sorry if this is a very stupid question and I'm missing something very trivial, though I could not solve it after thinking for a while.
In page 18-19 of http://www.springer.com/cda/content/document/cda_downloaddocument/9780817649586-c2.pdf?SGWID=0-0-45-993379-p174035557, the author concludes that $\check{H}^i (M, \mathbb{R}) = 0$ for $i = 1, 2$ in order to get an isomorphism $H^1 (M, S^1) = H^2 (M, \mathbb{Z})$.
Ok, I agree with the latter one, which should be given by assigning the Euler class, however the first one ($\check{H}^i (M, \mathbb{R}) = 0$) is an absurd. It's well known that \v{C}ech cohomology with values in the constant sheaf $\mathbb{R}$ is isomorphic to the de Rham cohomology, so clearly something is wrong.
Where's the mistake?
Thanks in advance.
You're thinking of two different things when you say "the sheaf $\Bbb R$". You mean the constant sheaf, wherein $\mathcal F(U)=\Bbb R$ for any open set $U$. The author is thinking of the sheaf of sections of the trivial bundle $\Bbb R$, also known as "The sheaf of smooth functions"; so in their case $\mathcal F(U)=C^\infty(U,\Bbb R)$. This is a fine sheaf, so indeed its higher cohomology does vanish.
In your setup, the sheaf that fits in the end of $\Bbb Z\to \Bbb R \to S^1$ is actually the constant sheaf over the group $S^1$. You can of course still take its cohomology but it doesn't calculate anything interesting (like, say, complex line bundles).