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!
 A: 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 $\pi$?
