Let $X$ be a topological space. Let $U_i$ be an open cover, with the index $i$ running through some set $I$. For $i_0$, $i_1$, ..., $i_n \in I$, write $U_{i_0 i_1 \cdots i_n}$ for $U_{i_0} \cap \cdots \cap U_{i_p}$. $\def\cE{\mathcal{E}}$Let $F$ be a sheaf of abelian groups on $X$ and $F_{i_0,i_1,...i_n}:=F(U_{i_0 i_1 \cdots i_n})$, we will write the group operation additively.
Let $I$ to be a set of indices, and $i_0, i_1, i_2... \in I$. For each array $(i_0,i_1,...i_n)$ associate a set $F_{i_0,i_1,...i_n}$ and take
$$(f_{\large i_0,i_1,...i_n})_{i_0,i_1,...i_n\in I}\in \prod_{\large i_0,i_1,...i_n}F_{\large i_0,i_1,...i_n}.$$
Define $d((f_{\large i_0,i_1,...i_n})_{\large i_0,i_1,...i_n\in I})=(\sum_{i=0}^{n+1}(-1)^kf_{\large i_0,...\hat{i_k},...i_{n+1}}|U_{i_0 i_1 \cdots i_{n+1}})_{\large i_0,i_1,...i_{n+1}\in I} $
I wonder if the following statement is true:
Assume $d((f_{\large i_0,i_1,...i_n})_{\large i_0,i_1,...i_n\in I})=0$. If any two of the indices ${k_0,k_1,...k_n}$ are equal, then $f_{\large k_0,k_1,...k_n}=0$
See here for the description pf the cohomology w.r.t. an open cover and this problem N for the original question(from Rick Miranda's book).
I have made no constructive approach to this problem so far. One idea is to prove it backwards, first prove that it's zero when all of the indices are equal, then it is zero. But there will be an evenness-oddness problem.
Update:
I referred to the wrong problem earlier. It should be problem N.