# Cohomological dimension of a non-compact interval

Iversen, in "Cohomology of Sheaves," proves a number of theorems about the cohomological dimension of a locally compact space. In particular, if $\mathcal{F}$ is any sheaf on $\mathbb{R}^1$, it is proved that the compactly supported cohomology of $\mathcal{F}$ vanishes in degree greater than one.

The argument is quite slick: namely, let $\mathcal{F}$ be a sheaf, and let $\alpha \in H^2_c(X, \mathcal{F})$ a non-vanishing cohomology class. By generalities about direct limits and sheaf cohomology, Iversen argues that there is a minimal closed set $Z \subset \mathbb{R}^1$ such that $H^2(Z, \mathcal{F})$ is not zero---namely, by Zorn it suffices to show that if $\alpha$ restricts to nonzero a cohomology class on a descending chain of closed subsets, then it restricts to a nonzero class on the intersection. (This is the fact alluded to about direct limits.) Now if $Z$ is this minimal set, the Mayer-Vietoris sequence(!) shows that $\alpha$ cannot map to zero in both $H^2_c(Z \cap [t, \infty))$ and $H^2_c(Z \cap (-\infty, t])$ (or it would be zero because the cohomology of a point is trivial). But this is a contradiction if $t$ is chosen appropriately.

Is this true in ordinary sheaf cohomology? The Mayer-Vietoris sequence is certainly still fine, but without the compactness the claim about cohomology and direct limits should fail. (E.g. in degree zero. For example, take $\bigoplus (i_n)_*(\mathbb{Z})$ where $i_n$ is the inclusion of $\{n\}$, so that global sections over $\mathbb{R}$ is an infinite product $\prod \mathbb{Z}$, clearly much bigger than the colimit (which is an infinite direct sum).) Nonetheless it would be intuitively very appealing if $\mathbb{R}$ had no sheaf cohomology in dimensions greater than one.

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Sheaf cohomology of a paracompact space can be computed using Čech cohomology. Now, the topological dimension of $\mathbb R$ is a one, so there is a cofinal family of open coverings of $\mathbb R$ for which the Čech complex is trivial above dimension $1$. That $\mathbb R$ has cohomological dimension one follows (well, one has to check that the dimension is not actually zero, of course)
Later: One can also use Godement's theorem 4.14.2 which states that if $X$ is a metrisable space with $\operatorname{cd}X\leq n$, then every subspace $Y\subseteq X$ has $\operatorname{cd}Y\leq n$.