  • $\begingroup$ No sweat. Cech cohomology's more commonly the thing of interest in such things, but it's good to know about singular homology's pathology. $\endgroup$ Oct 24, 2010 at 21:13

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