I am reading Rotman, Introduction to Homological Algebra, and $\textbf{Theorem 6.88.}$ reads:

Let $\mathcal{F}$ be a sheaf over a paracompact space $X$. If $X$ has an open cover $\mathcal{U}$ with $dim(\mathcal{N}(\mathcal{U}))<n$, then $\check{H}^q(\mathcal{F})=\{0\}$ for all $q\geq n+1$.

The proof of this theorem refers to Swan, The Theory of Sheaves, p.109, but I don't have this book.

However, $\mathcal{U}=\{X\}$ itself is an open cover where $dim(\mathcal{N}(\mathcal{U}))\leq 0$, and then by the above theorem, $\check{H}^q(\mathcal{F})=\{0\}$ for all $q\geq 1$, and this is definitely not true. This example even appears in Rotman too; it's $\textbf{Example 6.78.}$ (iii).

I think there is something wrong, but I don't know how to fix it. I think maybe there needs some kind of condition for an open cover $\mathcal{U}$: for example, every intersection of distinct sets in $\mathcal{U}$ is contractible. I am not sure if this is true if I add this condition. Or if every open cover $\mathcal{V}$ has a refinement $\mathcal{U}$ with $dim(\mathcal{N}(\mathcal{U}))<n$, then I am quite sure the theorem is true, but this is way too easier than the original theorem, so I don't think this is what Rotman wanted to state.

Thanks in advance.


There is indeed a mistake in the theorem. The correct version should be :

  • if $\dim\mathcal{N(U)}<n$, then $\check H^q(\mathcal{U,F})=0$ for all $q\geq n+1$.
  • in particular, if there exists $\mathcal{U}$ is a good cover for $\mathcal{F}$ (that is one such that any finite intersections $V=U_1\cap...\cap U_n$ satisfies $H^q(V,\mathcal{F})=0$ for all $q>0$) such that $\dim\mathcal{N(U)}<n$ then $\check H^q(\mathcal{F})=0$ for all $q\geq n+1$.
  • $\begingroup$ I am still in confusion. I think if the theorem is corrected in that way, it seems true. But the next theorem in Rotman's book claims that, as a corollary of this theorem, if X is compact, $H^p(X)=0$ for sufficiently large $p>n$. I am not sure that the corrected theorem implies this corollary. $\endgroup$ – J1U May 17 '18 at 1:01
  • 1
    $\begingroup$ Well there exists a good cover, since $X$ is compact then you can take a finite sub-cover which will still be a good cover. $\endgroup$ – Roland May 17 '18 at 6:45

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