Leray's theorem up to some degree

Question

I am interested in the proof of Leray's theorem that relates Čech cohomology and sheaf cohomology.

The theorem states that if we have a space $X$, a sheaf $\mathcal{F}$ and a covering of $X$ such that our sheaf is acyclic on the covering, i.e. Čech cohomology is $0$ on every finite intersection of elements in the covering, then these two cohomologies agree.

Now, my question. All the proves that I've found are by induction and assume that cohomology vanishes in every degree, so $\check{\mathrm{H}}^p(\mathcal{U}, \mathcal{F})=0$ for every $p>0$.

This looks like a rather strong assumption to check, especially if one is interested in computing cohomology just in low degrees (in particular I am interested in degree 0 and 1).

Looking at the proves that I found, indeed, it seems to me that we just need that Čech cohomology vanishes up to degree $p$ to prove the isomorphism in this degree (for example Climbing Mount Bourbaki).

I may be very wrong on this point, so the question is

• Can someone provide a counter example in which we really need that cohomology vanishes in all the degrees and not just up to what we are interested in?

or

• Can someone point out a reference which clearly states that we can just check up to degree $p$ (or $p+1$, depending on your indexing) to prove the isomorphism?

Thank you in advance,
Davide

PS: I thought that this question was at SE level and already asked there but, despite 5 "ups", none answered.

Answer 1

Well, you're perfectly correct. In fact, for the proof, you only need that $H^1$ is vanishing on all the intersections of all the open sets in your cover. However,...you need that vanishing for all quasi-coherent sheaves (or whatever category you're working in) and the higher cohomology of any sheaf is the $H^1$ of some syzygy (in fact, that's silently used in the proof), so at the end assuming the vanishing of $H^1$ for all sheaves is equivalent to assuming vanishing of all higher cohomology of all sheaves, so you're back where you started.

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ADDED IN RESPONSE TO QUESTION IN THE COMMENTS: If you look at the standard proof (say in Hartshorne), then it goes by taking your sheaf $F$, embedding it into a flasque quasi-coherent sheaf $G$ and looking at the corresponding short exact sequence $0\to F \to G\to G/F \to 0$. Then if $F$ has no $H^1$ on any of the intersections in the open cover, then you get that the sections of these sheaves on any of those open sets is still exact, so you get a long exact cohomology sequence for the Čech cohomology and of course you already get one for the usual sheaf cohmology from the original short exact sequence. $G$ has no higher cohmology in either realm because it is flasque, so we get two length 4 (+ two zeros at both ends) exact sequences with a natural map between them so by the 5-lemma and the fact that we have equality for $H^0$, indeed we get isomorphism for $H^1$ and that $H^p(F)\simeq H^{p-1}(G/F)$. Then to finish the proof one says that if we know the vanishing of $H^1$ for all sheaves, then we have isom for all sheaves and then the index shift we get between $F$ and $G/F$ gives us isom for all $H^p$.

So, indeed, if you are only interested in $H^1(F)$, then it is enough to know that $F$ itself has no cohomology on all those intersections.

Then again, the easiest way to prove this is if those open sets have some property, like being affine that implies no higher cohomology. Of course, if for some reason you know that your $F$ is special with respect to these open sets, then you are OK.