Let $A=\{z\in\mathbf{C}:1/2<|z|<1\}$ be an open annulus. Let us cover $A$ by 3 open sets:
$U_0,U_1$ and $U_2$ which we assume to be all homeomorphic to a 2 dimensional open disc. Moreover, we assume that $U_{ij}:=U_{i}\cap U_j$ are also homeomorphic to two dimensional open discs and that $U_0\cap U_1\cap U_2=\emptyset$. Let $\mathfrak{U}=\{U_0,U_1,U_2\}$.
Let $\mathcal{O}_A$ be the structure sheaf of $A$ where we think of $A$ as a complex analytic manifold. Since $A$, $U_i$'s and $U_{ij}$'s are connected open Riemann surfaces they are Stein manifolds and thus $H^q(A,\mathcal{O}_A)$, $H^q(U_i,\mathcal{O}_A|_{U_i})$ and
$H^q(U_{ij},\mathcal{O}_A|_{U_{ij}})$ all vanish for $q\geq 1$. Therefore I would expect
$$
\check{H}^{1}(\mathfrak{U},\mathcal{O}_A)\simeq H^1(A,\mathcal{O}_A)=0.
$$
Since
$$
C^1(\mathfrak{U},\mathcal{O}_A)=\mathcal{O}_A(U_{12})\times \mathcal{O}_A(U_{02})\times
\mathcal{O}_A(U_{01}).
$$
I would expect every 1-cochain to come from a 0-cochain. So for example, one could take
a triple of constant functions (which are clearly holomorphic) for example
$$
(1,2,3)\in C^1(\mathfrak{U},\mathcal{O}_A)
$$
Thus there should exist
$$
(f_0,f_1,f_2)\in \mathcal{O}_A(U_{0})\times \mathcal{O}_A(U_{1})\times
\mathcal{O}_A(U_{2})
$$
such that $df=((f_2-f_1)|_{U_{12}},(f_2-f_0)|_{U_{02}},(f_1-f_0)|_{U_{01}})=(1,2,3)$. But this is clearly impossible since $2-1\neq 3$!
Q: So why is Chech cohomology not computing $H^1(A,\mathcal{O}_A)$ here?
Best Answer
Functions $(f_0,f_1,f_2)$ as you requested do in fact exist. Your confusion comes from the fact that you are trying to impose the cocycle condition on the intersection $U_0\cap U_1\cap U_2$, which is empty. That is the $f_i$'s are such that $$ f_2(x) = f_1(x)+1\ \forall x\in U_1\cap U_2$$ $$ f_2(x) = f_0(x)+2\ \forall x\in U_0\cap U_2$$ $$ f_1(x) = f_0(x)+3\ \forall x\in U_0\cap U_1$$ To get a contradiction you'd need an $x$ such that these three things simultaneously hold. But such an $x$ doesn't exists.