# Why does the sheaf cohomology of the constant sheaf on $\mathbb{R}$ vanish?

Sorry if this is a very basic fact, but for some reason I am not able to solve it.

I am trying to prove for $X =\mathbb{R}$ that the sheaf cohomology groups $(i>0)$ of the constant sheaf $\mathbb{Z}_{X}$ vanish.

From what I gather online it has something to do with $\mathbb{R}$ being simply connected. I know it is even contractible and that these cohomology groups are isomorphic for homotopic spaces, but I am looking for an argument not using this homotopy argument.

I think I understand this sheaf, namely that $\mathbb{Z}_{X}(U) = \mathbb{Z}^{I}$ where I has cardinality equal to the number of connected components of $U$ and the restriction maps are just projections and diagonal maps. But somehow I can't seem to find a flasque/injective resolution or a sort of dimension shift argument. I can use excision and mayer-vietoris but I don't know if this helps.

Thanks in advance for any help!

• If you think of $\mathbb{Z}_X$ as the sheaf of locally constant functions $\mathbb{R} \to \mathbb{Z}$ (i.e. $\mathbb{Z}_X(U)$ is the set of locally constant functions $U \to \mathbb{Z}$) then maybe a natural flasque sheaf to try as the first step in a resolution would be the sheaf of all functions $\mathbb{R} \to \mathbb{Z}$. Or maybe, just try looking at the long exact sequence induced by the natural injection $\mathbb{Z}_X \hookrightarrow \mathbb{Z}^{\mathbb{R}}$. May 30, 2018 at 0:01
• @DanielSchepler yes, you mean its inbedding in the flasque sheaf of discontinuous functions, the problem I have is that their sheaf cokernel isn't very clear to me as I have to sheafify. Maybe their quotient presheaf is already a sheaf as it is simply connected? May 30, 2018 at 0:19
• This is probably an overkill, but you can also use the Cech to derived functor spectral sequence.
– loch
May 30, 2018 at 6:22
• @loch I don't know and I haven't proved such a spectral sequence, thus I'd rather find an argument more elementary :) May 30, 2018 at 9:11
• @loch How do you use Cech to derived functor spectral sequence ? In fact I really doubt there is an easy and purely algebraic argument for this problem, but if you have on I would love to read it ! May 30, 2018 at 9:35

Call a sheaf on $\mathbb{R}$ interval-flasque if for every inclusion of open intervals $(a, b) \subseteq (a', b')$, the restriction map $\Gamma((a', b'), \mathscr{F}) \to \Gamma((a, b), \mathscr{F})$ is surjective. (Here, we allow $a = -\infty$ and/or $b = \infty$ and similarly for $a', b'$.) We then claim that any interval-flasque sheaf of abelian groups on $\mathbb{R}$ is acyclic in $\mathfrak{Ab}(\mathbb{R})$. Since in particular, the constant sheaf $\mathbb{Z}$ on $\mathbb{R}$ is clearly interval-flasque, that will imply the desired result.

The proof proceeds similarly to the proof for generally flasque sheaves. Here I'll follow the outline of the facts that the proof from Hartshorne's Algebraic Geometry uses:

• If $0 \to \mathscr{F}' \to \mathscr{F} \to \mathscr{F}'' \to 0$ is a short exact sequence, and $\mathscr{F}'$ is interval-flasque, then $0 \to \Gamma(U, \mathscr{F}') \to \Gamma(U, \mathscr{F}) \to \Gamma(U, \mathscr{F}'') \to 0$ is exact for every open interval $U$. Here, in using Zorn's Lemma to show there exists a maximal preimage of $x \in \Gamma(U, \mathscr{F}'')$, you will need to use the fact that the union of a chain of intervals is again an interval. Then, in showing that a preimage on a strict subinterval can be extended (and therefore, a maximal preimage must be a section over the full larger interval), you will need to use the fact that a nonempty intersection of two intervals is an interval, and likewise for the union of two overlapping intervals.

• If $0 \to \mathscr{F}' \to \mathscr{F} \to \mathscr{F}'' \to 0$ is a short exact sequence, and $\mathscr{F}'$ and $\mathscr{F}$ are both interval-flasque, then so is $\mathscr{F}''$. Here, the proof is pretty much the same as for the generally flasque case: use the previous fact, along with the Snake Lemma on the obvious morphism between short exact sequences of abelian groups.

• Any injective object of $\mathfrak{Ab}(\mathbb{R})$ is interval-flasque. Well, we know it's flasque, so in particular it's interval-flasque.

So now, the standard proof that flasque sheaves are acyclic will also work for interval-flasque sheaves, with just a global replacement of "flasque" with "interval-flasque" in the proof. Note that we do need the fact here that $\mathbb{R} = (-\infty, \infty)$ is considered an open interval in the preceding facts.

Now, it might be interesting to see where this proof fails if you try to apply it on $S^1$ instead of $\mathbb{R}$. In particular, what goes wrong if you try to replace "open intervals in $\mathbb{R}$" with "open arcs in $S^1$"? What goes wrong if you try to fix this failure by restricting to open arcs of length less than $\pi$?

• Very nice ! Did you read this proof somewhere or did you make it ? Jun 1, 2018 at 20:50
• @Roland As far as I recall, I came up with it independently. Jun 1, 2018 at 21:51
• Indeed very nice! I was also thinking about filtering out the useful open subsets as we only have to say something globally about R, but I didn't know how to. I don't know if I am supposed to answer the questions, but it goes wrong for S^{1} with taking intersections. Do you mean that when the arcs are restricted the global arc S^{1} (2 \pi) is not obtained via Zorn? Jun 4, 2018 at 20:56
• @DanielSchepler do you still think your previous idea still holds: the quotient presheaf of the sheaf of discontinuous functions and the constant sheaf on R is a flasque sheaf? Jun 4, 2018 at 20:58
• @Lilolance Yes, I believe so - more generally, the presheaf quotient of a flasque sheaf by an interval-flasque sheaf will be a flasque sheaf. (The first fact above will apply to show the presheaf quotient coincides with the sheaf quotient on open intervals; and then any general open subset of $\mathbb{R}$ is a disjoint union of open intervals.) Jun 4, 2018 at 21:23