TITLE: About the definition of  Cech Cohomology
QUESTION [17 upvotes]: Let $X$ be a topological space with and open cover $\{U_i\}$ and let $\mathcal F$ be a sheaf of abelian groups on $X$. A $n$-cochain is a section $f_{i_0,\ldots,i_n}\in U_{i_0,\ldots,i_n}:= U_{i_0}\cap\ldots\cap U_{i_n}$; we can costruct the following abelian group (written in additive form):
$$
\check C^n(\mathcal U,\mathcal F):=\!\!\prod_{(i_o,\ldots,i_n)}\!\!\mathcal F(U_{i_0,\ldots,i_n})
$$
Now my question is the following:
we consider  oredered sequences $(i_o,\ldots,i_n)$ ? Because in this case in the direct product we have each group repetead $(n+1)!$ times, that is the number of permutations of the set $\{i_o,\ldots,i_n\}$.

REPLY [5 votes]: You can choose: both versions exist!
a) You can  take the product $\check C^n(\mathcal U,\mathcal F)=\!\!\prod_{(i_o,\ldots,i_n)}\!\!\mathcal F(U_{i_0,\ldots,i_n})$ over all $n+1$-tuples, so that indeed there will be much redundancy in your groups i.e. they will be repeated.  
b) Or you can put some total order on $I$ and consider the complex 
$$
\check C'^n(\mathcal U,\mathcal F):=\!\!\prod_{i_o\lt\ldots\lt i_n}\!\!\mathcal F(U_{i_0,\ldots,i_n})
$$
which is clearly more economical.  
b') There is a variant where you use the subcomplex $\check C_{alt}^\bullet(\mathcal U,\mathcal F)\subset \check C^n(\mathcal U,\mathcal F)$ of the complex in a) consisting of alternate families: for example if $n=1$ you require that $s_{ij}=-s_{ji}\in \mathcal F(U_i\cap U_j) \;\text {for } \;i\neq j$ and $s_{ii}=0.$  
These complexes give the same cohomology groups: it is pretty clear for b) and b') and it requires a calculation to show that the inclusion of complexes 
$$\check C_{alt}^\bullet(\mathcal U,\mathcal F)\hookrightarrow \check C^\bullet(\mathcal U,\mathcal F)$$ yields isomorphisms at the level of cohomology groups $$    \check H_{alt }^n(\mathcal U,\mathcal F)\hookrightarrow \check H^n(\mathcal U,\mathcal F)           $$
In that generality I must admit that all this is a bit boring.
For familiarizing oneself with effective calculations in low degrees, I recommend §12 of Forster's Lectures on Riemann Surfaces in which he very explicitly computes, for example, the first Čech cohomology group $\check H^1$ of some sheaves on Riemann surfaces.  
Edit
Let me say, as an answer to Galoisfan's question, that version b) is much more powerful.
Here is an example:     
Let  $X$ be  a Riemann surface and $\mathcal F$ a coherent sheaf.
If $\mathcal U=\lbrace U_0,U_1\rbrace$ is a covering of $X$ consisting of two open sets, then obviously $\check C'^n(\mathcal U,\mathcal F)=0$ for $n\geq 2$ since you cannot extract a sequence ot three strictly increasing numbers from $\lbrace 0,1\rbrace$ !
But if $U_0,U_1\subsetneq X$ are strict open subsets, they are Stein and Leray's theorem says that $ \check H^n(\mathcal U,\mathcal F) =  \check H^n(X,\mathcal F) $, the  genuine cohomology of $\mathcal F$ (i.e. the inductive limit over the coverings of $X$).
So version b) lets you prove  that all genuine cohomology groups $\check H^n(X,\mathcal F) \; (n\geq 2)$ are zero: quite a remarkable theorem!  
In the same vein, for every algebraic variety $X$ that can be covered by $n+1$ open affine subsets ($\mathbb P^n$ for example ) and every coherent algebraic sheaf $\mathcal F$ on $X$, the totally ordered  version of Čech cohomology shows that $\check H^k(X,\mathcal F) =0$ for $k\gt n$.